myresource.phoenix.edumyresource.phoenix.edu/secure/resource/MAT117R7/mat117_week1… · EXAMPLE 5...

55Polynomials:Factoring

5.1 Introduction to Factoring

5.2 Factoring Trinomials of the Type

5.3 Factoring The FOIL Method

5.4 Factoring The ac-Method

5.5 Factoring Trinomial Squares andDifferences of Squares

5.6 Factoring Sums or Differences of Cubes

5.7 Factoring: A General Strategy

5.8 Solving Quadratic Equations byFactoring

5.9 Applications of QuadraticEquations

a � 1:ax2 � bx � c,

a � 1:ax2 � bx � c,

x2 � bx � c

Real-World ApplicationAn outdoor-education ropes course includes a 25-ftcable that slopes downward from a height of 37 ft toa height of 30 ft. How far is it between the trees thatthe cable connects?

This problem appearsas Example 5 inSection 5.9.

Ch05pgs307-316 1/19/06 9:29 AM Page 307

ISB

N:0

-536

-477

42-6

Introductory and Intermediate Algebra, Third Edition, by Marvin L.Bittinger and Judith A.Beecher.Published by Addison Wesley.Copyright ©2007 by Pearson Education, Inc.

308

CHAPTER 5: Polynomials: Factoring

We introduce factoring with a review of factoring natural numbers. Con-sider the product We say that 3 and 5 are factors of 15 and that

is a factorization of 15. Since we also know that 15 and 1 arefactors of 15 and that is a factorization of 15.

Finding the Greatest Common Factor

The numbers 20 and 30 have several factors in common, among them 2 and5. The greatest of the common factors is called the greatest common factor,GCF. One way to find the GCF is by making a list of factors of each number.

List all the factors of 20: 1, 2, 4, 5, 10, and 20.

List all the factors of 30: 1, 2, 3, 5, 6, 10, 15, and 30.

Now list the numbers common to both lists, the common factors:

1, 2, 5, and 10.

Then the greatest common factor, the GCF, is 10, the largest number in thecommon list.

The preceding procedure gives meaning to the notion of a GCF, but thefollowing method, using prime factorizations, is generally faster.

EXAMPLE 1 Find the GCF of 20 and 30.

We find the prime factorization of each number. Then we draw lines be-tween the common factors.

The GCF



The GCF Note how we can use the expo-nents to determine the GCF. There are 2 lines for the 2’s, 1 line for the 3, 1 line for the 5, and no line for the 7.


We find the prime factorization of each number. Then we draw lines be-tween the common factors, if any exist.

Since there is no common prime factor, the GCF is 1.

77 � 7 � 11 � 71 � 111

30 � 2 � 3 � 5 � 21 � 31 � 51

� 2 � 2 � 3 � 5 � 22 � 31 � 51 � 60.

420 � 2 � 2 � 3 � 5 � 7 � 22 � 31 � 51 � 71

180 � 2 � 2 � 3 � 3 � 5 � 22 � 32 � 51

� 2 � 5 � 10.

30 � 2 � 3 � 5

20 � 2 � 2 � 5

15 � 115 � 15 � 1,3 � 5

15 � 3 � 5.

5.15.1 INTRODUCTION TO FACTORINGObjectivesFind the greatest commonfactor, the GCF, ofmonomials.

Factor polynomials when theterms have a common factor,factoring out the greatestcommon factor.

Factor certain expressionswith four terms usingfactoring by grouping.

Ch05pgs307-316 1/19/06 9:29 AM Page 308

ISB

N:0-536-47742-6


EXAMPLE 4 Find the GCF of 54, 90, and 252.


The GCF

Do Exercises 1–4.

Consider the product

To factor the polynomial on the right, we reverse the process of multiplication:

FACTOR; FACTORIZATION

To factor a polynomial is to express it as a product.

A factor of a polynomial P is a polynomial that can be used to expressP as a product.

A factorization of a polynomial is an expression that names thatpolynomial as a product.

In the factorization

the monomial is called the GCF of the terms, and Thefirst step in factoring polynomials is to find the GCF of the terms.

Consider the monomials

and

The GCF of these monomials is found by noting that the smallest expo-nent of x is 3.

Consider

and

The GCF of 20 and 30 is 10. The GCF of and is Then the GCF ofand is the product of the individual GCFs, 10x2.30x5

20x2x2.x5x2

30x5.20x2

x3,

x7.x6,x4,x3,

24x3.�72x4,12x5,12x3

12x5 � 72x4 � 24x3 � 12x3�x2 � 6x � 2�,

12x5 � 72x4 � 24x3 � 12x3�x2 � 6x � 2�.

12x3�x2 � 6x � 2� � 12x5 � 72x4 � 24x3.

� 21 � 32 � 18.

252 � 2 � 2 � 3 � 3 � 7 � 22 � 32 � 71

90 � 2 � 3 � 3 � 5 � 21 � 32 � 51,

54 � 2 � 3 � 3 � 3 � 21 � 33,

Find the GCF.

1. 40, 100

2. 7, 21

3. 72, 360, 432

4. 3, 5, 22

Answers on page A-20

309


Ch05pgs307-316 1/19/06 9:29 AM 09

ISB

N:0

-536

-477

42-6


EXAMPLE 5 Find the GCF of and

First, we find a prime factorization of the coefficients, attaching a factorof �1 for the negative coefficients.

The greatest positive common factor of the coefficients is 3.Then we find the GCF of the powers of x. That GCF is because 2 is the

smallest exponent of x. Thus the GCF of the set of monomials is

What about the �1 factors in Example 5? Strictly speaking, both 1 and �1are factors of any number or expression. We see this as follows:

Because the coefficient �3 is less than the coefficient 3, we consider andnot the GCF.

EXAMPLE 6 Find the GCF of 2py, and

We have

The greatest positive common factor of the coefficients is 2, the GCF of thepowers of p is p, and the GCF of the powers of y is 1 since there is no y-factorin the last monomial. Thus the GCF is 2p.

TO FIND THE GCF OF TWO ORMORE MONOMIALS

1. Find the prime factorization of the coefficients, including �1 as afactor if any coefficient is negative.

2. Determine any common prime factors of the coefficients. For eachone that occurs, include it as a factor of the GCF. If none occurs,use 1 as a factor.

3. Examine each of the variables as factors. If any appear as a factorof all the monomials, include it as a factor, using the smallestexponent of the variable. If none occurs in all the monomials, use1 as a factor.

4. The GCF is the product of the results of steps (2) and (3).

Do Exercises 5–8.

4p3 � 2 � 2 � p3.

2py � 2 � p � y,

�8py2 � �1 � 2 � 2 � 2 � p � y2,

14p2y3 � 2 � 7 � p2 � y3,

4p3.�8py2,14p2y3,

�3x2,3x2,

3x2 � 1 � 3x2 � ��1��3x2�.

3x2.x2,

�3x2 � �1 � 3 � x2

27x3 � 3 � 3 � 3 � x3,

�12x4 � �1 � 2 � 2 � 3 � x4,

15x5 � 3 � 5 � x5,

�3x2.27x3,�12x4,15x5,Find the GCF.

5.

6.

7.

8.


�63x3�14x5,�49x6,�35x7,

8m4n4�16m2n2,12mn3,�24m5n6,

2y2�5y3,3y6,

�16x312x2,

310


Ch05pgs307-316 1/19/06 9:29 AM Page 310

ISB

N:0-536-47742-6


Factoring When Terms Have a Common Factor

The polynomials we consider most when factoring are those with more thanone term. To multiply a monomial and a polynomial with more than oneterm, we multiply each term of the polynomial by the monomial using thedistributive laws:

and

To factor, we do the reverse. We express a polynomial as a product usingthe distributive laws in reverse:

and

Compare.

Multiply Factor

� 3x3 � 6x2 � 12x� 3x � x2 � 3x � 2x � 3x � 4

3x�x2 � 2x � 4�

ab � ac � a�b � c�.ab � ac � a�b � c�

a�b � c� � ab � ac.a�b � c� � ab � ac

9. a) Multiply:

b) Factor:

10. a) Multiply:

b) Factor:


2x3 � 10x2 � 8x.

2x�x2 � 5x � 4�.

3x � 6.

3�x � 2�.

311


� 3x�x2 � 2x � 4�� 3x � x2 � 3x � 2x � 3x � 4

3x3 � 6x2 � 12x

Caution!

Consider the following:

The terms of the polynomial, and have been factored butthe polynomial itself has not been factored. This is not what we mean by afactorization of the polynomial. The factorization is

A product

The expressions and are factors of

Do Exercises 9 and 10.

To factor, we first find the GCF of all terms. It may be 1.

EXAMPLE 7 Factor:

We have

Factoring each term

Factoring out the GCF, 7

Check: We multiply to check:

7�x2 � 2� � 7 � x2 � 7 � 2 � 7x2 � 14.

� 7�x2 � 2�. 7x2 � 14 � 7 � x2 � 7 � 2

7x2 � 14.

3x3 � 6x2 � 12x.x2 � 2x � 43x

3x�x2 � 2x � 4�.

�12x,6x2,3x3,

3x3 � 6x2 � 12x � 3 � x � x � x � 2 � 3 � x � x � 2 � 2 � 3 � x.

Ch05pgs307-316 1/19/06 9:29 AM Page 311

ISB

N:0

-536

-477

42-6


EXAMPLE 8 Factor:

Factoring each term

Factoring out the GCF,

Suppose in Example 8 that you had not recognized the GCF and removedonly part of it, as follows:

Note that still has a common factor of 2. You need not begin again.Just continue factoring out common factors, as follows, until finished:

EXAMPLE 9 Factor:

Factoring out theGCF,

Check: We multiply to check:

As you become more familiar with factoring, you will be able to spot theGCF without factoring each term. Then you can write just the answer.

EXAMPLES Factor.

10.

11.

12.

13.

Do Exercises 11–16.

45

x2 �15

x �25

�15

�4x2 � x � 2�

14p2y3 � 8py2 � 2py � 2py�7py2 � 4y � 1�8m3 � 16m � 8m�m2 � 2�24x2 � 12x � 36 � 12�2x2 � x � 3�

� 15x5 � 12x4 � 27x3 � 3x2.

� �3x2��5x3� � �3x2� �4x2� � �3x2� �9x� � �3x2� �1�3x2�5x3 � 4x2 � 9x � 1�

3x2 � 3x2�5x3 � 4x2 � 9x � 1�

15x5 � 12x4 � 27x3 � 3x2 � �3x2� �5x3� � �3x2� �4x2� � �3x2� �9x� � �3x2� �1�

15x5 � 12x4 � 27x3 � 3x2.

� 4x2�4x � 5�. � 2x2�2�4x � 5�� 2x2�2 � 4x � 2 � 5�

8x � 10

� 2x2�8x � 10�. 16x3 � 20x2 � �2x2� �8x� � �2x2� �10�

4x2 � 4x2�4x � 5� 16x3 � 20x2 � �4x2� �4x� � �4x2� �5�

16x3 � 20x2.Factor. Check by multiplying.

11.

12.

13.

14.

15.

16.

Factor.

17.

18.


x2�a � b� � 2�a � b�

x2�x � 7� � 3�x � 7�

84x2 � 56x � 28

35x7 � 49x6 � 14x5 � 63x3

34

t 3 �54

t 2 �74

t �14

9x4y2 � 15x3y � 3x2y

3y6 � 5y3 � 2y2

x2 � 3x

312


Caution!

Don’t forget the term �1.

Ch05pgs307-316 1/19/06 9:29 AM Page 312

ISB

N:0-536-47742-6


There are two important points to keep in mind as we study this chapter.

TIPS FOR FACTORING

• Before doing any other kind of factoring, first try to factor out the GCF.

• Always check the result of factoring by multiplying.

Factoring by Grouping: Four Terms

Certain polynomials with four terms can be factored using a method calledfactoring by grouping.

EXAMPLE 14 Factor:

The binomial is common to both terms:

The factorization is

Do Exercises 17 and 18 on the preceding page.

Consider the four-term polynomial

There is no factor other than 1 that is common to all the terms. We can, how-ever, factor and separately:

Factoring

Factoring

We have grouped certain terms and factored each polynomial separately:

as in Example 14. This method is called factoring by grouping. We began witha polynomial with four terms. After grouping and removing common factors,we obtained a polynomial with two parts, each having a common factor

Not all polynomials with four terms can be factored by this procedure,but it does give us a method to try.x � 1.

� �x2 � 2� �x � 1�, � x2�x � 1� � 2�x � 1�

x3 � x2 � 2x � 2 � �x3 � x2� � �2x � 2�

2x � 2 2x � 2 � 2�x � 1�.x3 � x2 x3 � x2 � x2�x � 1�;

2x � 2x3 � x2

x3 � x2 � 2x � 2.

�x2 � 2� �x � 1�.

x2�x � 1� � 2�x � 1� � �x2 � 2� �x � 1�.

x � 1

x2�x � 1� � 2�x � 1�.

Study Tips

CHECKLIST

The foundation of all yourstudy skills is TIME!

Are you staying onschedule and on time forclass and adapting yourstudy time and classschedule to yourpersonality?

Did you study theexamples in this chaptercarefully?

Are you asking questionsat appropriate times inclass and with yourtutors?

Are you doing exerciseswithout answers as part of every homeworkassignment to prepareyou for tests?

313


Ch05pgs307-316 1/19/06 9:29 AM Page 313

ISB

N:0

-536

-477

42-6


EXAMPLES Factor by grouping.

15.

Factoring each binomial

Factoring out the common factor

We think through this process as follows:

(1) Factor the first two terms.

(2) The factor gives us a hint to the factorization of the last two terms.

(3) Now we ask ourselves, “What times is ” The answer is �2.

16.


Factoring out the commonfactor

17.

Separating into two binomials

Check :


We can think through this process as follows.

(1) Factor the first two terms:

(2) The factor gives us a hint for factoring the last two terms:

(3) Now we ask ourselves, “What times is ” The answer is �1.

18.

19.

This polynomial is not factorable using factoring by grouping. It may befactorable, but not by methods that we will consider in this text.


x3 � x2 � 2x � 2 � x2�x � 1� � 2�x � 1�

� �4x2 � 7� �3x3 � 5� 12x5 � 20x2 � 21x3 � 35 � 4x2�3x3 � 5� � 7�3x3 � 5�

�x � 3?x � 3

�x � 3�.2x3 � 6x2 � x � 3 � 2x2�x � 3�

x � 3

2x3 � 6x2 � 2x2�x � 3�.

x � 3� �2x2 � 1� �x � 3��1�x � 3� � �x � 3.� 2x2�x � 3� � 1�x � 3�

� �2x3 � 6x2� � ��x � 3�2x3 � 6x2 � x � 3

x � 1 � �x2 � 1� �x � 1� � x2�x � 1� � 1�x � 1�

x3 � x2 � x � 1 � �x3 � x2� � �x � 1�

4x � 6?2x � 3

2x � 3

�2x � 3�6x3 � 9x2 � 4x � 6 � 3x2�2x � 3�

2x � 3� �3x2 � 2� �2x � 3�� 3x2�2x � 3� � 2�2x � 3�� 6x3 � 9x2� � �4x � 6�

6x3 � 9x2 � 4x � 6

Factor by grouping.

19.

20.

21.

22.

23.

24.


y4 � 2y3 � 2y � 10

4x3 � 6x2 � 6x � 9

3x3 � 6x2 � x � 2

3m5 � 15m3 � 2m2 � 10

8t 3 � 2t 2 � 12t � 3

x3 � 7x2 � 3x � 21

314


⎧⎨⎩ ⎧⎨⎩

⎧ ⎪ ⎨ ⎪ ⎩

Caution!

Don’t forget the 1.

Ch05pgs307-316 1/19/06 9:29 AM Page 314

ISB

N:0-536-47742-6


Find the GCF.

315

Exercise Set 5.1

EXERCISE SET For Extra Help5.15.1 Student’sSolutionsManual

Digital VideoTutor CD 3Videotape 5

Math TutorCenter

InterActMath

MyMathLabMathXL

1. �6x 2. 5x 3. x23x4,x2,x2,

4. 5. 2x, �8 6. �4x, �208x2,2x2,�24x28x4,

7. 8. 9. �5x, �20x3�x2,�48pq232p3q3,16p6q4,51xy34x3y2,�17x5y3,

Factor. Check by multiplying.

13. 14. 15. 2x2 � 6xx2 � 5xx2 � 6x

16. 17. 18. 3x4 � x2x3 � 6x28y2 � 8y

19. 20. 21. 2x2 � 2x � 85x5 � 10x38x4 � 24x2

22. 23. 24. 16p6q4 � 32p5q3 � 48pq217x5y3 � 34x3y2 � 51xy8x2 � 4x � 20

25. 26. 27. x5y5 � x4y3 � x3y3 � x2y25x5 � 10x2 � 8x6x4 � 10x3 � 3x2

28. 29. 30. 8y3 � 20y2 � 12y � 162x7 � 2x6 � 64x5 � 4x3x9y6 � x7y5 � x4y4 � x3y3

31. 32. 2.5x6 � 0.5x4 � 5x3 � 10x21.6x4 � 2.4x3 � 3.2x2 � 6.4x

33. 34.59

x7 �29

x5 �49

x3 �19

x53

x6 �43

x5 �13

x4 �13

x3

Factor.

35. 36. 3z2�2z � 1� � �2z � 1�x2�x � 3� � 2�x � 3�

37. 38. m4�8 � 3m� � 7�8 � 3m�5a3�2a � 7� � �2a � 7�

10. �6x, 11. 12. x3y3x4y4,�x7y5,�x9y6,�x2y2x3y3,x4y3,x5y5,�24x5�x2,

Ch05pgs307-316 1/19/06 9:30 AM Page 315

ISB

N:0

-536

-477

42-6


Factor by grouping.

316


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

39. 40. 41. 2x3 � 6x2 � x � 36z3 � 3z2 � 2z � 1x3 � 3x2 � 2x � 6

42. 43. 44. 10x3 � 25x2 � 4x � 108x3 � 12x2 � 6x � 93x3 � 2x2 � 3x � 2

45. 46. 47. 5x3 � 5x2 � x � 118x3 � 21x2 � 30x � 3512p3 � 16p2 � 3p � 4

48. 49. 50. 2x3 � 12x2 � 5x � 30x3 � 8x2 � 3x � 247x3 � 14x2 � x � 2

51. 52. 20g 3 � 4g 2 � 25g � 52x3 � 8x2 � 9x � 36

53. Josh says that there is no need to print answers forExercises 13–52 at the back of the book. Is he correct insaying this? Why or why not?

54. Explain how one could construct a polynomialwith four terms that can be factored by grouping.DWDW

Solve.

55. [2.7d] 56. [2.7e]4x � 8x � 16 � 6�x � 2��2x � 48

57. Divide: [1.6a] 58. Solve for p. [2.4b]A �p � q

2�108

�4.

Multiply. [4.6d]

59. 60. 61. 62. � y � 7�2� y � 7� � y � 7�� y � 7�2� y � 5� � y � 7�

Find the intercepts of the equation. Then graph the equation. [3.3a]

Factor.

67. 68. 69. x12 � x7 � x5 � 1x6 � x4 � x2 � 14x5 � 6x3 � 6x2 � 9

70. 71. p3 � p2 � 3p � 10x3 � x2 � 2x � 5

SKILL MAINTENANCE

SYNTHESIS

63. x � y � 4 64. x � y � 3 65. 5x � 3y � 15 66. y � 3x � 6

Ch05pgs307-316 1/19/06 9:30 AM Page 316

ISB

N:0-536-47742-6


1. Consider the trinomial

a) Complete the following table.

b) Explain why you need toconsider only positive factors,as in the following table.

c) Factor:

2. Factor:


x 2 � 13x � 36.

x 2 � 7x � 12.

x 2 � 7x � 12.

317

5.2 Factoring Trinomials of the Type x2 � bx � c

Factoring

We now begin a study of the factoring of trinomials. We first factor trinomi-als like

by a refined trial-and-error process. In this section, we restrict our attention totrinomials of the type where The coefficient a is oftencalled the leading coefficient.

To understand the factoring that follows, compare the followingmultiplications:

F O I L

Note that for all four products:

• The product of the two binomials is a trinomial.

• The coefficient of x in the trinomial is the sum of the constant terms inthe binomials.

• The constant term in the trinomial is the product of the constant terms inthe binomials.

These observations lead to a method for factoring certain trinomials. The firsttype we consider has a positive constant term, just as in the first two multi-plications above.

CONSTANT TERM POSITIVETo factor we think of FOIL in reverse. We multiplied x times xto get the first term of the trinomial, so we know that the first term of each bi-nomial factor is x. Next, we look for numbers p and q such that

To get the middle term and the last term of the trinomial, we look for twonumbers p and q whose product is 10 and whose sum is 7. Those numbers are2 and 5. Thus the factorization is

Check:

� x 2 � 7x � 10.

�x � 2� �x � 5� � x 2 � 5x � 2x � 10

�x � 2� �x � 5�.

x 2 � 7x � 10 � �x � p� �x � q�.

x 2 � 7x � 10,

� x 2 � 4x � 21.

�x � 3� �x � 7� � x 2 � 7x � 3x � ��3�7

� x 2 � 4x � 21;

�x � 3� �x � 7� � x 2 � 7x � 3x � 3��7�

� x 2 � 7x � 10;

�x � 2� �x � 5� � x 2 � 5x � 2x � 2 � 5

� x 2 � 7x � 10;

�x � 2� �x � 5� � x 2 � 5x � 2x � 2 � 5

a � 1.ax 2 � bx � c,

x 2 � 5x � 6 and x 2 � 3x � 10

x2 � bx � c

5.25.2 FACTORING TRINOMIALS OF THETYPE x2 � bx � c

ObjectiveFactor trinomials of the type

by examiningthe constant term c.x2 � bx � c

1, 12 13

�1, �12

2, 6

�2, �6

3, 4

�3, �4

PAIRS OF FACTORS

SUMS OF FACTORS

1, 12

2, 6

3, 4

PAIRS OF FACTORS

SUMS OF FACTORS

Ch05pgs317-326 1/19/06 9:32 AM Page 317

ISB

N:0

-536

-477

42-6


EXAMPLE 1 Factor:

Think of FOIL in reverse. The first term of each factor is x: Next, we look for two numbers whose product is 6 and

whose sum is 5. All the pairs of factors of 6 are shown in the table on the leftbelow. Since both the product, 6, and the sum, 5, of the pair of numbers mustbe positive, we need consider only the positive factors, listed in the table onthe right.

The numbers we need are 2 and 3.

The factorization is We can check by multiplying to seewhether we get the original trinomial.

Check:


Compare these multiplications:

TO FACTOR WHEN IS POSITIVE

When the constant term of a trinomial is positive, look for twonumbers with the same sign. The sign is that of the middle term:

EXAMPLE 2 Factor:

Since the constant term, 12, is positive and the coefficient of the middleterm, �8, is negative, we look for a factorization of 12 in which both factorsare negative. Their sum must be �8.

The factorization is The student should check by multiplying.

Do Exercises 3–5.

� y � 2� � y � 6�.

y 2 � 8y � 12.

x 2 � 7x � 10 � �x � 2� �x � 5�.

x 2 � 7x � 10 � �x � 2� �x � 5�;

cx 2 � bx � c

�x � 2� �x � 5� � x 2 � 5x � 2x � 10 � x 2 � 7x � 10.

�x � 2� �x � 5� � x 2 � 5x � 2x � 10 � x 2 � 7x � 10;

�x � 2� �x � 3� � x 2 � 3x � 2x � 6 � x 2 � 5x � 6.

�x � 2� �x � 3�.

�x � � �x � �.

x 2 � 5x � 6.3. Explain why you would notconsider the pairs of factorslisted below in factoring

Factor.

4.

5.


t 2 � 9t � 20

x2 � 8x � 15

y2 � 8y � 12.

318


1, 6 7

�1, �6 �7

2, 3 5

�2, �3 �5

PAIRS OF FACTORS SUMS OF FACTORS

1, 6 7

2, 3 5

PAIRS OF FACTORS

SUMS OF FACTORS

�1, �12 �13

�2, �6 �8

�3, �4 �7


1, 12

2, 6

3, 4

PAIRS OF FACTORS

SUMS OF FACTORS

The numbers we need are�2 and �6.

Ch05pgs317-326 1/19/06 9:32 AM Page 318

ISB

N:0-536-47742-6


CONSTANT TERM NEGATIVEAs we saw in two of the multiplications earlier in this section, the product oftwo binomials can have a negative constant term:

and

Note that when the signs of the constants in the binomials are reversed, onlythe sign of the middle term in the product changes.

EXAMPLE 3 Factor:

The constant term, �20, must be expressed as the product of a negativenumber and a positive number. Since the sum of these two numbers must benegative (specifically, �8), the negative number must have the greater ab-solute value.

The numbers that we are looking for are 2 and �10. The factorization is

Check:

TO FACTOR WHEN IS NEGATIVE

When the constant term of a trinomial is negative, look for twonumbers whose product is negative. One must be positive and theother negative:

Consider pairs of numbers for which the number with the largerabsolute value has the same sign as b, the coefficient of the middle term.

Do Exercises 6 and 7. (Exercise 7 is on the following page.)

x 2 � 4x � 21 � �x � 3� �x � 7�.

x 2 � 4x � 21 � �x � 3� �x � 7�;

cx 2 � bx � c

� x 2 � 8x � 20.

�x � 2� �x � 10� � x 2 � 10x � 2x � 20

�x � 2� �x � 10�.

x 2 � 8x � 20.

�x � 3� �x � 7� � x 2 � 4x � 21.

�x � 3� �x � 7� � x 2 � 4x � 21

6. Consider

a) Explain why you would notconsider the pairs of factorslisted below in factoring

b) Explain why you wouldconsider the pairs of factorslisted below in factoring

c) Factor:


x2 � 5x � 24.

x2 � 5x � 24.

x2 � 5x � 24.

x2 � 5x � 24.

319


1, �20 �19

2, �10 �8

4, �5 �1

5, �4 1

10, �2 8

20, �1 19


⎫⎪⎬⎪⎭

Because these sums areall positive, for thisproblem all of thecorresponding pairscan be disregarded.Note that in all threepairs, the positivenumber has the greaterabsolute value.

The numbers we needare 2 and �10.

1, �24

2, �12

3, �8

4, �6

PAIRS OF FACTORS

SUMS OF FACTORS

�1, 24

�2, 12

�3, 8

�4, 6

PAIRS OF FACTORS

SUMS OF FACTORS

Ch05pgs317-326 1/19/06 9:32 AM Page 319

ISB

N:0

-536

-477

42-6


EXAMPLE 4 Factor:

It helps to first write the trinomial in descending order:Since the constant term, �24, is negative, we look for a factorization of �24 inwhich one factor is positive and one factor is negative. Their sum must be 5,so the positive factor must have the larger absolute value. Thus we consideronly pairs of factors in which the positive term has the larger absolute value.

The factorization is The check is left to the student.


EXAMPLE 5 Factor:

Consider this trinomial as We look for numbers p andq such that

Since the constant term, �110, is negative, we look for a factorization of �110in which one factor is positive and one factor is negative. Their sum must be�1. The middle-term coefficient, �1, is small compared to �110. This tells usthat the desired factors are close to each other in absolute value. The numberswe want are 10 and �11. The factorization is

EXAMPLE 6 Factor:

We consider the trinomial in the equivalent form

This way we think of as the “constant” term and 4b as the “coefficient”of the middle term. Then we try to express as a product of two factorswhose sum is 4b. Those factors are and 7b. The factorization is

Check:

There are polynomials that are not factorable.

EXAMPLE 7 Factor:

Since 5 has very few factors, we can easily check all possibilities.

x 2 � x � 5.

� a2 � 4ab � 21b2.

�a � 3b� �a � 7b� � a2 � 7ab � 3ba � 21b2

�a � 3b� �a � 7b�.�3b

�21b2�21b2

a2 � 4ba � 21b2.

a2 � 4ab � 21b2.

�x 2 � 10� �x 2 � 11�.

x 4 � x 2 � 110 � �x 2 � p� �x 2 � q�.

�x 2�2 � x 2 � 110.

x 4 � x 2 � 110.

�t � 3� �t � 8�.

The numbers we needare �3 and 8.

t 2 � 5t � 24.

t 2 � 24 � 5t.7. Consider

a) Explain why you would notconsider the pairs of factorslisted below in factoring

b) Explain why you wouldconsider the pairs of factorslisted below in factoring

c) Factor:

Factor.

8.

9.


�24 � 10t � t 2

a2 � 24 � 10a

x2 � 10x � 24.

x2 � 10x � 24.

x2 � 10x � 24.

x2 � 10x � 24.

320


�1, 24 23

�2, 12 10

�3, 8 5

�4, 6 2


5, 1 6

�5, �1 �6


1, �24

2, �12

3, �8

4, �6

PAIRS OF FACTORS

SUMS OF FACTORS

�1, 24

�2, 12

�3, 8

�4, 6

PAIRS OF FACTORS

SUMS OF FACTORS

Ch05pgs317-326 1/19/06 9:32 AM Page 320

ISB

N:0-536-47742-6


There are no factors whose sum is �1. Thus the polynomial is not factorableinto factors that are polynomials.

In this text, a polynomial like that cannot be factored furtheris said to be prime. In more advanced courses, polynomials like can be factored and are not considered prime.


Often factoring requires two or more steps. In general, when told to fac-tor, we should factor completely. This means that the final factorizationshould not contain any factors that can be factored further.

EXAMPLE 8 Factor:

Always look first for a common factor. This time there is one, 2x, which wefactor out first:

Now consider Since the constant term is positive and the co-efficient of the middle term is negative, we look for a factorization of 25 inwhich both factors are negative. Their sum must be �10.

The factorization of is or The finalfactorization is We check by multiplying:


Once any common factors have been factored out, the following sum-mary can be used to factor

TO FACTOR

1. First arrange in descending order.2. Use a trial-and-error process that looks for factors of c whose

sum is b.3. If c is positive, the signs of the factors are the same as the sign

of b.4. If c is negative, one factor is positive and the other is negative.

If the sum of two factors is the opposite of b, changing the sign of each factor will give the desired factors whose sum is b.

5. Check by multiplying.

x 2 � bx � c

x 2 � bx � c.

� 2x 3 � 20x 2 � 50x.

� �2x� �x 2� � �2x� �10x� � �2x� �25� 2x�x � 5�2 � 2x�x 2 � 10x � 25�

2x�x � 5�2.�x � 5�2.�x � 5� �x � 5�,x 2 � 10x � 25

The numbers we needare �5 and �5.

x 2 � 10x � 25.

2x 3 � 20x 2 � 50x � 2x�x 2 � 10x � 25�.

2x 3 � 20x 2 � 50x.

x 2 � x � 5x 2 � x � 5

Factor.

10.

11.

12.

Factor.

13.

14.

15.


3x3 � 24x2 � 48x

p2 � pq � 3pq2

x3 � 4x2 � 12x

x2 � 2x � 7

t 4 � 5t 2 � 14

y2 � 12 � 4y

321


�25, �1 �26

�5, �5 �10


Ch05pgs317-326 1/19/06 9:32 AM Page 321

ISB

N:0

-536

-477

42-6


LEADING COEFFICIENT �1

EXAMPLE 9 Factor:

Note that the polynomial is written in ascending order. When we write itin descending order, we get

which has a leading coefficient of �1. Before factoring in such a case, we canfactor out a �1, as follows:

Then we proceed to factor We get

We can also express this answer in two other ways by multiplying either bino-mial by �1. Thus each of the following is a correct answer:

Multiplying by �1

Multiplying by �1


x � 2 � �x � 5� ��x � 2�.x � 5 � ��x � 5� �x � 2�

�x 2 � 3x � 10 � �1�x � 5� �x � 2�

�x 2 � 3x � 10 � �1�x 2 � 3x � 10� � �1�x � 5� �x � 2�.

x 2 � 3x � 10.

�x 2 � 3x � 10 � �1�x 2 � 3x � 10�.

�x 2 � 3x � 10,

10 � 3x � x 2.

Factor.

16.

17.


�x2 � 3x � 18

14 � 5x � x2

Study Tips TIME MANAGEMENT (PART 2)

Here are someadditional tips to help

you with timemanagement. (See alsothe Study Tips on time

management inSections 2.2 and 5.7.)

� Are you a morning or an evening person? If you are an evening person, it might bebest to avoid scheduling early-morning classes. If you are a morning person, do theopposite, but go to bed earlier to compensate. Nothing can drain your study time andeffectiveness like fatigue.

� Keep on schedule. Your course syllabus provides a plan for the semester’s schedule.Use a write-on calendar, daily planner, Palm Pilot or other PDA, or laptop computer tooutline your time for the semester. Be sure to note deadlines involving term papersand exams so you can begin a task early, breaking it down into smaller segments thatcan be accomplished more easily.

� Balance your class schedule. You may be someone who prefers large blocks of timefor study on the off days. In that case, it might be advantageous for you to take coursesthat meet only three days a week. Keep in mind, however, that this might be a problemwhen tests in more than one course are scheduled for the same day.

“Time is our most important asset, yet we tend to waste it, kill it, and spend it rather thaninvest it.”

Jim Rohn, motivational speaker

322


Ch05pgs317-326 1/19/06 9:32 AM Page 322

ISB

N:0-536-47742-6


323

Exercise Set 5.2



Math TutorCenter

InterActMath

MyMathLabMathXL

Factor. Remember that you can check by multiplying.

1. 2. 3. x2 � 7x � 12x2 � 5x � 6x2 � 8x � 15

4. 5. 6. y2 � 11y � 28x2 � 6x � 9x2 � 9x � 8

7. 8. 9. b2 � 5b � 4a2 � 7a � 30x2 � 5x � 14

10. 11. 12. x2 �25

x �1

25x2 �

23

x �19

z2 � 8z � 7

PAIRS OF FACTORS SUMS OF FACTORS PAIRS OF FACTORS SUMS OF FACTORS PAIRS OF FACTORS SUMS OF FACTORS




Ch05pgs317-326 1/19/06 9:32 AM Page 323

ISB

N:0

-536

-477

42-6


324


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

13. 14. 15. 16. x2 � 4x � 21y2 � 11y � 10t 2 � 12t � 35d2 � 7d � 10

17. 18. 19. 20. y2 � 3y � 28x2 � 7x � 18x2 � 5x � 3x2 � x � 1

21. 22. 23. 24. x3 � 7x2 � 60xy3 � 4y2 � 45yx3 � x2 � 42xx3 � 6x2 � 16x

25. 26. 27. 28. b4 � 5b2 � 24c4 � c2 � 56x2 � 72 � 6x�2x � 99 � x2

29. 30. 31. 32. x2 � 2x � 15x2 � x � 42x4 � x2 � 6a4 � 2a2 � 35

33. 34. 35. 36. a2 � 19a � 88x2 � 20x � 10011 � 3w � w 27 � 2p � p2

37. 38. 39. 40. �z2 � 36 � 9z24 � a2 � 10a45 � 4x � x230 � 7x � x2

Ch05pgs317-326 1/19/06 9:32 AM Page 324

ISB

N:0-536-47742-6


325

Exercise Set 5.2

41. 42. 43. 44. 4x2 � 40x � 100x2 � 21x � 72x4 � 20x3 � 96x2x4 � 21x3 � 100x2

45. 46. 47. 48. a2 � 9a � 90a2 � a � 132y2 � 21y � 108x2 � 25x � 144

49. 50. 51. 52. 112 � 9y � y2108 � 3x � x296 � 22d � d2120 � 23x � x2

53. 54. 55. 56. a2 � 2ab � 3b2p2 � 3pq � 10q2t 2 � 0.3t � 0.10y2 � 0.2y � 0.08

57. 58. 59. 60. x2 � 11xy � 24y2m2 � 5mn � 4n272 � 6m � m284 � 8t � t 2

61. 62. 63. 64. 7x9 � 28x8 � 35x76a10 � 30a9 � 84a8p2 � 5pq � 24q2s2 � 2st � 15t 2

65. Gwyneth factors as Is she wrong? Why or why not? What

advice would you offer?

66. When searching for a factorization, why do we listpairs of numbers with the correct product instead ofpairs of numbers with the correct sum?

DW�x2 � 5x� �x � 3�.

x3 � 8x2 � 15xDW

67. Without multiplying explain why it cannot possibly be a factorization of

68. What is the advantage of writing out the prime factorization of c when factoring with alarge value of c?

x2 � bx � c

DWx2 � 35x � 306.

�x � 17� �x � 18�,DW

Ch05pgs317-326 1/19/06 9:32 AM Page 325

ISB

N:0

-536

-477

42-6


326


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

Multiply. [4.6d]

81. Find all integers m for which can befactored.

82. Find all integers b for which can befactored.

a2 � ba � 50y2 � my � 50

69. 70. 71. 72. �4w � 11�2�7w � 6�2�7w � 6� �4w � 11�8x�2x2 � 6x � 1�

73. 74. 75. �3x � 5y� �2x � 7y��y��y2 � 3y � 5��4w � 11� �4w � 11�

76. Simplify: [4.2a, b]�3x4�3.

77. 78. 2x � 7 � 03x � 8 � 0

Solve. [2.3a]

79. Arrests for Counterfeiting. In a recent year,29,200 people were arrested for counterfeiting. Thisnumber was down 1.2% from the preceding year. Howmany people were arrested the preceding year? [2.5a]

80. The first angle of a triangle is four times as large asthe second. The measure of the third angle is 30°greater than that of the second. Find the anglemeasures. [2.6a]

Solve.

Factor completely.

83. 84. 85. x2 �307 x �

257x2 �

14 x �

18x2 �

12 x �

316

86. 87. 88. a2m � 11am � 28b2n � 7bn � 1013 x3 �

13 x2 � 2x

Find a polynomial in factored form for the shaded area. (Leave answers in terms of �.)

89. 90.

x

x xx

x

x

SKILL MAINTENANCE

SYNTHESIS

Ch05pgs317-326 1/19/06 9:32 AM Page 326

ISB

N:0-536-47742-6


327


ax2 � bx � c, a � 1:

In Section 5.2, we learned a trial-and-error method to factor trinomials ofthe type In this section, we factor trinomials in which the coeffi-cient of the leading term is not 1. The procedure we learn is a refined trial-and-error method.

The FOIL Method

We want to factor trinomials of the type Consider the followingmultiplication:

F O I L

� � � 20

� � 20

F O I L

To factor we reverse the above multiplication, using what wemight call an “unFOIL” process. We look for two binomials and whose product is The product of theFirst terms must be The product of the Outside terms plus the productof the Inside terms must be The product of the Last terms must be 20.We know from the preceding discussion that the answer is Generally, however, finding such an answer is a refined trial-and-errorprocess. It turns out that is also a correct answer, butwe generally choose an answer in which the first coefficients are positive.

We will use the following trial-and-error method.

THE FOIL METHOD

To factor using the FOIL method:

1. Factor out the largest common factor, if one exists.2. Find two First terms whose product is

FOIL

3. Find two Last terms whose product is c:

FOIL

4. Look for Outer and Inner products resulting from steps (2) and (3)for which the sum is bx:

I FOILO

5. Always check by multiplying.

� x � � � x � � � ax2 � bx � c.

� x � � � x � � � ax2 � bx � c.

� x � � � x � � � ax2 � bx � c.

ax2.

ax2 � bx � c, a � 1,

��2x � 5� ��3x � 4�

�2x � 5� �3x � 4�.23x.

6x2.�rx � p� �sx � q� � 6x2 � 23x � 20.

sx � qrx � p6x2 � 23x � 20,

5 � 45 � 32 � 42 � 3

�

23x6x2�

15x8x6x2�2x � 5� �3x � 4� �

ax2 � bx � c.

x2x2 � bx � c.

5.35.3 FACTORING THE FOIL METHOD

ax2 � bx � c, a � 1:


, , usingthe FOIL method.

a � 1ax2 � bx � c

}

The ac-method in Section 5.4

To the student : In Section 5.4,we will consider an alternativemethod for the same kind offactoring. It involves factoringby grouping and is called theac-method.

To the instructor : We presenttwo ways to factor generaltrinomials in Sections 5.3 and5.4: the FOIL method inSection 5.3 and the ac-methodin Section 5.4. You can teachboth methods and let thestudent use the one that he orshe prefers or you can selectjust one.

Ch05pgs327-335 1/19/06 9:35 AM Page 327

ISB

N:0

-536

-477

42-6


EXAMPLE 1 Factor:

1) First, we check for a common factor. Here there is none (other than 1 or�1).

2) Find two First terms whose product is

The only possibilities for the First terms are and x, so any factori-zation must be of the form

3) Find two Last terms whose product is �8.

Possible factorizations of �8 are

and

Since the First terms are not identical, we must also consider

and

4) Inspect the Outer and Inner products resulting from steps (2) and (3).Look for a combination in which the sum of the products is the middleterm,

Trial Product

Wrong middle term

Wrong middle term

Wrong middle term

Correct middle term!

Wrong middle term

Wrong middle term

Wrong middle term

Wrong middle term

The correct factorization is

5) Check:

Two observations can be made from Example 1. First, we listed all pos-sible trials even though we could have stopped after having found the correctfactorization. We did this to show that each trial differs only in the middleterm of the product. Second, note that as in Section 5.2, only the sign of themiddle term changes when the signs in the binomials are reversed:

Plus Minus

Minus Plus Middle term changes sign


�3x � 4� �x � 2� � 3x2 � 2x � 8.

�3x � 4� �x � 2� � 3x2 � 2x � 8

�3x � 2� �x � 4� � 3x2 � 10x � 8.

�3x � 2� �x � 4�.

� 3x2 � 2x � 83x2 � 6x � 4x � 8�3x � 4� �x � 2�� 3x2 � 2x � 83x2 � 6x � 4x � 8�3x � 4� �x � 2�� 3x2 � 23x � 83x2 � 24x � x � 8�3x � 1� �x � 8�� 3x2 � 23x � 83x2 � 24x � x � 8�3x � 1� �x � 8�� 3x2 � 10x � 83x2 � 12x � 2x � 8�3x � 2� �x � 4�� 3x2 � 10x � 83x2 � 12x � 2x � 8�3x � 2� �x � 4�� 3x2 � 5x � 83x2 � 3x � 8x � 8�3x � 8� �x � 1�� 3x2 � 5x � 83x2 � 3x � 8x � 8�3x � 8� �x � 1�

�10x:

��4� � 2.4 � ��2�,��1� � 8,1 � ��8�,

2 � ��4�.��2� � 4,8 � ��1�,��8� � 1,

�3x � ��x � �.

3x

3x2.

3x2 � 10x � 8.Factor.

1.

2.


12x2 � 17x � 5

2x2 � x � 15

Study Tips

READING EXAMPLES

A careful study of theexamples in these sections onfactoring is critical. Read themcarefully to ensure success!

328


Ch05pgs327-335 1/19/06 9:35 AM Page 328

ISB

N:0-536-47742-6


EXAMPLE 2 Factor:

1) First, we factor out the largest common factor, 4:

Now we factor the trinomial

2) Because can be factored as or we have these possibilitiesfor factorizations:

or

3) There are four pairs of factors of 10 and they each can be listed in twoways:

10, 1 �10, �1 5, 2 �5, �2

and

1, 10 �1, �10 2, 5 �2, �5.

4) The two possibilities from step (2) and the eight possibilities fromstep (3) give or 16 possibilities for factorizations. We look for Outerand Inner products resulting from steps (2) and (3) for which the sum isthe middle term, Since the sign of the middle term is negative, butthe sign of the last term, 10, is positive, the two factors of 10 must bothbe negative. This means only four pairings from step (3) need be consid-ered. We first try these factors with

If none gives the correct factorization, we will consider

Trial Product

Wrong middle term

Wrong middle term

Wrong middle term

Correct middle term!

Since we have a correct factorization, we need not consider

The factorization of is but do not forgetthe common factor ! We must include it in order to factor the original trinomial:

5) Check:

Caution!

When factoring any polynomial, always look for a common factor. Failureto do so is such a common error that this caution bears repeating.

4�3x � 2� �2x � 5� � 4�6x2 � 19x � 10� � 24x2 � 76x � 40.

� 4�3x � 2� �2x � 5�. 24x2 � 76x � 40 � 4�6x2 � 19x � 10�

�3x � 2� �2x � 5�,6x2 � 19x � 10

�6x � � �x � �.

� 6x2 � 19x � 106x2 � 15x � 4x � 10�3x � 2� �2x � 5�� 6x2 � 16x � 106x2 � 6x � 10x � 10�3x � 5� �2x � 2�� 6x2 � 32x � 106x2 � 30x � 2x � 10�3x � 1� �2x � 10�� 6x2 � 23x � 106x2 � 3x � 20x � 10�3x � 10� �2x � 1�

�6x � � �x � �.

�3x � � �2x � �.

�19x.

2 � 8,

�6x � � �x � �.�3x � � �2x � �

6x � x,3x � 2x6x2

6x2 � 19x � 10.

4�6x2 � 19x � 10�.

24x2 � 76x � 40.

329


ax2 � bx � c, a � 1:

Ch05pgs327-335 1/19/06 9:35 AM Page 329

ISB

N:0

-536

-477

42-6


In Example 2, look again at the possibility Without mul-tiplying, we can reject such a possibility. To see why, consider the following:

The expression has a common factor, 2. But we removed the largestcommon factor in the first step. If were one of the factors, then 2 wouldhave to be a common factor in addition to the original 4. Thus, can-not be part of the factorization of the original trinomial.

Given that the largest common factor is factored out at the outset, weneed not consider factorizations that have a common factor.


EXAMPLE 3 Factor:

1) There is no common factor (other than 1 or �1).

2) Because factors as or we have these possibilities forfactorizations:

or

3) There are two pairs of factors of 7 and they each can be listed in two ways:

1, 7 �1, �7 and 7, 1 �7, �1.

4) From steps (2) and (3), we see that there are 8 possibilities for factoriza-tions. Look for Outer and Inner products for which the sum is the middleterm. Because all coefficients in are positive, we need con-sider only positive factors of 7. The possibilities are

Correct middle term

The factorization is

5) Check:

Do Exercise 5.

TIPS FOR FACTORING

• Always factor out the largest common factor, if one exists.

• Once the common factor has been factored out of the originaltrinomial, no binomial factor can contain a common factor(other than 1 or �1).

• If c is positive, then the signs in both binomial factors mustmatch the sign of b. (This assumes that )

• Reversing the signs in the binomials reverses the sign of themiddle term of their product.

• Organize your work so that you can keep track of whichpossibilities have or have not been checked.

• Always check by multiplying.

a � 0.

ax2 � bx � c, a �/ 1

�5x � 1� �2x � 7� � 10x2 � 37x � 7.

�5x � 1� �2x � 7�.

�5x � 1� �2x � 7� � 10x2 � 37x � 7.

�5x � 7� �2x � 1� � 10x2 � 19x � 7,

�10x � 7� �x � 1� � 10x2 � 17x � 7,

�10x � 1� �x � 7� � 10x2 � 71x � 7,

10x2 � 37x � 7

�5x � � �2x � �.�10x � � �x � �

5x � 2x,10x � x10x2

10x2 � 37x � 7.

�2x � 2�2x � 2

2x � 2

�3x � 5� �2x � 2� � 2�3x � 5� �x � 1�.

�3x � 5� �2x � 2�.Factor.

3.

4.

5. Factor:


6x2 � 7x � 2.

20x2 � 46x � 24

3x2 � 19x � 20

330


Ch05pgs327-335 1/19/06 9:35 AM Page 330

ISB

N:0-536-47742-6


EXAMPLE 4 Factor:

An important problem-solving strategy is to find a way to make newproblems look like problems we already know how to solve. (See Example 9 inSection 5.2.) The factoring tips above apply only to trinomials of the form

with This leads us to rewrite in descend-ing order:

Writing in descending order

Although looks similar to the trinomials we have factored,the tips above require a positive leading coefficient. This can be attained byfactoring out �1:

Factoring out �1 changes the signs of the coefficients.

Using the result from Example 1

The factorization of is Other correct an-swers are

Multiplying by �1

Multiplying by �1


EXAMPLE 5 Factor:

1) Factor out a common factor, if any.

There is none (other than 1 or �1).

2) Factor the first term,

Possibilities are 2p, 3p and 6p, p. We have these as possibilities forfactorizations:

or

3) Factor the last term, which has a negative coefficient.

There are six pairs of factors and each can be listed in two ways:

and

4) The coefficient of the middle term is negative, so we look for combinationsof factors from steps (2) and (3) such that the sum of their products has anegative coefficient. We try some possibilities:

Correct middle term

The factorization of is

5) The check is left to the student.


�2p � 7q� �3p � 4q�.6p2 � 13pq � 28q2

�2p � 7q� �3p � 4q� � 6p2 � 13pq � 28q2.

�2p � q� �3p � 28q� � 6p2 � 53pq � 28q2,

�4q, 7q.4q, �7q�2q, 14q2q, �14q�q, 28qq, �28q

7q, �4q�7q, 4q14q, �2q�14q, 2q28q, �q�28q, q

�28q2,

�6p � � � p � �.�2p � � �3p � �

6p2.

6p2 � 13pq � 28q2.

3x � 2 � ��3x � 2� �x � 4�.x � 4 10x � 8 � 3x2 � �3x � 2� ��x � 4�

�1�3x � 2� �x � 4�.10x � 8 � 3x2

� �1�3x � 2� �x � 4�.

�3x2 � 10x � 8 � �1�3x2 � 10x � 8�

�3x2 � 10x � 8

10x � 8 � 3x2 � �3x2 � 10x � 8.

10x � 8 � 3x2a � 0.ax2 � bx � c,

10x � 8 � 3x2. Factor.

6.

7.

Factor.

8.

9.


6x2 � 15xy � 9y2

6a2 � 5ab � b2

2x � 8 � 6x2

2 � x � 6x2

331


ax2 � bx � c, a � 1:

Ch05pgs327-335 1/19/06 9:35 AM Page 331

ISB

N:0

-536

-477

42-6


332


CALCULATOR CORNER

Checking Factorizations A partial check of a factorization can beperformed using a table or a graph. To check the factorization

for example, we enterand on the

equation-editor screen (see page 175). Then we set up a table in AUTO

mode (see page 180). If the factorization is correct, the values of and will be the same regardless of the table settings used.

We can also graph andIf the graphs appear to coincide, the

factorization is probably correct.

Keep in mind that these procedures provide only a partial check sincewe cannot view all possible values of x in a table nor can we see the entiregraph.

Exercises: Use a table or a graph to determine whether thefactorization is correct.

1.

2.

3.

4.

5.

6.

7.

8. x2 � 4 � �x � 2� �x � 2�x2 � 4 � �x � 2� �x � 2�12x2 � 17x � 5 � �4x � 1� �3x � 5�12x2 � 17x � 5 � �6x � 1� �2x � 5�10x2 � 37x � 7 � �5x � 1� �2x � 7�5x2 � 17x � 12 � �5x � 3� �x � 4�4x2 � 5x � 6 � �4x � 3� �x � 2�24x2 � 76x � 40 � 4�3x � 2� �2x � 5�

y1 � 6x 3 � 9x 2 � 4x � 6,y2 � (3x 2 � 2)(2x � 3)

5�5

�14

10

Yscl = 2

y2 � �3x 2 � 2� �2x � 3�.y1 � 6x 3 � 9x 2 � 4x � 6

�261�98�25�6�51487

�261�98�25�6�51487

X

X � �3

Y1 Y2

�3�2�10123

y2y1

y2 � �3x 2 � 2� �2x � 3�y1 � 6x 3 � 9x 2 � 4x � 66x 3 � 9x 2 � 4x � 6 � �3x 2 � 2� �2x � 3�,

Ch05pgs327-335 1/19/06 9:35 AM Page 332

ISB

N:0-536-47742-6


333

Exercise Set 5.3



Math TutorCenter

InterActMath

MyMathLabMathXL

Factor.

1. 2. 3. 4. 4x2 � 17x � 155x2 � x � 183x2 � x � 42x2 � 7x � 4

5. 6. 7. 8. 7x2 � 15x � 23x2 � 4x � 16x2 � 23x � 76x2 � 23x � 7

9. 10. 11. 12. 15x2 � 19x � 102x2 � x � 19x2 � 6x � 84x2 � 4x � 15

13. 14. 15. 16. 18x2 � 3x � 103x2 � 5x � 22x2 � 5x � 29x2 � 18x � 16

17. 18. 19. 20. 35x2 � 34x � 814x2 � 19x � 315x2 � 19x � 1012x2 � 31x � 20

21. 22. 23. 24. 16 � 36x2 � 48x49 � 42x � 9x26 � 13x � 6x29x2 � 18x � 8

25. 26. 27. 28. 9a2 � 12a � 535x2 � 57x � 4416p2 � 78p � 2724x2 � 47x � 2

29. 30. 31. 32. 6x2 � 33x � 1512x2 � 28x � 2415 � x � 2x220 � 6x � 2x2

Ch05pgs327-335 1/19/06 9:35 AM Page 333

ISB

N:0

-536

-477

42-6


334


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

33. 34. 35. 36. �9 � 18x2 � 21x4y � 6y2 � 1018t 2 � 24t � 630x2 � 24x � 54

37. 38. 39. 40. 6x2 � 33x � 1512x2 � 28x � 246t 2 � 13t � 63x2 � 4x � 1

41. 42. 43. 44. 14y2 � 35y � 149x2 � 18x � 16�19x � 15x2 � 6�1 � 2x2 � x

45. 46. 47. 48. 15x3 � 19x2 � 10x12p3 � 31p2 � 20p18x2 � 3x � 1015x2 � 25x � 10

49. 50. 51. 52. 1 � p � 2p2�15x2 � 19x � 633t � 15 � 6t 216 � 18x � 9x2

53. 54. 55. 56. 144x5 � 168x4 � 48x3168x3 � 45x2 � 3x70x4 � 68x3 � 16x214x4 � 19x3 � 3x2

57. 58. 59. 60. 9x2 � 42x � 4925t 2 � 80t � 649x4 � 18x2 � 815x4 � 19x2 � 6

61. 62. 63. 64. 9y2 � 42y � 4725x2 � 79x � 6418x3 � 21x2 � 9x6x3 � 4x2 � 10x

Ch05pgs327-335 1/19/06 9:35 AM Page 334

ISB

N:0-536-47742-6


335

Exercise Set 5.3

65. 66. 67. 68. 12a2 � 17ab � 6b212m2 � mn � 20n22x2 � 11x � 96x2 � 19x � 5

69. 70. 71. 72. 10s2 � 4st � 6t 29a2 � 18ab � 8b23p2 � 16pq � 12q26a2 � ab � 15b2

73. 74. 75. 76. 15a2 � 5ab � 20b218x2 � 6xy � 24y230a2 � 87ab � 30b235p2 � 34pq � 8q2

77. Explain how the factoring in Exercise 21 can beused to aid the factoring in Exercise 71.

78. A student presents the following work:

Is it correct? Explain.

� 2�x � 3� �x � 4�. 4x2 � 28x � 48 � �2x � 6� �2x � 8�

DWDW

Solve. [2.4b]

79. for q 80. for x 81. for y 82. for qp � q � r � 2,3x � 2y � 6,y � mx � b,A � pq � 7,

Solve. [2.7e]

83. 84. 2x � 4�x � 3x� � 6x � 8 � 9x5 � 4x � �11

85. Graph: [3.2a] 86. Divide: [4.1e]y12

y4 .y �25

x � 1.

Find the intercepts of the equation. [3.3a]

87. 88. 89. x � 1.3y � 6.54x � 16y � 644x � 16y � 64

90. 91. 92. y � 2x � 5y � 4 � 5x23 x �

58 y � 5

12

Factor.

93. 94. �15x2m � 26xm � 820x2n � 16xn � 3

95. 96. x2n�1 � 2xn�1 � x3x6a � 2x3a � 1

97.–106. Use the TABLE feature to check the factoring in Exercises 15–24.

SKILL MAINTENANCE

SYNTHESIS

Ch05pgs327-335 1/19/06 9:35 AM Page 335

ISB

N:0

-536

-477

42-6


336


The ac-Method

Another method for factoring trinomials of the type , , in-volves the product, ac, of the leading coefficient a and the last term c. It iscalled the ac-method. Because it uses factoring by grouping, it is also referredto as the grouping method.

We know how to factor the trinomial . We look for factors ofthe constant term, 6, whose sum is the coefficient of the middle term, 5. Whathappens when the leading coefficient is not 1? To factor a trinomial like

, we can use a method similar to what we used for That method is outlined as follows.

THE ac-METHOD

To factor , , using the ac-method:

1. Factor out a common factor, if any.2. Multiply the leading coefficient a and the constant c.3. Try to factor the product ac so that the sum of the factors is b.

That is, find integers p and q such that and .4. Split the middle term. That is, write it as a sum using the factors

found in step (3).5. Factor by grouping.6. Check by multiplying.

EXAMPLE 1 Factor:

1) First, we factor out a common factor, if any. There is none (other than 1 or ).

2) We multiply the leading coefficient, 3, and the constant, :

3) Then we look for a factorization of in which the sum of the factors isthe coefficient of the middle term, .

4) Next, we split the middle term as a sum or a difference using the factorsfound in step (3): �10x � 2x � 12x.

�10�24

3��8� � �24.

�8

�1

3x2 � 10x � 8.

p � q � bpq � ac

a � 1ax2 � bx � c

x2 � 5x � 6.3x2 � 10x � 8

x2 � 5x � 6

a � 1ax2 � bx � c

5.45.4 FACTORING THE ac-METHOD

ax2 � bx � c, a � 1:


, , usingthe ac-method.

a � 1ax2 � bx � c

�1, 24 23

1, �24 �23

�2, 12 10

2, �12 �10

�3, 8 5

3, �8 �5

�4, 6 2

4, �6 �2


2 � ��12� � �10

Ch05pgs336-341 1/19/06 9:40 AM Page 336

ISB

N:0-536-47742-6


5) Finally, we factor by grouping, as follows:

Substituting for

Factoring by grouping

We can also split the middle term as . We still get the same factorization, although the factors may be in a different order. Note the following:

Substituting for


6) Check:


EXAMPLE 2 Factor:

1) First, we factor out a common factor, if any. The number 2 is common toall three terms, so we factor it out:

2) Next, we factor the trinomial . We multiply the leading coef-ficient and the constant, 4 and :

3) We try to factor so that the sum of the factors is 4.

4) Then we split the middle term, , as follows:

5) Finally, we factor by grouping:

Substituting for 4x


The factorization of is . But don’t forget thecommon factor! We must include it to get a factorization of the original tri-nomial:

6) Check:


2�2x � 3� �2x � 1� � 2�4x2 � 4x � 3� � 8x2 � 8x � 6.

8x2 � 8x � 6 � 2�2x � 3� �2x � 1�.

�2x � 3� �2x � 1�4x2 � 4x � 3

� �2x � 3� �2x � 1�. � 2x�2x � 1� � 3�2x � 1� � �4x2 � 2x� � �6x � 3�

�2x � 6x 4x2 � 4x � 3 � 4x2 � 2x � 6x � 3

4x � �2x � 6x.4x

�12

4��3� � �12.�34x2 � 4x � 3

2�4x2 � 4x � 3�.

8x2 � 8x � 6.

�3x � 2� �x � 4� � 3x2 � 10x � 8.

� �3x � 2� �x � 4�. � 3x�x � 4� � 2�x � 4� � �3x2 � 12x� � �2x � 8�

�10x�12x � 2x 3x2 � 10x � 8 � 3x2 � 12x � 2x � 8

�12x � 2x

� �x � 4� �3x � 2�. � x�3x � 2� � 4�3x � 2� � �3x2 � 2x� � ��12x � 8�

�10x2x � 12x 3x2 � 10x � 8 � 3x2 � 2x � 12x � 8

Factor.

1.

2.

Factor.

3.

4.


20x2 � 46x � 24

6x2 � 15x � 9

12x2 � 17x � 5

6x2 � 7x � 2

337

5.4 Factoring , : The ac-Method

a � 1ax2 � bx � c

�1, 12 11

1, �12 �11

�2, 6 4

2, �6 �4

�3, 4 1

3, �4 �1


�2 � 6 � 4

Ch05pgs336-341 1/19/06 9:40 AM Page 337

ISB

N:0

-536

-477

42-6


338


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

5.45.4 Student’sSolutionsManual


Math TutorCenter

InterActMath

MyMathLabMathXLEXERCISE SET For Extra Help

Factor. Note that the middle term has already been split.

1. 2. 3. x2 � 4x � x � 4x2 � 3x � x � 3x2 � 2x � 7x � 14

4. 5. 6. 3x2 � 2x � 3x � 26x2 � 4x � 9x � 6a2 � 5a � 2a � 10

7. 8. 9. 35x2 � 40x � 21x � 2424 � 18y � 20y � 15y23x2 � 4x � 12x � 16

10. 11. 12. 2x4 � 6x2 � 5x2 � 154x2 � 6x � 6x � 98x2 � 6x � 28x � 21

13. 14. 9x4 � 6x2 � 6x2 � 42x4 � 6x2 � 5x2 � 15

Factor by grouping.

15. 16. 17. 18. 3x2 � x � 43x2 � 4x � 155x2 � x � 182x2 � 7x � 4

19. 20. 21. 22. 7x2 � 15x � 23x2 � 4x � 16x2 � 13x � 66x2 � 23x � 7

23. 24. 25. 26. 15x2 � 19x � 102x2 � x � 19x2 � 6x � 84x2 � 4x � 15

Ch05pgs336-341 1/19/06 9:40 AM Page 338

ISB

N:0-536-47742-6


339

Exercise Set 5.4

27. 28. 29. 30. 18x2 � 3x � 103x2 � 5x � 22x2 � 5x � 29x2 � 18x � 16

31. 32. 33. 34. 35x2 � 34x � 814x2 � 19x � 315x2 � 19x � 1012x2 � 31x � 20

35. 36. 37. 38. 25x2 � 40x � 1649 � 42x � 9x26 � 13x � 6x29x2 � 18x � 8

39. 40. 41. 42. 17x � 4x2 � 155 � 9a2 � 12a16a2 � 78a � 2724x2 � 47x � 2

43. 44. 45. 46. 6x2 � 33x � 1512x2 � 28x � 2415 � x � 2x220 � 6x � 2x2

47. 48. 49. 50. �9 � 18x2 � 21x4y � 6y2 � 1018t 2 � 24t � 630x2 � 24x � 54

51. 52. 53. 54. 6x2 � 33x � 1512x2 � 28x � 246t 2 � t � 153x2 � 4x � 1

55. 56. 57. 58. 14y2 � 35y � 149x2 � 18x � 16�19x � 15x2 � 6�1 � 2x2 � x

Ch05pgs336-341 1/19/06 9:40 AM 9

ISB

N:0

-536

-477

42-6


340


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

59. 60. 61. 62. 15x3 � 19x2 � 10x12p3 � 31p2 � 20p18x2 � 3x � 1015x2 � 25x � 10

63. 64. 65. 66. �15x2 � 19x � 633t � 15 � 6t 21 � p � 2p24 � x � 5x2

67. 68. 69. 70. 144x5 � 168x4 � 48x3168x3 � 45x2 � 3x70x4 � 68x3 � 16x214x4 � 19x3 � 3x2

71. 72. 73. 74. 9x2 � 42x � 4925t 2 � 80t � 649x4 � 18x2 � 815x4 � 19x2 � 6

75. 76. 77. 78. 9y2 � 42y � 4725x2 � 79x � 6418x3 � 21x2 � 9x6x3 � 4x2 � 10x

79. 80. 81. 82. 12a2 � 17ab � 6b212m2 � mn � 20n22x2 � 11x � 96x2 � 19x � 5

83. 84. 85. 86. 10s2 � 4st � 6t 29a2 � 18ab � 8b23p2 � 16pq � 12q26a2 � ab � 15b2

Ch05pgs336-341 1/19/06 9:40 AM 0

ISB

N:0-536-47742-6


341

Exercise Set 5.4

87. 88. 89. 90. 15a2 � 5ab � 20b218x2 � 6xy � 24y230a2 � 87ab � 30b235p2 � 34pq � 8q2

91. 92. 93. 94. 15x3 � 33x4 � 6x535x5 � 57x4 � 44x360x � 4x2 � 8x360x � 18x2 � 6x3

95. If you have studied both the FOIL and the ac-methods of factoring , , decidewhich method you think is better and explain why.

96. Explain factoring , , using theac-method as though you were teaching a fellowstudent.

a � 1ax2 � bx � cDWa � 1ax2 � bx � c

DW

Solve. [2.7d, e]

Factor.

97. 98. 99. 6 � 3x � �18�3.8x � �824.6�10x � 1000

100. 101. 102. �2�x � 7� � �4�x � 5�12 x � 6x � 10 � x � 5x3 � 2x � 4x � �9

103. 3x � 6x � 2�x � 4� � 2�9 � 4x�

Solve. [2.6a]

105. The earth is a sphere (or ball) that is about 40,000 kmin circumference. Find the radius of the earth, inkilometers and in miles. Use 3.14 for . (Hint :

)

106. The second angle of a triangle is 10° less than twice thefirst. The third angle is 15° more than four times thefirst. Find the measure of the second angle.

1 km � 0.62 mi.�

107. 108. 24x2n � 22xn � 39x10 � 12x5 � 4

109. 110. �a � 4�2 � 2�a � 4� � 116x10 � 8x5 � 1

111.–120. Use the TABLE feature to check the factoring in Exercises 15–24.

SKILL MAINTENANCE

104. �6�x � 4� � 8�4 � x� � 3�x � 7�

SYNTHESIS

Ch05pgs336-341 1/19/06 9:40 AM Page 341

ISB

N:0

-536

-477

42-6


342


In this section, we first learn to factor trinomials that are squares of bino-mials. Then we factor binomials that are differences of squares.

Recognizing Trinomial Squares

Some trinomials are squares of binomials. For example, the trinomialis the square of the binomial To see this, we can

calculate It is or A trinomial thatis the square of a binomial is called a trinomial square, or a perfect-squaretrinomial.

In Chapter 4, we considered squaring binomials as special-product rules:

We can use these equations in reverse to factor trinomial squares.

TRINOMIAL SQUARES

How can we recognize when an expression to be factored is a trinomialsquare? Look at and In order for an expres-sion to be a trinomial square:

a) The two expressions and must be squares, such as

4,

When the coefficient is a perfect square and the power(s) of the variable(s)is (are) even, then the expression is a perfect square.

b) There must be no minus sign before or

c) If we multiply A and B and double the result, we get either the re-maining term or its opposite.

EXAMPLE 1 Determine whether is a trinomial square.

a) We know that and 9 are squares.

b) There is no minus sign before or 9.

c) If we multiply the square roots, x and 3, and double the product, we get theremaining term:

Thus, is the square of a binomial. In fact,


The answer is no, because only one term, is a square.x2,

x2 � 6x � 11

x2 � 6x � 9 � �x � 3�2.x2 � 6x � 9

2 � x � 3 � 6x.

x2

x2

x2 � 6x � 9

2 � AB,

B2.A2

16t 2.25x4,x2,

B2A2

A2 � 2AB � B2.A2 � 2AB � B2

A2 � 2AB � B2 � �A � B�2

A2 � 2AB � B2 � �A � B�2;

�A � B�2 � A2 � 2AB � B2.

�A � B�2 � A2 � 2AB � B2;

x2 � 10x � 25.x2 � 2 � x � 5 � 52,�x � 5�2.x � 5.x2 � 10x � 25

5.55.5 FACTORING TRINOMIAL SQUARESAND DIFFERENCES OF SQUARES

It would be helpful to memorize thistable of perfect squares.

ObjectivesRecognize trinomial squares.

Factor trinomial squares.

Recognize differences of squares.

Factor differences of squares,being careful to factorcompletely.

1 1

2 4

3 9

4 16

5 25

6 36

7 49

8 64

9 81

10 100

11 121

12 144

13 169

14 196

15 225

16 256

17 289

18 324

19 361

20 400

21 441

22 484

23 529

24 576

25 625

NUMBER,N

PERFECTSQUARE,

N2

Ch05pgs342-351 1/19/06 9:42 AM Page 342

ISB

N:0-536-47742-6



It helps to first write the trinomial in descending order:

a) We know that and 49 are squares.

b) There is no minus sign before or 49.

c) If we multiply the square roots, 4x and 7, and double the product, we getthe opposite of the remaining term: 56x is the opposite of �56x.

Thus, is a trinomial square. In fact,

Do Exercises 1–8.

Factoring Trinomial Squares

We can use the factoring methods from Sections 5.2–5.4 to factor trinomialsquares, but there is a faster method using the following equations.

FACTORING TRINOMIAL SQUARES

We consider 3 to be a square root of 9 because Similarly, A is asquare root of We use square roots of the squared terms and the sign of theremaining term to factor a trinomial square.

EXAMPLE 4 Factor:

The sign of themiddle term is positive.

EXAMPLE 5 Factor:

Changing to descending order

The sign of the middle termis negative.

EXAMPLE 6 Factor:


A2 � 2 A B � B2 � �A � B�2

16x2 � 40x � 25 � �4x�2 � 2 � 4x � 5 � 52 � �4x � 5�2

16x2 � 40x � 25.

� �x � 7�2

� x2 � 2 � x � 7 � 72

x2 � 49 � 14x � x2 � 14x � 49

x2 � 49 � 14x.

A2 � 2 A B � B2 � �A � B�2

x2 � 6x � 9 � x2 � 2 � x � 3 � 32 � �x � 3�2

x2 � 6x � 9.

A2.32 � 9.

A2 � 2AB � B2 � �A � B�2

A2 � 2AB � B2 � �A � B�2;

�4x � 7�2.16x2 � 56x � 49 �16x2 � 49 � 56x

2 � 4x � 7 � 56x;

16x2

16x2

16x2 � 56x � 49.

16x2 � 49 � 56x Determine whether each is atrinomial square. Write “yes” or “no.”

1.

2.

3.

4.

5.

6.

7.

8.

Factor.

9.

10.

11.

12.

13.


49 � 56y � 16y2

25x2 � 70x � 49

4 � t 2 � 4t

1 � 2x � x2

x2 � 2x � 1

25a2 � 9 � 30a

p2 � 6p � 9

16x2 � 40x � 25

5x2 � 16 � 14x

25 � 20y � 4y2

t 2 � 12t � 4

25 � x2 � 10x

x2 � 8x � 16

343


Ch05pgs342-351 1/19/06 9:42 AM Page 343

ISB

N:0

-536

-477

42-6


EXAMPLE 7 Factor:

EXAMPLE 8 Factor:

Always look first for a common factor. This time there is one, 3m:

EXAMPLE 9 Factor:


Recognizing Differences of Squares

The following polynomials are differences of squares:

To factor a difference of squares such as think about the formula weused in Chapter 4:

Equations are reversible, so we also know the following.

DIFFERENCE OF SQUARES

Thus,

To use this formula, we must be able to recognize when it applies. A differ-ence of squares is an expression like the following:

How can we recognize such expressions? Look at In order for a bino-mial to be a difference of squares:

a) There must be two expressions, both squares, such as

9, 1,

b) The terms must have different signs.

49y8.x6,25t 4,4x2,

A2 � B2.

A2 � B2.

x2 � 9 � �x � 3� �x � 3�.

A2 � B2 � �A � B� �A � B�

�A � B� �A � B� � A2 � B2.

x2 � 9,

a2 � 25b2.4t 2 � 49,x2 � 9,

� �2p � 3q�2

4p2 � 12pq � 9q2 � �2p�2 � 2�2p� �3q� � �3q�2

4p2 � 12pq � 9q2.

� 3m�5m � 7�2.

� 3m��5m�2 � 2�5m� �7� � 72� 75m3 � 210m2 � 147m � 3m�25m2 � 70m � 49�

75m3 � 210m2 � 147m.

� �t 2 � 10�2

t 4 � 20t 2 � 100 � �t 2�2 � 2�t 2� �10� � 102

t 4 � 20t 2 � 100.Factor.

14.

15.

16.

17.


9a2 � 30ab � 25b2

4z5 � 20z4 � 25z3

p4 � 18p2 � 81

48m2 � 75 � 120m

344


Ch05pgs342-351 1/19/06 9:42 AM Page 344

ISB

N:0-536-47742-6


EXAMPLE 10 Is a difference of squares?

a) The first expression is a square:The second expression is a square:

b) The terms have different signs, and

Thus we have a difference of squares,


a) The expression is not a square.

The expression is not a difference of squares.


a) The expressions and 16 are squares: and

b) The terms have different signs, and

Thus we have a difference of squares. We can also see this by rewriting in theequivalent form:


Factoring Differences of Squares

To factor a difference of squares, we use the following equation.

FACTORING A DIFFERENCE OF SQUARES

To factor a difference of squares we find A and B, which aresquare roots of the expressions and We then use A and B to form twofactors. One is the sum and the other is the difference

EXAMPLE 13 Factor:

EXAMPLE 14 Factor:

A2 � B2 � �A � B� �A � B�

9 � 16t 4 � 32 � �4t 2�2 � �3 � 4t 2� �3 � 4t 2�

9 � 16t 4.

A2 � B2 � �A � B� �A � B�

x2 � 4 � x2 � 22 � �x � 2� �x � 2�

x2 � 4.

A � B.A � B,B2.A2

A2 � B2,

A2 � B2 � �A � B� �A � B�

16 � 4x2.

�16.�4x2

16 � 42.4x2 � �2x�24x2

�4x2 � 16

t 3

25 � t 3

�3x�2 � 82.

�64.�9x2

64 � 82.9x2 � �3x�2.

9x2 � 64 Determine whether each is adifference of squares. Write “yes” or “no.”

18.

19.

20.

21.

22.

23.

24.


�49 � 25t 2

9w 6 � 1

16x4 � 49

4x2 � 15

y2 � 36

t 2 � 24

x2 � 25

345


Ch05pgs342-351 1/19/06 9:42 AM Page 345

ISB

N:0

-536

-477

42-6


EXAMPLE 15 Factor:

EXAMPLE 16 Factor:

EXAMPLE 17 Factor:

Always look first for a factor common to all terms. This time there is one,

EXAMPLE 18 Factor:


Caution!

Note carefully in these examples that a difference of squares is not thesquare of the difference; that is,

For example,

but

Similarly,

For example,

but

� 49.

� 100 � 60 � 9

102 � 2 � 10 � 3 � 32 � 100 � 2 � 10 � 3 � 9

�10 � 3� �10 � 3� � 7 � 13 � 91,

A2 � 2AB � B2 � �A � B� �A � B�.

452 � 52 � 2025 � 25 � 2000.

�45 � 5�2 � 402 � 1600,

A2 � B2 � �A � B�2.

� x4�7 � 3x� �7 � 3x� 49x4 � 9x6 � x4�49 � 9x2�

49x4 � 9x6.

� 2x2�3 � 5x2� �3 � 5x2� � 2x2�32 � �5x2�2�

18x2 � 50x6 � 2x2�9 � 25x4�

2x2.

18x2 � 50x6.

x2 �19

� x2 � � 13�2

� �x �13��x �

13�

x2 �19

.

m2 � 4p2 � m2 � �2p�2 � �m � 2p� �m � 2p�

m2 � 4p2.Factor.

25.

26.

27.

28.

29.[Hint:


t 6 � �t 3�2.]1 � 12,5 � 20t 6

64x4 � 25x6

a2 � 25b2

4t 2 � 64

x2 � 9

346


Ch05pgs342-351 1/19/06 9:42 AM Page 346

ISB

N:0-536-47742-6


FACTORING COMPLETELYIf a factor with more than one term can still be factored, you should do so.When no factor can be factored further, you have factored completely. Al-ways factor completely whenever told to factor.

EXAMPLE 19 Factor:

Factoring a difference of squares

Factoring further; is adifference of squares.

The polynomial cannot be factored further into polynomials withreal coefficients.

Caution!

Apart from possibly removing a common factor, you cannot factor a sumof squares. In particular,

Consider Here a sum of squares has a common factor, 25.Factoring, we get where is prime. For example,

EXAMPLE 20 Factor:

factor is a differenceof squares.

The polynomial cannot be factored further into polynomials withreal coefficients.

EXAMPLE 21 Factor:

Factoring a differenceof squares

Factoring further. Thefactor is adifference of squares.

TIPS FOR FACTORING

• Always look first for a common factor! If there is one, factor it out.

• Be alert for trinomial squares and differences of squares. Oncerecognized, they can be factored without trial and error.

• Always factor completely.

• Check by multiplying.


y2 � 4x6 � � y2 � 4x6� � y � 2x3� � y � 2x3�

y4 � 16x12 � � y2 � 4x6� � y2 � 4x6�

y4 � 16x12.

14 x4 � 9

14 x4 � 9

Factoringfurther. The � � 1

4x4 � 9�� 1

2x2 � 3�� 1

2x2 � 3�

Factoring adifference ofsquares

1

16x8 � 81 � � 1

4x4 � 9�� 1

4x4 � 9�

116

x8 � 81.

x2 � 4 � �x � 2�2.

x2 � 425�x2 � 4�,25x2 � 100.

A2 � B2 � �A � B�2.

p2 � 4

p2 � 4 � � p2 � 4� � p � 2� � p � 2� � � p2 � 4� � p2 � 4�

p4 � 16 � � p2�2 � 42

p4 � 16.

Factor completely.

30.

31.

32.


49p4 � 25q6

16 �1

81y8

81x4 � 1

Study Tips

SKILL MAINTENANCEEXERCISES

It is never too soon to beginreviewing for the finalexamination. The SkillMaintenance exercises foundin each exercise set review andreinforce skills taught inearlier sections. Include all ofthese exercises in your weeklypreparation. Answers to bothodd-numbered and even-numbered exercises appear atthe back of the book.

347


Ch05pgs342-351 1/19/06 9:42 AM Page 347

ISB

N:0

-536

-477

42-6


348


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.



Math TutorCenter

InterActMath


Determine whether each of the following is a trinomial square.

1. 2. 3. 4. x2 � 14x � 49x2 � 16x � 64x2 � 16x � 64x2 � 14x � 49

5. 6. 7. 8. 36x2 � 24x � 169x2 � 36x � 24x2 � 3x � 9x2 � 2x � 4

Factor completely. Remember to look first for a common factor and to check by multiplying.

9. 10. 11. 12. x2 � 20x � 100x2 � 16x � 64x2 � 20x � 100x2 � 14x � 49

13. 14. 15. 16. 4 � x2 � 4x4 � 4x � x2x2 � 2x � 1x2 � 2x � 1

17. 18. 19. 49 � 56y � 16y264 � 16a2 � a4q4 � 6q2 � 9

20. 21. 22. 2x2 � 40x � 2002x2 � 4x � 275 � 48a2 � 120a

23. 24. 25. 12q2 � 36q � 27x3 � 24x2 � 144xx3 � 18x2 � 81x

26. 27. 28. 64 � 112x � 49x249 � 42x � 9x220p2 � 100p � 125

29. 30. 31. 1 � 4x4 � 4x2a4 � 14a2 � 495y4 � 10y2 � 5

Ch05pgs342-351 1/19/06 9:43 AM Page 348

ISB

N:0-536-47742-6


349

Exercise Set 5.5

32. 33. 34. 25m2 � 20mn � 4n24p2 � 12pq � 9q21 � 2a5 � a10

35. 36. 37. 81a2 � 18ab � b2x2 � 14xy � 49y2a2 � 6ab � 9b2

38. 39. 40. 16m2 � 40mn � 25n236a2 � 96ab � 64b264p2 � 16pq � q2

Determine whether each of the following is a difference of squares.

41. 42. 43. 44. x2 � 9x2 � 25x2 � 36x2 � 4

45. 46. 47. 48. �1 � 36x216x2 � 25y2x2 � 80y2x2 � 45

49. 50. 51. 52. x2 � 36p2 � 9q2 � 1y2 � 4

53. 54. 55. 56. p2 � q2a2 � b2�64 � m2�49 � t 2

57. 58. 59. 60. 81 � w 2100 � k2w 2 � 49z225t 2 � m2

61. 62. 63. 64. 9a2 � 16b24x2 � 25y225x2 � 416a2 � 9

Factor completely. Remember to look first for a common factor.

65. 66. 67. 68. 16x � 81x336x � 49x324x2 � 548x2 � 98

Ch05pgs342-351 1/19/06 9:43 AM Page 349

ISB

N:0

-536

-477

42-6


350


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

73. 74. 75. 76. y4 � 1a4 � 1625a4 � 949a4 � 81

77. 78. 79. 80. x8 � 11 � y84x4 � 645x4 � 405

94. 95. 96. �256 � 1.6�64 � ��32�8.1 � ��9�

81. 82. 83. 84. x2 �1

25y2 �

116

x8 � 81x12 � 16

85. 86. 87. 88. p4q4 � 116m4 � t 414

� 9q225 �1

49x2

91. 92. 93. ��23� �

45

�1000 � ��2.5��110� � 10

89. Explain in your own words how to determinewhether a polynomial is a trinomial square.

90. Spiro concludes that since it must follow that What mistake is he making? How

would you go about correcting the misunderstanding?�x � 3� �x � 3�.

x2 � 9 ��x � 3� �x � 3�,x2 � 9 �DWDW

Divide. [1.6a, c]

Find a polynomial for the shaded area. (Leave results in terms of where appropriate.) [4.4d]�

97. 98.

x x

yy y

y

y

x

x

69. 70. 71. 72. �0.16p2 � 0.0025��0.09y2 � 0.0004�� 1625

x8 � 49�� 116

� 49x8�

SKILL MAINTENANCE

Ch05pgs342-351 1/19/06 9:43 AM Page 350

ISB

N:0-536-47742-6


351

Exercise Set 5.5

Simplify.

99. [4.1d] 100. [4.2a, b]�5a2b3�2y5 � y7

Find the intercepts. Then graph the equation. [3.3a]

101. 102. 3x � 5y � 15y � 6x � 6

Factor completely, if possible.

103. 104. 105. 106. x2 � 5x � 25x2 � 22x � 12127x3 � 13x49x2 � 216

107. 108. 109. 110. 4x4 � 4x2x8 � 28162x2 � 8218x3 � 12x2 � 2x

115. 116. 117. 118. 1.28x2 � 20.64x2 � 1.213.24x2 � 0.810.49p � p3

123. 124. 125. 126. �x � 7�2 � 4x � 249b2n � 12bn � 49x18 � 48x9 � 6481 � b4k

127. 128. 49�x � 1�2 � 42�x � 1� � 9� y � 3�2 � 2� y � 3� � 1

119. 120. 121. 122. a2n � 49b2nx2 � � 1x �2

� y � 5�2 � 36q2�x � 3�2 � 9

111. 112. 113. 114. x2 � 2.2518x3 �8

25 x3x2 �133x5 � 12x3

Find c such that the polynomial is the square of a binomial.

129. 130. cy2 � 24y � 9cy2 � 6y � 1

Use the TABLE feature to determine whether the factorization is correct.

131. 132. x2 � 49 � �x � 7� �x � 7�x2 � 9 � �x � 3� �x � 3�

133. 134. x2 � 49 � �x � 7�2x2 � 9 � �x � 3�2

SYNTHESIS

Ch05pgs342-351 1/19/06 9:43 AM Page 351

ISB

N:0

-536

-477

42-6


352


Sums or Differences of Cubes

We can factor the sum or the difference of two expressions that are cubes.Consider the following products:

and

.

The above equations (reversed) show how we can factor a sum or a differenceof two cubes.

SUM OR DIFFERENCE OF CUBES

;

Note that what we are considering here is a sum or a difference of cubes. We are not cubing a binomial. For example, is not the same as . The table of cubes in the margin is helpful.

EXAMPLE 1 Factor: .

We have

.

In one set of parentheses, we write the cube root of the first term, x. Then we write the cube root of the second term, �3. This gives us the expres-sion :

.

To get the next factor, we think of and do the following:

Square the first term: .

Multiply the terms, , and thenchange the sign: 3x.

Square the second term: .

.

Note that we cannot factor . It is not a trinomial square nor can itbe factored by trial and error. Check this on your own.

x2 � 3x � 9

�A � B��A2 � AB � B2�

�x � 3��x2 � 3x � 9�

��3�2 � 9

x��3� � �3x

x � x � x2

x � 3

�x � 3� � �

x � 3

x3 � 27 � x3 � 33

A3 � B3

x3 � 27

A3 � B3�A � B�3

A3 � B3 � �A � B� �A2 � AB � B2�A3 � B3 � �A � B� �A2 � AB � B2�

� A3 � B3

� A3 � A2B � AB2 � A2B � AB2 � B3

�A � B� �A2 � AB � B2� � A�A2 � AB � B2� � B�A2 � AB � B2�

� A3 � B3

� A3 � A2B � AB2 � A2B � AB2 � B3

�A � B� �A2 � AB � B2� � A�A2 � AB � B2� � B�A2 � AB � B2�

5.65.6 FACTORING SUMS OR DIFFERENCES OF CUBES

ObjectiveFactor sums or differences of cubes.

0.2 0.008

0.1 0.001

0 0

1 1

2 8

3 27

4 64

5 125

6 216

7 343

8 512

9 729

10 1000

N N 3

⎫ ⎬ ⎭

Ch05pgs352-356 1/19/06 9:44 AM Page 352

ISB

N:0-536-47742-6



EXAMPLE 2 Factor: .

We have

.

In one set of parentheses, we write the cube root of the first term, 5x. Then wewrite a plus sign, and then the cube root of the second term, y :

.

To get the next factor, we think of and do the following:

Square the first term: .

Multiply the terms, , and thenchange the sign: �5xy.

Square the second term: .

.


EXAMPLE 3 Factor: .

We first look for the largest common factor:

.

EXAMPLE 4 Factor: .

We can express this polynomial as a difference of squares:

.

We factor as follows:

.

One factor is a sum of two cubes, and the other factor is a difference of twocubes. We factor them:

.

We have now factored completely.

In Example 4, had we thought of factoring first as a difference of twocubes, we would have had

.

In this case, we might have missed some factors; can be fac-tored as , but we probably would not haveknown to do such factoring.

�a2 � ab � b2� �a2 � ab � b2�a4 � a2b2 � b4

� �a � b� �a � b� �a4 � a2b2 � b4� �a2�3 � �b2�3 � �a2 � b2� �a4 � a2b2 � b4�

a6 � b6 � �a � b� �a2 � ab � b2� �a � b� �a2 � ab � b2�

a6 � b6 � �a3 � b3� �a3 � b3�

a6 � b6 � �a3�2 � �b3�2

a6 � b6

� 2y�4y2 � 5x2� �16y4 � 20x2y2 � 25x4� � 2y��4y2�3 � �5x2�3�

128y7 � 250x6y � 2y�64y6 � 125x6�

128y7 � 250x6y

� A � B� � A2 � AB � B2�

�5x � y� �25x2 � 5xy � y2�

y � y � y2

5x � y � 5xy

�5x� �5x� � 25x2

5x � y

�5x � y� � �

125x3 � y3 � �5x�3 � y3

125x3 � y3

Factor.

1.

2.

Factor.

3.

4.


8y3 � z3

27x3 � y3

64 � y3

x3 � 8

353

5.6 Factoring Sums or Differences of Cubes

⎫ ⎬ ⎭

Ch05pgs352-356 1/19/06 9:45 AM Page 353

ISB

N:0

-536

-477

42-6


EXAMPLE 5 Factor: .

We have

Factoring a differenceof squares

.

Each factor is a sum or a difference of cubes. We factor each:

.

FACTORING SUMMARY

Sum of cubes: ;

Difference of cubes: ;

Difference of squares: ;

Sum of squares: cannot be factored using realnumbers if the largest common factor has been removed.

Do Exercises 5–8.

A2 � B2

A2 � B2 � �A � B� �A � B�A3 � B3 � �A � B� �A2 � AB � B2�A3 � B3 � �A � B� �A2 � AB � B2�

� �2a � 3b� �4a2 � 6ab � 9b2� �2a � 3b� �4a2 � 6ab � 9b2�

� ��2a�3 � �3b�3� ��2a�3 � �3b�3�

� �8a3 � 27b3� �8a3 � 27b3� 64a6 � 729b6 � �8a3�2 � �27b3�2

64a6 � 729b6Factor.

5.

6.

7.

8.


x3 � 0.027

729x6 � 64y6

16x7y � 54xy7

m6 � n6

Study Tips BETTER TEST TAKING

� Treat every homework exercise as if it were a test question. If you had to work aproblem at your job with no backup answer provided, what would you do? You wouldprobably work it very deliberately, checking and rechecking every step. You might workit more than one time, or you might try to work it another way to check the result. Tryto use this approach when doing your homework. Treat every exercise as though itwere a test question with no answer at the back of the book.

� Be sure that you do questions without answers as part of every homework assignmentwhether or not the instructor has assigned them! One reason a test may seem such adifferent task is that questions on a test lack answers. That is the reason for taking atest: to see if you can do the questions without assistance. As part of your testpreparation, be sure you do some exercises for which you do not have the answers.Thus when you take a test, you are doing a more familiar task.

The purpose of doing your homework using these approaches is to give you more test-taking practice beforehand. Let’s use a sports analogy: At a basketball game, the playerstake lots of practice shots before the game. They play the first half, go to the locker room,and come out for the second half. What do they do before the second half, even thoughthey have just played 20 minutes of basketball? They shoot baskets again! We suggest thesame approach here. Create more and more situations in which you practice taking testquestions by treating each homework exercise like a test question and by doing exercisesfor which you have no answers. Good luck!

“He who does not venture has no luck.”

Mexican proverb

How often do you makethe following statement

after taking a test: “Iwas able to do the

homework, but I frozeduring the test”? This

can be an excuse forpoor study habits. Hereare two tips to help you

with this difficulty.Both are intended tomake test taking less

stressful by getting youto practice good

test-taking habits on adaily basis.

354


Ch05pgs352-356 1/19/06 9:45 AM Page 354

ISB

N:0-536-47742-6


Factor.

355

Exercise Set 5.6



Math TutorCenter

InterActMath

MyMathLabMathXL

1. 2. 3. x3 � 1a3 � 8z3 � 27

4. 5. 6. x3 � 1y3 � 125c3 � 64

7. 8. 9. y3 � 827x3 � 18a3 � 1

10. 11. 12. 64 � 125x38 � 27b3p3 � 27

16. 17. 18. x3 � y3a3 � b327y3 � 64

19. 20. 21. 2y3 � 128b3 �1

27a3 �18

13. 14. 15. 8x3 � 27125x3 � 164y3 � 1

22. 23. 24. 54x3 � 224a3 � 33z3 � 3

25. 26. 27. 5x3 � 40z3ab3 � 125ars3 � 64r

Ch05pgs352-356 1/19/06 9:45 AM Page 355

ISB

N:0

-536

-477

42-6


356


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

28. 29. 30. y3 � 0.125x3 � 0.0012y3 � 54z3

31. 32. 33. 2y4 � 128y125c6 � 8d664x6 � 8t 6

34. 35. 36. t 6 � 1z6 � 13z5 � 3z2

39. Explain how someone could construct a binomialthat is both a difference of two cubes and a difference oftwo squares.

40. Explain how you could use factoring orgraphing to explain why .x3 � 8 � �x � 2�3DWDW

37. 38. p6 � q6t 6 � 64y6

SKILL MAINTENANCE

Simplify. [4.2b]

41. 42. 43. �x3

4 ��2

�a�4b�9��2�7y�5�3

Multiply.

44. [4.6b] 45. [4.6c] 46. [4.6a]�x � 0.1� �x � 0.5��w �13�2

�2y5 � 3� �2y5 � 3�

SYNTHESIS

Factor. Assume that variables in exponents represent positive integers.

47. 48. 49. 50. 827 x3 �

164 y33x3a � 24y3ba3x3 � b3y3x6a � y3b

51. 52. 53. 54. �1 � x�3 � �x � 1�6�x � y�3 � x37x3 �78

124 x3y3 �

13 z3

55. 56. y4 � 8y3 � y � 8�a � 2�3 � �a � 2�3

Ch05pgs352-356 1/19/06 9:45 AM Page 356

ISB

N:0-536-47742-6


357


We now combine all of our factoring techniques and consider a gen-eral strategy for factoring polynomials. Here we will encounter poly-

nomials of all the types we have considered, in random order, so you will havethe opportunity to determine which method to use.

FACTORING STRATEGY

To factor a polynomial:

a) Always look first for a common factor. If there is one, factor outthe largest common factor.

b) Then look at the number of terms.

Two terms: Determine whether you have a difference of squares,Do not try to factor a sum of squares:

Three terms: Determine whether the trinomial is a square. If it is,you know how to factor. If not, try trial and error, using FOIL orthe ac-method.

Four terms: Try factoring by grouping.

c) Always factor completely. If a factor with more than one term canstill be factored, you should factor it. When no factor can befactored further, you have finished.

d) Check by multiplying.

EXAMPLE 1 Factor:

a) We look for a common factor:

b) The factor has only two terms. It is a difference of squares:. We factor and then include the common factor:

c) We see that one of the factors, is again a difference of squares. Wefactor it:

This is a sum of squares. It cannot be factored!

We have factored completely because no factor with more than one termcan be factored further.

d) Check:

� 5t 4 � 80. � 5�t 4 � 16�

5�t 2 � 4� �t � 2� �t � 2� � 5�t2 � 4� �t 2 � 4�

5�t 2 � 4� �t � 2� �t � 2�.

t2 � 4,

5�t 2 � 4� �t 2 � 4�.

t 4 � 16�t 2�2 � 42t 4 � 16

5t 4 � 80 � 5�t4 � 16�.

5t 4 � 80.

A2 � B2.A2 � B2.

5.75.7 FACTORING: A GENERAL STRATEGYObjectiveFactor polynomialscompletely using any of themethods considered in thischapter.

Ch05pgs357-365 1/19/06 9:46 AM Page 357

ISB

N:0

-536

-477

42-6


EXAMPLE 2 Factor:

a) We look for a common factor. There isn’t one.

b) There are four terms. We try factoring by grouping:

Separating into two binomials



c) None of these factors can be factored further, so we have factoredcompletely.

d) Check:

EXAMPLE 3 Factor:

a) We look first for a common factor. This time there is one, :

b) The factor has three terms, but it is not a trinomial square.We factor it using trial and error:

c) No factor with more than one term can be factored further, so we have fac-tored completely.

d) Check:

EXAMPLE 4 Factor:

a) We look first for a common factor. There isn’t one.

b) There are three terms. We see that this polynomial is a trinomial square. Wefactor it:

We could use trial and error if we have not recognized that we have a tri-nomial square.

c) Since cannot be factored further, we have factored completely.

d) Check:

Do Exercises 1–5.

�x2 � 5�2 � �x2�2 � 2�x2� �5� � 52 � x4 � 10x2 � 25.

x2 � 5

x4 � 10x2 � 25 � �x2�2 � 2 � x2 � 5 � 52 � �x2 � 5�2.

x4 � 10x2 � 25.

x3�x � 7� �x � 5� � x3�x2 � 2x � 35� � x5 � 2x4 � 35x3.

x5 � 2x4 � 35x3 � x3�x2 � 2x � 35� � x3�x � 7� �x � 5�.

x2 � 2x � 35

x5 � 2x4 � 35x3 � x3�x2 � 2x � 35�.

x3

x5 � 2x4 � 35x3.

� 2x3 � 10x2 � x � 5. �2x2 � 1� �x � 5� � 2x2 � x � 2x2 � 5 � 1 � x � 1 � 5

x � 5� �2x2 � 1� �x � 5�.

� 2x2�x � 5� � 1�x � 5�� 2x3 � 10x2� � �x � 5�

2x3 � 10x2 � x � 5

2x3 � 10x2 � x � 5.Factor.

1.

2.

3.

4.

5.


8x3 � 200x

3x3 � 12x2 � 2x � 8

2x4 � 8x3 � 6x2

x6 � 8x3 � 16

3m4 � 3

358


Don’t forget to include the commonfactor in the final answer!

Ch05pgs357-365 1/19/06 9:46 AM Page 358

ISB

N:0-536-47742-6


EXAMPLE 5 Factor:

a) We look first for a common factor:

b) There are three terms in . We determine whether the trino-mial is a square. Since only is a square, we do not have a trinomialsquare. Can the trinomial be factored by trial and error? A key to theanswer is that x is only in the term . The polynomial might be in aform like , but there would be no x in the middle term.Thus, cannot be factored.

c) Have we factored completely? Yes, because no factor with more thanone term can be factored further.

d) The check is left to the student.

EXAMPLE 6 Factor:

a) We look for a common factor:

b) There are three terms in , but this trinomial cannot be factoredfurther.

c) Neither factor can be factored further, so we have factored completely.

d) The check is left to the student.

EXAMPLE 7 Factor:


b) There are four terms. We try factoring by grouping:

c) Have we factored completely? Since neither factor can be factored further,we have factored completely.

d) Check:

EXAMPLE 8 Factor:


b) There are three terms. We determine whether the trinomial is a square. Thefirst term and the last term are squares:

Since twice the product of 5x and 2y is the other term,

the trinomial is a perfect square.We factor by writing the square roots of the square terms and the sign

of the middle term:

c) Since cannot be factored further, we have factored completely.5x � 2y

25x2 � 20xy � 4y2 � �5x � 2y�2.

2 � 5x � 2y � 20xy,

25x2 � �5x�2 and 4y2 � �2y�2.

25x2 � 20xy � 4y2.

� p � q� �x � y� � px � py � qx � qy.

� � p � q� �x � y�. px � py � qx � qy � p�x � y� � q�x � y�

px � py � qx � qy.

2x � y � 2

� � p � q� �2x � y � 2�. � p � q� �x � 2� � � p � q� �x � y� � � p � q� ��x � 2� � �x � y��

�p � q� �x � 2� � �p � q� �x � y�.

2 � 7xy � y2�1 � y� �2 � y�

�7xy

y22 � 7xy � y2

6x2y4 � 21x3y5 � 3x2y6 � 3x2y4�2 � 7xy � y2�.

6x2y4 � 21x3y5 � 3x2y6.

Study Tips

TIME MANAGEMENT (PART 3)

Here are some additional tipsto help you with time manage-ment. (See also the Study Tipson time management inSections 2.2 and 5.2.)

� Avoid “time killers.” Welive in a media age, andthe Internet, e-mail,television, and movies areall time killers. Keep trackof the time you spend onsuch activities andcompare it to the timeyou spend studying.

� Prioritize your tasks. Becareful about taking ontoo many collegeactivities that fall outsideof academics. Theseactivities are importantbut keep them to aminimum to be sure thatyou have enough time foryour studies.

� Be aggressive about yourstudy tasks. Instead ofworrying over your mathhomework or test prepa-ration, do something toget yourself started. If thetask is large, break it downinto smaller parts, and doone at a time. You will besurprised at how quicklythe large task can then becompleted.

“Time is more valuable thanmoney. You can get moremoney, but you can’t get moretime.”

Jim Rohn, motivational speaker

359


Ch05pgs357-365 1/19/06 9:47 AM Page 359

ISB

N:0

-536

-477

42-6


d) Check:

EXAMPLE 9 Factor:


b) There are three terms. We determine whether the trinomial is a square. Thefirst term is a square, but neither of the other terms is a square, so we donot have a trinomial square. We factor, thinking of the product pq as a sin-gle variable. We consider this possibility for factorization:

We factor the last term, 12. All the signs are positive, so we consider onlypositive factors. Possibilities are 1, 12 and 2, 6 and 3, 4. The pair 3, 4 givesa sum of 7 for the coefficient of the middle term. Thus,

c) No factor with more than one term can be factored further, so we have fac-tored completely.

d) Check:

EXAMPLE 10 Factor:


b) There are three terms in . We determine whether the tri-nomial is a square. Since none of the terms is a square, we do not have a trinomial square. We factor . Possibilities are , and , andothers. We also factor the last term, . Possibilities are , and

and others. We look for factors such that the sum of their productsis the middle term. The in the middle term, , should lead us to try

We try some possibilities:

c) No factor with more than one term can be factored further, so we have fac-tored completely. The factorization, including the common factor, is

d) Check:

� 8x4 � 20x2y � 12y2. � 4�2x4 � 5x2y � 3y2� � 4�2x4 � 6x2y � x2y � 3y2�

4�2x2 � y� �x2 � 3y� � 4��2x2� �x2� � 2x2��3y� � yx2 � y ��3y��

4�2x2 � y� �x2 � 3y�.

�2x2 � y� �x2 � 3y� � 2x4 � 5x2y � 3y2.

�2x2 � y� �x2 � 3y� � 2x4 � 5x2y � 3y2,

�2x2� �x2�.�5x2yx2

y�3y,�y3y�3y2

x32xx22x22x4

2x4 � 5x2y � 3y2

8x4 � 20x2y � 12y2 � 4�2x4 � 5x2y � 3y2�.

8x4 � 20x2y � 12y2.

� p2q2 � 7pq � 12. � pq � 3�� pq � 4� � � pq� � pq� � 4 � pq � 3 � pq � 3 � 4

p2q2 � 7pq � 12 � � pq � 3� � pq � 4�.

� pq � � � pq � �.

p2q2 � 7pq � 12.

� 25x2 � 20xy � 4y2. �5x � 2y�2 � �5x�2 � 2�5x� �2y� � �2y�2Factor.

6.

7.

8.

9.

10.


x4 � 2x2y2 � y4

ax2 � ay � bx2 � by

�a � b��x � y2��a � b��x � 5� �

10p6q2 � 4p5q3 � 2p4q4

15x4 � 5x2y � 10y2

360


Ch05pgs357-365 1/19/06 9:47 AM Page 360

ISB

N:0-536-47742-6


EXAMPLE 11 Factor:


b) There are two terms. Since and , we see that we dohave a difference of squares. Thus,

c) The last factor can be factored further. It is also a difference of squares.

d) Check:

EXAMPLE 12 Factor:


b) The factor has only two terms. It is a difference of two cubes. Wefactor as follows:

c) No factor with more than one term can be factored further, so we have fac-tored completely. The factorization, including the common factor, is

Do Exercises 6–13. (Exercises 6–10 are on the preceding page.)

5�2t � s� �4t 2 � 2ts � s2�.

�2t � s� �4t 2 � 2ts � s2�.

8t 3 � s3

40t 3 � 5s3 � 5�8t 3 � s3�.

40t 3 � 5s3.

� a4 � 16b4.

�a2 � 4b2� �a � 2b� �a � 2b� � �a2 � 4b2� �a2 � 4b2�

a4 � 16b4 � �a2 � 4b2� �a � 2b� �a � 2b�

a4 � 16b4 � �a2 � 4b2� �a2 � 4b2�

16b4 � �4b2�2a4 � �a2�2

a4 � 16b4. Factor.

11.

12.

13.


15a3 � 120b3

p4 � 81q4

x2y2 � 5xy � 4

Study Tips SPECIAL VIDEOTAPES

In addition to the videotaped lecturesfor the books, there is a special VHStape, Math Problem Solving in the RealWorld. Check with your instructor tosee whether this tape is available onyour campus.

There is also a special Math StudySkills for Students Video on CD that isdesigned to help you make better useof your math study time and improveyour retention of concepts and pro-cedures taught in classes from basicmathematics through intermediatealgebra. (See the Preface for moreinformation.)

361


Ch05pgs357-365 1/19/06 9:47 AM Page 361

ISB

N:0

-536

-477

42-6


362


Factor completely.

Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.



Math TutorCenter

InterActMath


1. 2. 3. a2 � 25 � 10a2t 2 � 183x2 � 192

4. 5. 6. 8y2 � 18y � 52x2 � 11x � 12y2 � 49 � 14y

7. 8. 9. x3 � 3x2 � 4x � 12x3 � 18x2 � 81xx3 � 24x2 � 144x

10. 11. 12. 50x2 � 3248x2 � 3x3 � 5x2 � 25x � 125

13. 14. 15. x2 � 420x3 � 4x2 � 72x9x3 � 12x2 � 45x

16. 17. 18. m4 � 8m3 � 8m2 � 64mx4 � 7x2 � 3x3 � 21xt 2 � 25

19. 20. 21. 20 � 6x � 2x22x6 � 8x5 � 8x4x5 � 14x4 � 49x3

22. 23. 24. x2 � 8x � 5x2 � 6x � 145 � 3x � 6x2

25. 26. 27. 1 � y85x5 � 80x4x4 � 64

Ch05pgs357-365 1/19/06 9:47 AM Page 362

ISB

N:0-536-47742-6


363

Exercise Set 5.7

28. 29. 30. x6 � 2x5 � 7x4x5 � 4x4 � 3x3t 8 � 1

34. 35. 36. x2y � xy29x2y2 � 36xy125a3 � 8b3

37. 38. 39. �a � b��x � 3� � �a � b��x � 4�10p4q4 � 35p3q3 � 10p2q22�rh � 2�r 2

40. 41. 42. 3� p � q� � q2� p � q��x � 1��x � 1� � y �x � 1�5c�a3 � b� � �a3 � b�

43. 44. 45. 6q2 � 3q � 2pq � pa2 � 3a � ay � 3yn2 � 2n � np � 2p

46. 47. 48. x2 � y2 � 2xy4b2 � a2 � 4ab2x2 � 4x � xy � 2y

49. 50. 51. 49m4 � 112m2n � 64n29c2 � 6cd � d216x2 � 24xy � 9y2

52. 53. 54. 0.01x4 � 0.1x2y2 � 0.25y4y4 � 10y2z2 � 25z44x2y2 � 12xyz � 9z2

31. 32. 33. mx2 � my236a2 � 15a �2516

181

x6 �8

27x3 �

169

Ch05pgs357-365 1/19/06 9:47 AM Page 363

ISB

N:0

-536

-477

42-6


364


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

55. 56. 57. a2 � ab � 2b24p2q � pq2 � 4p314

a2 �13

ab �19

b2

58. 59. 60. 15 � x2y2 � 8xy2mn � 360n2 � m23b2 � 17ab � 6a2

61. 62. 63. r 5s2 � 10r 4s � 16r 3p2q2 � 7pq � 6m2n2 � 4mn � 32

64. 65. 66. 2s6t 2 � 10s3t 3 � 12t 4a5 � 4a4b � 5a3b2p5q2 � 3p4q � 10p3

67. 68. 69. 7x6 � 7y6p2 �1

49b2a2 �

125

b2

70. 71. 72. 15a4 � 15b416 � p4q416p3 � 54q3

73. 74. 75. q3 � 8q2 � q � 881a4 � b41 � 16x12y12

76. 77. 78. 4ab5 � 32b4 � a2b6112xy � 49x2 � 64y2m3 � 7m2 � 4m � 28

79. Kelly factored as , while Tony factored it as . Evaluate each expressionfor several values of x. Then explain why both answersare correct.

80. Describe in your own words a strategy that can beused to factor polynomials.DW

�4 � x�2�x � 4�216 � 8x � x2DW

Ch05pgs357-365 1/19/06 9:47 AM Page 364

ISB

N:0-536-47742-6


365

Exercise Set 5.7

Factor completely.

81. 82. 83. 84. �3x � 5y� �2x � 7y��3x � 5y�2�3x � 5y�2�3x � 5y� �3x � 5y�

85. Divide: [1.6c] 86. Simplify: . [1.4a]15 � ��3� � 3 � ��21� � 4275

� ��1110�.

87. Solve for X. [2.4b] 88. Solve: [2.7e]4�x � 9� � 2�x � 7� � 14.A � aX � bX � 7

89. 90. 91. 92.15

x2 � x �45

12.25x2 � 7x � 1x4 � 9a4 � 2a2 � 1

93. 94. 95. 96. ��x4 � 7x2 � 18�18 � y3 � 9y � 2y2x3 � �x � 3x2� � 35x2 � 13x � 7.2

97. 98. 99. 100. 3x4 � 15x2 � 12x3 � x2 � 4x � 4x3 � x2 � �4x � 4�a3 � 4a2 � a � 4

101. 102. y2� y � 1� � 4y� y � 1� � 21� y � 1�y2� y � 1� � 2y� y � 1� � � y � 1�

103. 104. 6�x � 1�2 � 7y�x � 1� � 3y2� y � 4�2 � 2x� y � 4� � x2

SKILL MAINTENANCE

Multiply. [4.6d], [4.7f]

SYNTHESIS

Ch05pgs357-365 1/19/06 9:47 AM Page 365

ISB

N:0

-536

-477

42-6


Second-degree equations like and are ex-amples of quadratic equations.

QUADRATIC EQUATION

A quadratic equation is an equation equivalent to an equation of the type

In order to solve quadratic equations, we need a new equation-solvingprinciple.

The Principle of Zero Products

The product of two numbers is 0 if one or both of the numbers is 0. Further-more, if any product is 0, then a factor must be 0. For example:

If then we know that

If then we know that or

If then we know that or

Caution!

In a product such as we cannot conclude with certainty that ais 24 or that b is 24, but if we can conclude that or

EXAMPLE 1 Solve:

We have a product of 0. This equation will be true when either factor is 0.Thus it is true when

or

Here we have two simple equations that we know how to solve:

or

Each of the numbers �3 and 2 is a solution of the original equation, as we cansee in the following checks.

Check: For �3:

TRUE 0 0��5�

��3 � 3� ��3 � 2� ? 0

�x � 3� �x � 2� � 0

x � 2.x � �3

x � 2 � 0.x � 3 � 0

�x � 3� �x � 2� � 0.

b � 0.a � 0ab � 0,ab � 24,

x � 2 � 0.x � 3 � 0�x � 3� �x � 2� � 0,

2x � 9 � 0.x � 0x�2x � 9� � 0,

x � 0.7x � 0,

a � 0.ax2 � bx � c � 0,

9 � x2 � 0x2 � x � 156 � 0

5.85.8 SOLVING QUADRATIC EQUATIONSBY FACTORING

ObjectivesSolve equations (alreadyfactored) using the principleof zero products.

Solve quadratic equations byfactoring and then using theprinciple of zero products.

Study Tips

WORKING WITH A CLASSMATE

If you are finding it difficult tomaster a particular topic orconcept, try talking about itwith a classmate. Verbalizingyour questions about thematerial might help clarify it.If your classmate is alsofinding the material difficult, itis possible that the majority ofthe people in your class areconfused and you can askyour instructor to explain theconcept again.

366


For 2:

TRUE 0 5�0�

�2 � 3� �2 � 2� ? 0

�x � 3� �x � 2� � 0

Ch05pgs366-374 1/19/06 9:50 AM Page 366

ISB

N:0-536-47742-6


We now have a principle to help in solving quadratic equations.

THE PRINCIPLE OF ZERO PRODUCTS

An equation is true if and only if is true or is true,or both are true. (A product is 0 if and only if one or both of thefactors is 0.)

EXAMPLE 2 Solve:

We have

Using the principle of zero products

Solving the two equations separately

Check: For

TRUE 0 0��7 1

5�

��1 � 1� ��7 15�

�5��15� � 1� ��

15 � 7� ? 0

�5x � 1� �x � 7� � 0

�15 :

x � �15 or x � 7.

5x � �1 or x � 7

5x � 1 � 0 or x � 7 � 0

�5x � 1� �x � 7� � 0

�5x � 1� �x � 7� � 0.

b � 0a � 0ab � 0

Solve using the principle of zeroproducts.

1.

2.

3.

4. Solve:


y�3y � 17� � 0.

�4t � 1� �3t � 2� � 0

�x � 7� �x � 3� � 0

�x � 3� �x � 4� � 0

367

5.8 Solving Quadratic Equations by Factoring

For 7:

TRUE 0 36 � 0

�35 � 1� � 0 �5�7� � 1� �7 � 7� ? 0

�5x � 1� �x � 7� � 0

The solutions are and 7.

When you solve an equation using the principle of zero products, a checkby substitution, as in Examples 1 and 2, will detect errors in solving.

Do Exercises 1–3.

When some factors have only one term, you can still use the principle ofzero products.

EXAMPLE 3 Solve:

We have


Check: For 0:

TRUE 0 0 � ��9�

0 � �2 � 0 � 9� ? 0

x�2x � 9� � 0

x � 0 or x �92

.

x � 0 or 2x � 9

x � 0 or 2x � 9 � 0

x�2x � 9� � 0

x�2x � 9� � 0.

�15

For

TRUE 0 92 � 0

92 � �9 � 9�

92 � �2 �92 � 9� ? 0

x�2x � 9� � 0

92 :

Do Exercise 4.

Ch05pgs366-374 1/19/06 9:50 AM Page 367

ISB

N:0

-536

-477

42-6


Using Factoring to Solve Equations

Using factoring and the principle of zero products, we can solve some newkinds of equations. Thus we have extended our equation-solving abilities.

EXAMPLE 4 Solve:

There are no like terms to collect, and we have a squared term. We firstfactor the polynomial. Then we use the principle of zero products.

Factoring


Check: For �2:

TRUE 0 �6 � 6

4 � 10 � 6 ��2�2 � 5��2� � 6 ? 0

x2 � 5x � 6 � 0

x � �2 or x � �3

x � 2 � 0 or x � 3 � 0

�x � 2� �x � 3� � 0

x2 � 5x � 6 � 0

x2 � 5x � 6 � 0.

5. Solve:

Solve.

6.

7.

Solve.

8.

9.


9x2 � 16

x2 � 4x � 0

x2 � 6x � 9

x2 � 3x � 28

x2 � x � 6 � 0.

368


For �3:

TRUE 0 �6 � 6

9 � 15 � 6 ��3�2 � 5��3� � 6 ? 0

x2 � 5x � 6 � 0

The solutions are �2 and �3.

Caution!

Keep in mind that you must have 0 on one side of the equation before youcan use the principle of zero products. Get all nonzero terms on one sideand 0 on the other.

Do Exercise 5.

EXAMPLE 5 Solve:

We first add 16 to get a 0 on one side:

Adding 16

Factoring


Solving each equation

There is only one solution, 4. The check is left to the student.


EXAMPLE 6 Solve:

Factoring out a common factor


The solutions are 0 and �5. The check is left to the student.

x � 0 or x � �5

x � 0 or x � 5 � 0

x�x � 5� � 0

x2 � 5x � 0

x2 � 5x � 0.

x � 4 or x � 4.

x � 4 � 0 or x � 4 � 0

�x � 4� �x � 4� � 0

x2 � 8x � 16 � 0

x2 � 8x � �16

x2 � 8x � �16.

Ch05pgs366-374 1/19/06 9:50 AM Page 368

ISB

N:0-536-47742-6


EXAMPLE 7 Solve:

Subtracting 25 on both sides to get 0 onone side


Solving each equation

The solutions are and The check is left to the student.


EXAMPLE 8 Solve:

In this case, the leading coefficient of the trinomial is negative. Thus wefirst multiply by �1 and then proceed as we have in Examples 1–7.

Multiplying by �1

Simplifying

Factoring

Using the principle ofzero products

The solutions are and 1. The check is left to the student.


EXAMPLE 9 Solve:

Be careful with an equation like this one! It might be tempting to set eachfactor equal to 5. Remember: We must have a 0 on one side. We first carry outthe product on the left. Then we subtract 5 on both sides to get 0 on one side.Then we proceed with the principle of zero products.

Multiplying on the left

Subtracting 5

Simplifying

Factoring


The solutions are �3 and 3. The check is left to the student.

Do Exercise 12.

x � �3 or x � 3

x � 3 � 0 or x � 3 � 0

�x � 3� �x � 3� � 0

x2 � 9 � 0

x2 � 4 � 5 � 5 � 5

x2 � 4 � 5

�x � 2� �x � 2� � 5

�x � 2� �x � 2� � 5.

�35

x � �35

or x � 1

5x � �3 or x � 1

5x � 3 � 0 or x � 1 � 0

�5x � 3� �x � 1� � 0

5x2 � 2x � 3 � 0

�1��5x2 � 2x � 3� � �1 � 0

�5x2 � 2x � 3 � 0

�5x2 � 2x � 3 � 0.

�52 .5

2

x �52

or x � �52

2x � 5 or 2x � �5

2x � 5 � 0 or 2x � 5 � 0

�2x � 5� �2x � 5� � 0

4x2 � 25 � 0

4x2 � 25

4x2 � 25. Solve.

10.

11.

12. Solve:


�x � 1� �x � 1� � 8.

10 � 3x � x2 � 0

�2x2 � 13x � 21 � 0

369


Ch05pgs366-374 1/19/06 9:50 AM Page 369

ISB

N:0

-536

-477

42-6


ALGEBRAIC–GRAPHICAL CONNECTION

In Chapter 3, we graphed linear equations of the type andRecall that to find the x-intercept, we replaced y with 0 and

solved for x. This procedure can also be used to find the x-interceptswhen an equation of the form is to be graphed.Although the details of creating such graphs will be left to Chapter 11, we consider them briefly here from the standpoint of finding the x-intercepts. The graphs are shaped like the following curves. Note thateach x-intercept represents a solution of

EXAMPLE 10 Find the x-intercepts of thegraph of shown at right. (Thegrid is intentionally not included.)

To find the x-intercepts, we let andsolve for x :

Substituting 0for y

Factoring

Using the principleof zero products

The solutions of the equation are 5 and The x-interceptsof the graph of are and We can now label themon the graph.


x

y y � x 2 � 4x � 5

(�1, 0) (5, 0)

��1, 0�.�5, 0�y � x2 � 4x � 5�1.0 � x2 � 4x � 5

x � 5 or x � �1.

x � 5 � 0 or x � 1 � 0

0 � �x � 5� �x � 1�

0 � x2 � 4x � 5

y � x2 � 4x � 5

y � 0

y � x2 � 4x � 5

x

y

x

y

x

y

x

y

a �� 0 a �� 0

Nox-intercept

Twox-intercepts

Onex-intercept

ax2 � bx � c � 0.

a � 0,y � ax2 � bx � c,

Ax � By � C.y � mx � b

13. Find the x-intercepts of thegraph shown below.

14. Use only the graph shown belowto solve


x

y

�4 �2 2 4

�4

2

4

�5 �3 �1 1 3 5

�5

1

3

5

y � 3x � x 2

3x � x2 � 0.

x

y

y � x 2 � 4x � 5

370


AG

x

y y � x 2 � 4x � 5

Ch05pgs366-374 1/19/06 9:50 AM Page 370

ISB

N:0-536-47742-6


371


CALCULATOR CORNER

Solving Quadratic Equations We can solve quadratic equationsgraphically. Consider the equation First, we must write theequation with 0 on one side. To do this, we subtract 8 on both sides of theequation; we get Next, we graph in awindow that shows the x-intercepts. The standard window works well inthis case.

The solutions of the equation are the values of x for whichThese are also the first coordinates of the x-intercepts

of the graph. We use the ZERO feature from the CALC menu to find thesenumbers. To find the solution corresponding to the leftmost x-intercept,we first press F m 2 to select the ZERO feature. The prompt “LeftBound?” appears. Next, we use the f or the g key to move the cursor tothe left of the intercept and press [. Now the prompt “Right Bound?”appears. Then we move the cursor to the right of the intercept and press[. The prompt “Guess?” appears. We move the cursor close to theintercept and press [ again. We now see the cursor positioned at theleftmost x-intercept and the coordinates of that point, aredisplayed. Thus, when This is one solution ofthe equation.

We can repeat this procedure to find the first coordinate of the otherx-intercept. We see that at that point. Thus the solutions of theequation are �4 and 2. Note that the x-intercepts ofthe graph of are and

Exercises:1. Solve each of the equations in Examples 4–8 graphically.

�2, 0�.��4, 0�y � x 2 � 2x � 8x 2 � 2x � 8 � 0

x � 2

x � �4.x 2 � 2x � 8 � 0y � 0,x � �4,

x 2 � 2x � 8 � 0.

y � x 2 � 2x � 8

10�10

�10

10

y � x 2 � 2x � 8x 2 � 2x � 8 � 0.

x 2 � 2x � 8.

y � x 2 � 2x � 8

10�10

�10

10

ZeroX � �4 Y � 0

y � x 2 � 2x � 8

10�10

�10

10

ZeroX � 2 Y � 0

Ch05pgs366-374 1/19/06 9:50 AM Page 371

ISB

N:0

-536

-477

42-6


372


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.



Math TutorCenter

InterActMath


Solve using the principle of zero products.

1. 2. 3. �x � 3� �x � 8� � 0�x � 2� �x � 7� � 0�x � 4� �x � 9� � 0

4. 5. 6. �x � 13� �x � 53� � 0�x � 12� �x � 11� � 0�x � 6� �x � 8� � 0

7. 8. 9. 0 � y� y � 18�y� y � 5� � 0x�x � 3� � 0

10. 11. 12. �2x � 9� �x � 8� � 0�2x � 5� �x � 4� � 00 � x�x � 19�

13. 14. 15. �7x � 28� �28x � 7� � 0 �4x � 9� �14x � 7� � 0�5x � 1� �4x � 12� � 0

16. 17. 18. 55x�8x � 9� � 02x�3x � 2� � 0�13x � 14� �6x � 5� � 0

22. 23. 24. �x � 5� �x � 75� �5x � 1� � 09x�3x � 2� �2x � 1� � 0�0.1x � 0.3� �0.4x � 20� � 0

19. 20. 21. �0.3x � 0.1� �0.05x � 1� � 0�74 x �

116� �2

3 x �1615� � 0�1

5 � 2x� �19 � 3x� � 0

Solve by factoring and using the principle of zero products. Remember to check.

25. 26. 27. 28. x2 � 4x � 21 � 0x2 � 7x � 18 � 0x2 � 7x � 6 � 0x2 � 6x � 5 � 0

29. 30. 31. 32. x2 � 3x � 0x2 � 8x � 0x2 � 9x � 14 � 0x2 � 8x � 15 � 0

33. 34. 35. 36. 100 � x2x2 � 16x2 � 16x � 0x2 � 18x � 0

37. 38. 39. 40. 0 � 25 � x2 � 10x0 � 6x � x2 � 94x2 � 9 � 09x2 � 4 � 0

41. 42. 43. 44. 7x2 � 8x5x2 � 6x1 � x2 � 2xx2 � 16 � 8x

45. 46. 47. 48. 2y2 � 12y � �1012y2 � 5y � 23x2 � 7x � 206x2 � 4x � 10

Ch05pgs366-374 1/19/06 9:51 AM Page 372

ISB

N:0-536-47742-6


373

Exercise Set 5.8

49. 50. 51. 52. 64a2 � 81100y2 � 49x�x � 5� � 14t�3t � 1� � 2

53. 54. 55. 56. 12x2 � 17x � 5 � 010x2 � 23x � 12 � 03x2 � 8x � 9 � 2xx2 � 5x � 18 � 2x

60. 61. 62.

x

yy � x 2 � 2x � 8

x

yy � x 2 � 2x � 15

x

y

y � 2x 2 � 3x � 9

63. Use the following graph to solve 64. Use the following graph to solve

4

2

�2

�5�1

1

3

1 3 5�4 �2�1

�3�4

�7

�5

x

y

y � x2 � x � 6

x2 � x � 6 � 0.

2

�5

1

3

1 2 3 5�4 �2�3

�4

�7

�5�6

x

y

y � x2 � 3x � 4

x2 � 3x � 4 � 0.

Find the x-intercepts for the graph of the equation. (The grids are intentionally not included.)

57. 58. 59.

x

y

y � 2x 2 � x � 10

x

y

y � x 2 � x � 6

x

y

y � x 2 � 3x � 4

Ch05pgs366-374 1/19/06 9:51 AM Page 373

ISB

N:0

-536

-477

42-6


374


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

65. Use the following graph to solve 66. Use the following graph to solve

y � �x2 � x � 6

2 4

2

4

�2

�5�1

1

3

56

1 3 5�4 �1�2�3

�3�4

x

y

�x2 � x � 6 � 0.

2 4

2

4

�2

�5�1

1

3

5

1 3 5�4 �1�2�3

�3�4�5

x

y

y � �x2 � 2x � 3

�x2 � 2x � 3 � 0.

67. What is wrong with the following? Explain thecorrect method of solution.

68. What is incorrect about solving by dividingby x on both sides?

x2 � 3xDW

x � 11 or x � 4

x � 3 � 8 or x � 4 � 8

�x � 3� �x � 4� � 8

DW

Translate to an algebraic expression. [1.1b]

69. The square of the sum of a and b 70. The sum of the squares of a and b

Divide. [1.6c]

71. 72. 73. 74. �3

16 � ��58��

58 �

316�24.3 � 5.4144 � ��9�

Solve.

75. 76. 77. 78. �t � 5�2 � 2�5 � t��t � 3�2 � 36y � y � 8� � 16� y � 1�b �b � 9� � 4�5 � 2b�

79. 80. 81. 82. 2725 x2 � 1

35

16 x2 � 5x2 �2536 � 0x2 �

164 � 0

83. Find an equation that has the given numbers assolutions. For example, 3 and �2 are solutions of

a) �3, 4 b) �3, �4 c)d) 5, �5 e) 0, 0.1,

84. Matching. Match each equation in the first columnwith the equivalent equation in the second column.

2x2 � 20x � 4 � 03x2 � 4x � 8 � 0

x2 � 3x � 18 � 0x2 � 2x � 9 � 0

�x � 1� �5x � 5� � 0�2x � 5� �x � 4� � 0

9x2 � 12x � 24 � 05x2 � 5 � 0

�2x � 8� �2x � 5� � 0�x � 6� �x � 3� � 0

4x2 � 8x � 36 � 0x2 � 10x � 2 � 0

14

12 , 1

2

x2 � x � 6 � 0.

Use a graphing calculator to find the solutions of the equation. Round solutions to the nearest hundredth.

85. 86. �x2 � 0.63x � 0.22 � 0x2 � 9.10x � 15.77 � 0

87. 88. 6.4x2 � 8.45x � 94.06 � 00.84x2 � 2.30x � 0

SKILL MAINTENANCE

SYNTHESIS

Ch05pgs366-374 1/19/06 9:51 AM Page 374

ISB

N:0-536-47742-6


1. Dimensions of Picture. Arectangular picture is twice aslong as it is wide. If the area ofthe picture is , find itsdimensions.

Answer on page A-23

w

2w

288 in2

375

5.9 Applications of Quadratic Equations

Applied Problems, Quadratic Equations, and Factoring

We can now use our new method for solving quadratic equations and the fivesteps for solving problems.

EXAMPLE 1 Manufacturing Marble Slabs. Marble Supreme sells rectan-gular marble slabs used for tempering fudge in candy shops. The most popu-lar slab that Marble Supreme sells is twice as long as it is wide and has an areaof What are the dimensions of the slab?

1. Familiarize. We first make a drawing. Recall that the area of any rectan-gle is Length Width. We let the width of the slab, in inches. Thelength is then .

2. Translate. We reword and translate as follows:

Rewording : The area of the rectangle is .

Translating : 7200

3. Solve. We solve the equation as follows:

Subtracting 7200 to get 0 on one side

Removing a common factor of 2


Dividing by 2

or Using the principle of zero products

or Solving each equation

4. Check. The solutions of the equation are 60 and . Since the widthmust be positive, cannot be a solution. To check 60 in., we note thatif the width is 60 in., then the length is and the area is

Thus the solution 60 checks.

5. State. The slab is 60 in. wide and 120 in. long.

Do Exercise 1.

60 in. � 120 in. � 7200 in2.2 � 60 in. � 120 in.,

�60�60

x � �60.x � 60

x � 60 � 0x � 60 � 0

�x � 60��x � 60� � 0

2�x � 60� �x � 60� � 0

2�x2 � 3600� � 0

2x2 � 7200 � 0

2x2 � 7200

2x � x � 7200

�2x � x

7200 cm2

x2x

2xx ��

7200 in2.

5.95.9 APPLICATIONS OF QUADRATIC EQUATIONS

ObjectivesSolve applied problemsinvolving quadraticequations that can be solved by factoring.

Solve applied problemsinvolving the Pythagoreantheorem and quadraticequations that can be solvedby factoring.

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

Ch05pgs375-387 1/19/06 9:53 AM Page 375

ISB

N:0

-536

-477

42-6


EXAMPLE 2 Racing Sailboat. The height of a triangular sail on a racingsailboat is 9 ft more than the base. The area of the triangle is . Find theheight and the base of the sail.Source: Whitney Gladstone, North Graphics, San Diego, CA

1. Familiarize. We first make a drawing. If you don’t remember the for-mula for the area of a triangle, look it up in the list of formulas at the backof this book or in a geometry book. The area is (base)(height).

We let the base of the triangle. Then the height.

2. Translate. It helps to reword this problem before translating:

times Base times Height is 110. Rewording

b 110 Translating


Multiplying

Multiplying by 2

Simplifying


Factoring


or

4. Check. The base of a triangle cannot have a negative length, so cannot be a solution. Suppose the base is 11 ft. The height is 9 ft morethan the base, so the height is or 20 ft, and the area is

(11)(20), or . These numbers check in the original problem.

5. State. The height is 20 ft and the base is 11 ft.

Do Exercise 2.

110 ft212

11 ft � 9 ft,

�20

b � �20.b � 11

b � 20 � 0b � 11 � 0

�b � 11� �b � 20� � 0

b2 � 9b � 220 � 0

b2 � 9b � 220 � 220 � 220

b2 � 9b � 220

2 �12

�b2 � 9b� � 2 � 110

12

�b2 � 9b� � 110

12

� b � �b � 9� � 110

��b � 9��12

12

b + 9

b

b � 9 �b �

12

110 ft22. Dimensions of a Sail. The

mainsail of Stacey’s lightning-styled sailboat has an area of125 . The sail is 15 ft tallerthan it is wide. Find the heightand the width of the sail.

Answer on page A-23

b � 15

b

ft2

376


⎧ ⎨ ⎩

Ch05pgs375-387 1/19/06 9:53 AM Page 376

ISB

N:0-536-47742-6


EXAMPLE 3 Games in a Sports League. In a sports league of x teams inwhich each team plays every other team twice, the total number N of gamesto be played is given by

Maggie’s basketball league plays a total of 240 games. How many teams are inthe league?

1., 2. Familiarize and Translate. We are given that x is the number of teams in a league and N is the number of games. To familiarizeyourself with this problem, reread Example 4 in Section 4.3 wherewe first considered it. To find the number of teams x in a league inwhich 240 games are played, we substitute 240 for N in theequation:

Substituting 240 for N



Factoring


or

4. Check. The solutions of the equation are 16 and . Since thenumber of teams cannot be negative, cannot be a solution.But 16 checks, since .

5. State. There are 16 teams in the league.

Do Exercise 3.

162 � 16 � 256 � 16 � 240�15

�15

x � �15.x � 16

x � 15 � 0x � 16 � 0

�x � 16� �x � 15� � 0

x2 � x � 240 � 0

x2 � x � 240 � 240 � 240

x2 � x � 240

x2 � x � 240.

x2 � x � N.

3. Use for thefollowing.

a) Volleyball League. Amy’svolleyball league has 19 teams. What is the totalnumber of games to beplayed?

b) Softball League. Barry’sslow-pitch softball leagueplays a total of 72 games. How many teams are in the league?


N � x2 � x

Study Tips FIVE STEPS FOR PROBLEM SOLVING

1. Familiarize yourself with the situation.

a) Carefully read and reread until you understandwhat you are being asked to find.

b) Draw a diagram or see if there is a formula thatapplies.

c) Assign a letter, or variable, to the unknown.

2. Translate the problem to an equation using theletter or variable.

3. Solve the equation.

4. Check the answer in the original wording of theproblem.

5. State the answer to the problem clearly withappropriate units.

“Most worthwhile achievements are the result of manylittle things done in a simple direction.”

Nido Quebin, speaker/entrepreneur

Are youremembering to use

the five steps forproblem solving

that were developedin Section 2.6?

377


Ch05pgs375-387 1/19/06 9:53 AM Page 377

ISB

N:0

-536

-477

42-6


EXAMPLE 4 Athletic Numbers. The product of the numbers of twoconsecutive entrants in a marathon race is 156. Find the numbers.

1. Familiarize. The numbers are consecutive integers. Recall that consecu-tive integers are next to each other, such as 49 and 50, or and . Let

the smaller integer; then the larger integer.

2. Translate. It helps to reword the problem before translating:

First integer times Second integer is 156. Rewording

x 156 Translating


Multiplying


Simplifying

Factoring

or Using the principle of zeroproducts

or

4. Check. The solutions of the equation are 12 and . When x is 12, thenis 13, and . The numbers 12 and 13 are consecutive in-

tegers that are solutions to the problem. When x is , then is and . The numbers and are consecutive inte-gers, but they are not solutions of the problem because negative numbersare not used as entry numbers.

5. State. The entry numbers are 12 and 13.

Do Exercise 4.

The Pythagorean Theorem

The following problems involve the Pythagorean theorem, which relates thelengths of the sides of a right triangle. A triangle is a right triangle if it has a90°, or right, angle. The side opposite the 90° angle is called the hypotenuse.The other sides are called legs.

�12�13��13� ��12� � 156�12,x � 1�13

12 � 13 � 156x � 1�13

x � �13.x � 12

x � 13 � 0x � 12 � 0

�x � 12� �x � 13� � 0

x2 � x � 156 � 0

x2 � x � 156 � 156 � 156

x2 � x � 156

x�x � 1� � 156

��x � 1��

x � 1 �x ��5�6

x � 1x

4. Page Numbers. The product ofthe page numbers on two facingpages of a book is 506. Find thepage numbers.

Answer on page A-23

x

x � 1

⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

378


Ch05pgs375-387 1/19/06 9:54 AM Page 378

ISB

N:0-536-47742-6


THE PYTHAGOREAN THEOREM

In any right triangle, if a and b are the lengths of the legs and c is thelength of the hypotenuse, then

EXAMPLE 5 Physical Education. An outdoor-education ropes course in-cludes a 25-ft cable that slopes downward from a height of 37 ft to a height of30 ft. How far is it between the trees that the cable connects?

1. Familiarize. We make a drawing as above, noting that when we subtract30 ft from 37 ft, we get the height of the right triangle that is formed. We let

the distance between the trees.

2. Translate. A right triangle is formed, so we can use the Pythagorean theorem:

Substituting 7 for the length of a leg and 25 for the length of the hypotenuse


Squaring 7 and 25

Subtracting 625

Factoring


or

4. Check. Since the distance between the trees cannot be negative, cannot be a solution. If the distance is 24 ft, we have

which is Thus, 24 checks and is a solution.

5. State. The distance between the trees is 24 ft.

Do Exercise 5.

252.49 � 576 � 625,72 � 242 �

�24

b � �24.b � 24

b � 24 � 0b � 24 � 0

�b � 24� �b � 24� � 0

b2 � 576 � 0

49 � b2 � 625

72 � b2 � 252

72 � b2 � 252.

a2 � b2 � c2

7 ft25 ft

b

b �

25 ft

?

37 ft

30 ft

ac

b

The symbol denotes a 90° angle.

a2 � b2 � c2.

5. Reach of a Ladder. Twila has a26-ft ladder leaning against herhouse. If the bottom of theladder is 10 ft from the base ofthe house, how high does theladder reach?

Answer on page A-23

379


Ch05pgs375-387 1/19/06 9:54 AM Page 379

ISB

N:0

-536

-477

42-6


EXAMPLE 6 Ladder Settings. A ladder of length 13 ft is placedagainst a building in such a way that the distance from the top of theladder to the ground is 7 ft more than the distance from the bottom ofthe ladder to the building. Find both distances.

1. Familiarize. We first make a drawing. The ladder and the missingdistances form the hypotenuse and legs of a right triangle. We let

the length of the side (leg) across the bottom. Then thelength of the other side (leg). The hypotenuse has length 13 ft.

2. Translate. Since a right triangle is formed, we can use thePythagorean theorem:

Substituting


Squaring the binomial and 13

Collecting like terms


Simplifying

Factoring out a common factor

Dividing by 2

Factoring


or

4. Check. The negative integer cannot be the length of a side.When , , and . So 5 and 12 check.

5. State. The distance from the top of the ladder to the ground is 12 ft. The distance from the bottom of the ladder to the building is 5 ft.

Do Exercise 6.

52 � 122 � 132x � 7 � 12x � 5�12

x � 5.x � �12

x � 5 � 0x � 12 � 0

�x � 12� �x � 5� � 0

x2 � 7x � 60 � 0

2�x2 � 7x � 60� � 0

2x2 � 14x � 120 � 0

2x2 � 14x � 49 � 169 � 169 � 169

2x2 � 14x � 49 � 169

x2 � �x2 � 14x � 49� � 169

x2 � �x � 7�2 � 132.

a2 � b2 � c2

x � 7

x

13 ft

x � 7 �x �

6. Right-Triangle Geometry. Thelength of one leg of a righttriangle is 1 m longer than theother. The length of thehypotenuse is 5 m. Find thelengths of the legs.

Answer on page A-23

380


Ch05pgs375-387 1/19/06 9:54 AM Page 380

ISB

N:0-536-47742-6


The goal of these matchingquestions is to practice step (2),Translate, of the five-step problem-solving process. Translate eachword problem to an equation andselect a correct translation fromequations A–O.

A.

B.

C.

D.

E.

F.

G.

H.

I.

J.

K.

L.

M.

N.

O.


x � �x � 1� � �x � 2� � 180

x2 � 60 � 7021

x�x � 2� � 3599

2x2 � x � 288

x � 6% � x � 40,704

59% � x � 60

x �23

x � �x � 2� � 180

12

x�x � 1� � 1770

2�x � 2� � 2x � 240

6% � x � 40,704

x2 � �x � 70�2 � 1302

x2 � �x � 2�2 � 3599

59 � x � 60

x�x � 60� � 7021

2x � x � 288

Translatingfor Success

6. Cell-Phone Tower. A guy wireon a cell-phone tower is 130 ftlong and is attached to the topof the tower. The height of thetower is 70 ft longer than thedistance from the point on the ground where the wire isattached to the bottom of thetower. Find the height of thetower.

7. Sales Meeting Attendance.PTQ Corporation holds a salesmeeting in Tucson. Of the 60 employees, 59 of them attendthe meeting. What percentattend the meeting?

8. Dimensions of a Pool. Arectangular swimming pool istwice as long as it is wide. Thearea of the surface is Find the dimensions of the pool.

9. Dimensions of a Triangle. Theheight of a triangle is 1 cm lessthan the length of the base. Thearea of the triangle is Find the height and the lengthof the base.

10. Width of a Rectangle. Thelength of a rectangle is 60 ftlonger than the width. Find thewidth if the area of the rectangleis 7021 ft2.

1770 cm2.

288 ft2.

1. Angle Measures. The measuresof the angles of a triangle arethree consecutive integers. Findthe measures of the angles.

2. Rectangle Dimensions. Thearea of a rectangle is The length is 2 ft longer than thewidth. Find the dimensions ofthe rectangle.

3. Sales Tax. Claire paid $40,704for a new SUV. This included 6%for sales tax. How much did theSUV cost before tax?

4. Wire Cutting. A 180-m wire iscut into three pieces. The thirdpiece is 2 m longer than the first.The second is two-thirds as longas the first. How long is eachpiece?

5. Perimeter. The perimeter of arectangle is 240 ft. The length is2 ft greater than the width. Findthe length and the width.

3599 ft2.

Ch05pgs375-387 1/19/06 9:54 AM Page 381

ISB

N:0

-536

-477

42-6


382


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.



Math TutorCenter

InterActMath


, Solve.

3. Furnishings. A rectangular table in Arlo’s House ofTunes is six times as long as it is wide. The area of thetable is . Find the length and the width of the table.

4. Dimensions of Picture. A rectangular picture is threetimes as long as it is wide. The area of the picture is

. Find the dimensions of the picture.

3w

w

588 in2

w 6w

24 ft2

1. Design. The screen of the TI-84 Plus graphingcalculator is nearly rectangular. The length of therectangle is 2 cm more than the width. If the area of therectangle is find the length and the width.

2. Area of a Garden. The length of a rectangular garden is 4 m greater than the width. The area of the garden is . Find the length and the width.

ww + 4

96 m2

w + 2

w

24 cm2,

5. Dimensions of a Triangle. A triangle is 10 cm widerthan it is tall. The area is . Find the height and thebase.

6. Dimensions of a Triangle. The height of a triangle is3 cm less than the length of the base. The area of thetriangle is . Find the height and the length of thebase.

b

b � 3

35 cm2

h �10

h

28 cm2

Ch05pgs375-387 1/19/06 9:54 AM Page 382

ISB

N:0-536-47742-6


383

Exercise Set 5.9

Games in a League. Use for Exercises 9–12.x2 � x � N

9. A chess league has 14 teams. What is the total numberof games to be played if each team plays every otherteam twice?

10. A women’s volleyball league has 23 teams. What is thetotal number of games to be played if each team playsevery other team twice?

11. A slow-pitch softball league plays a total of 132 games.How many teams are in the league if each team playsevery other team twice?

12. A basketball league plays a total of 90 games. How manyteams are in the league if each team plays every otherteam twice?

7. Road Design. A triangular traffic island has a base halfas long as its height. The island has an area of .Find the base and the height.

8. Dimensions of a Sail. The height of the jib sail on aLightning sailboat is 5 ft greater than the length of its“foot.” The area of the sail is . Find the length ofthe foot and the height of the sail.

x + 5

x

42 ft2

h

h

12

64 m2

Handshakes. A researcher wants to investigate the potential spread of germs by contact. She knows that the number ofpossible handshakes within a group of x people, assuming each person shakes every other person’s hand only once, is given by

Use this formula for Exercises 13–16.

N � 12 �x2 � x�.

13. There are 100 people at a party. How many handshakesare possible?

14. There are 40 people at a meeting. How manyhandshakes are possible?

Ch05pgs375-387 1/19/06 9:54 AM Page 383

ISB

N:0

-536

-477

42-6


384


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

17. Toasting. During a toast at a party, there were 190 “clicks” of glasses. How many people took part in the toast?

18. High-fives. After winning the championship, allDetroit Pistons teammates exchanged “high-fives.”Altogether there were 66 high-fives. How many playerswere there?

19. Consecutive Page Numbers. The product of the pagenumbers on two facing pages of a book is 210. Find thepage numbers.

20. Consecutive Page Numbers. The product of the pagenumbers on two facing pages of a book is 420. Find thepage numbers.

21. The product of two consecutive even integers is 168.Find the integers. (See Section 2.6.)

22. The product of two consecutive even integers is 224.Find the integers. (See Section 2.6.)

23. The product of two consecutive odd integers is 255.Find the integers.


25. Right-Triangle Geometry. The length of one leg of aright triangle is 8 ft. The length of the hypotenuse is 2 ftlonger than the other leg. Find the length of thehypotenuse and the other leg.

26. Right-Triangle Geometry. The length of one leg of aright triangle is 24 ft. The length of the other leg is 16 ftshorter than the hypotenuse. Find the length of thehypotenuse and the other leg.

27. Roadway Design. Elliott Street is 24 ft wide when itends at Main Street in Brattleboro, Vermont. A 40-ft longdiagonal crosswalk allows pedestrians to cross MainStreet to or from either corner of Elliott Street (see thefigure). Determine the width of Main Street.

28. Sailing. The mainsail of a Lightning sailboat is a righttriangle in which the hypotenuse is called the leech. If a24-ft tall mainsail has a leech length of 26 ft and ifDacron® sailcloth costs $10 per square foot, find thecost of a new mainsail.

24 ft26 ft

24 ft

40 ft

Elliott St.

Main St.

15. Everyone at a meeting shook hands with each other.There were 300 handshakes in all. How many peoplewere at the meeting?

16. Everyone at a party shook hands with each other. Therewere 153 handshakes in all. How many people were atthe party?

Ch05pgs375-387 1/19/06 9:54 AM Page 384

ISB

N:0-536-47742-6


385

Exercise Set 5.9

29. Lookout Tower. The diagonal braces in a lookout towerare 15 ft long and span a distance of 12 ft. How highdoes each brace reach vertically?

30. Aviation. Engine failure forced Geraldine to pilot herCessna 150 to an emergency landing. To land, Geraldine’splane glided 17,000 ft over a 15,000-ft stretch of desertedhighway. From what altitude did the descent begin?

12 ft

15 ft h

31. Architecture. An architect has allocated a rectangularspace of for a square dining room and a 10-ftwide kitchen, as shown in the figure. Find thedimensions of each room.

32. Guy Wire. The guy wire on a TV antenna is 1 m longerthan the height of the antenna. If the guy wire isanchored 3 m from the foot of the antenna, how tall isthe antenna?

3 m

10 ft

264 ft2

Rocket Launch. A model rocket is launched with an initialvelocity of . Its height h, in feet, after t seconds isgiven by the formula

33. After how many seconds will the rocket first reach aheight of 464 ft?

34. After how many seconds from launching will the rocketagain be at that same height of 464 ft? (See Exercise 33.)

h

t

h

h � 180t � 16t 2.

180 ft�sec

Ch05pgs375-387 1/19/06 9:54 AM Page 385

ISB

N:0

-536

-477

42-6


386


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

35. The sum of the squares of two consecutive odd positiveintegers is 74. Find the integers.

36. The sum of the squares of two consecutive odd positiveintegers is 130. Find the integers.

37. An archaeologist has measuring sticks of 3 ft, 4 ft,and 5 ft. Explain how she could draw a 7-ft by 9-ftrectangle on a piece of land being excavated.

38. Look closely at the problem-solving techniquesdeveloped in this chapter. What kinds of equations dowe use? In order to solve these equations, whatadditional new skill do we need? Compare the skillslearned in this chapter with those of Chapter 2.

DWDW

SKILL MAINTENANCE

39. To a polynomial is to express it as aproduct. [5.1b]

40. A(n) of a polynomial P is a polynomialthat can be used to express P as a product. [5.1b]

41. A factorization of a polynomial is an expression that namesthat polynomial as a(n) . [5.1b]

42. When factoring, always look first for the .[5.1b]

43. The expression is an example of a. [4.3i]

44. The asserts that when dividing withexponential notation, if the bases are the same, keep thebase and subtract the exponent of the denominator fromthe exponent of the numerator. [4.1e]

45. For the graph of the equation the pair is known as the . [3.3a]

46. For the graph of the equation the number is known as the . [3.4c]

434x � 3y � 12,

�0, �4�4x � 3y � 12,

�5x2 � 8x � 7

quotient rule

product rule

slope

common factor

common multiple

factor

x-intercept

y-intercept

binomial

trinomial

quotient

product

i VOCABULARY REINFORCEMENT

In each of Exercises 39–46, fill in the blank with the correct term from the given list. Some of the choices may not be used andsome may be used more than once.

Ch05pgs375-387 1/19/06 9:54 AM Page 386

ISB

N:0-536-47742-6


387

Exercise Set 5.9

SYNTHESIS

47. Telephone Service. Use the information in the figurebelow to determine the height of the telephone pole.

48. Roofing. A square of shingles covers of surfacearea. How many squares will be needed to reshingle thehouse shown?

24 ft

16 ft

25 ft

32 ft

100 ft2

34 ft

5 ft

x

x � 112

49. Pool Sidewalk. A cement walk of constant width isbuilt around a 20-ft by 40-ft rectangular pool. The totalarea of the pool and the walk is . Find the widthof the walk.

50. Rain-Gutter Design. An open rectangular gutter ismade by turning up the sides of a piece of metal 20 in.wide. The area of the cross-section of the gutter is .Find the depth of the gutter.

20 in.

50 in2

50 in2

20 ft

40 ft

x

x

1500 ft2

51. Dimensions of an Open Box. A rectangular piece ofcardboard is twice as long as it is wide. A 4-cm square iscut out of each corner, and the sides are turned up tomake a box with an open top. The volume of the box is

. Find the original dimensions of the cardboard.

52. Solve for x.

63 cm

60 cm36 cm

x

44

V = 616 cm3

616 cm3

53. Dimensions of a Closed Box. The total surface area of aclosed box is . The box is 9 ft high and has asquare base and lid. Find the length of a side of the base.

54. The ones digit of a number less than 100 is 4 greaterthan the tens digit. The sum of the number and theproduct of the digits is 58. Find the number.

350 ft2

Ch05pgs375-387 1/19/06 9:54 AM Page 387

ISB

N:0

-536

-477

42-6


IMPORTANT PROPERTIES AND FORMULAS

Factoring Formulas:

The Principle of Zero Products: An equation is true if and only if is true or is true, or bothare true.

The Pythagorean Theorem: c a

b

a2 � b2 � c2

b � 0a � 0ab � 0

A2 � 2AB � B2 � �A � B�2

A2 � 2AB � B2 � �A � B�2,

A2 � B2 � �A � B� �A � B�,

Review Exercises

Find the GCF. [5.1a] 7. 8. 6x3 � 12x2 � 3xx2 � 14x � 49

The review that follows is meant to prepare you for a chapter exam. It consists of three parts. The first part, ConceptReinforcement, is designed to increase understanding of the concepts through true/false exercises. The second part isa list of important properties and formulas. The third part is the Review Exercises. These provide practice exercises forthe exam, together with references to section objectives so you can go back and review. Before beginning, stop andlook back over the skills you have obtained. What skills in mathematics do you have now that you did not have beforestudying this chapter?

388


Summary and Review55

i CONCEPT REINFORCEMENT

Determine whether the statement is true or false. Answers are given at the back of the book.

1. The greatest common factor (GCF) of a set of natural numbers is at least 1 andalways less than or equal to the smallest number in the set.

2. To factor use a trial-and-error process that looks for factors of bwhose sum is c.

3.

4. A product is 0 if and only if all factors are 0.

5. In order that the principle of zero products can be used, one side of theequation must be 0.

�x � a�2 � ��a � x�2

x2 � bx � c,

1. 25y6�15y2,

2. 36xy�60x2y,12x3,

Factor completely. [5.7a]

3. 4. x2 � 3x5 � 20x6

5. 6. x2 � 4x � 129x2 � 4

9. 10. 6x2 � 5x � 1x3 � x2 � 3x � 3

11. 12. 9x3 � 12x2 � 45xx4 � 81

Ch05pgs388-392 1/19/06 9:59 AM Page 388

ISB

N:0-536-47742-6


389

Summary and Review: Chapter 5

13. 14. x4 � 4x3 � 2x � 82x2 � 50

15. 16. 8x6 � 32x5 � 4x416x4 � 1

17. 18. x2 � 975 � 12x2 � 60x

19. 20. 8x3 � 125x3 � x2 � 30x

21. 22. 6x2 � 28x � 489x2 � 25 � 30x

23. 24. 2x2 � 7x � 4x2 � 6x � 9

25. 26. 3x2 � 2718x2 � 12x � 2

27. 28. 25x2 � 20x � 415 � 8x � x2

29. 30. x2y2 � xy � 1249b10 � 4a8 � 28a4b5

31. 32. m2 � 5m � mt � 5t12a2 � 84ab � 147b2

33. 32x4 � 128y4z4

Solve. [5.8a, b]

34. 35. x2 � 2x � 35 � 0�x � 1� �x � 3� � 0

36. 37. 3x2 � 2 � 5xx2 � x � 12 � 0

38. 39. 16 � x�x � 6�2x2 � 5x � 12

Solve. [5.9a, b]

40. Sharks’ Teeth. Sharks’ teeth are shaped like triangles.The height of a tooth of a great white shark is 1 cmlonger than the base. The area is Find the heightand the base.

b � 1

b

15 cm2.

41. The product of two consecutive even integers is 288.Find the integers.


43. Tree Supports. A duckbill-anchor system is used tosupport a newly planted Bradford pear tree. The cable is2 ft longer than the distance from the base of the tree towhere the cable is attached to the tree. The cables areanchored 4 ft from the tree. How far from the groundare the cables attached to the tree?

4 ft

Ch05pgs388-392 1/19/06 9:59 AM Page 389

ISB

N:0

-536

-477

42-6


390


44. If the sides of a square are lengthened by 3 km, the areaincreases to Find the length of a side of theoriginal square.

81 km2.

Find the x-intercepts for the graph of the equation. [5.8b]

45.

x

yy � x 2 � 9x � 20

46.

x

y y � 2x 2 � 7x � 15

47. Write a problem for a classmate to solve such thatonly one of the two solutions of a quadratic equationcan be used as an answer. [5.9a, b]

DW

48. On a quiz, Sheri writes the factorization of as Explain Sheri’s

mistake. [5.5d]�2x � 10� �2x � 10�.4x2 � 100

DW

SYNTHESIS

50. The cube of a number is the same as twice the square ofthe number. Find all such numbers.

51. The length of a rectangle is two times its width. Whenthe length is increased by 20 and the width decreased by1, the area is 160. Find the original length and width.

Solve. [5.8b]

52. x2 � 25 � 0

53. �x � 2� �x � 3� �2x � 5� � 0

54. �x � 3�4x2 � 3x�x � 3� � �x � 3�10 � 0

55. Find a polynomial for the shaded area in the figurebelow. [5.2a]

x

Solve. [5.9a]

49. The pages of a book measure 15 cm by 20 cm. Marginsof equal width surround the printing on each page andconstitute one-half of the area of the page. Find thewidth of the margins.

15 cm

20 cm

Ch05pgs388-392 1/19/06 9:59 AM Page 390

ISB

N:0-536-47742-6


391

Test: Chapter 5

Chapter Test551. Find the GCF: 28x3, 48x7.

Factor completely.

2. 3. 4. 6y2 � 8y3 � 4y4x2 � 25 � 10xx2 � 7x � 10

5. 6. 7. x3 � 2x2 � 3xx2 � 5xx3 � x2 � 2x � 2

8. 9. 10. x2 � x � 124x2 � 928x � 48 � 10x2

11. 12. 13. 60x � 45x2 � 203w 2 � 756m3 � 9m2 � 3m

14. 15. 16. 5x2 � 26x � 549x2 � 84x � 363x4 � 48

17. 18. 19. 4x2 � 4x � 1580 � 5x4x4 � 2x3 � 3x � 6

20. 21. 22. 1000a3 � 27b33m2 � 9mn � 30n26t 3 � 9t 2 � 15t

Solve.

23. 24. 25. x�x � 3� � 282x2 � 7x � 15x2 � x � 20 � 0

Work It Out!Chapter Test Video

on CD

For Extra Help

Ch05pgs388-392 1/19/06 9:59 AM Page 391

ISB

N:0

-536

-477

42-6


392


Co

pyr

igh

t ©

200

7 P

ears

on

Ed

uca

tio

n, I

nc.

Solve.

26. The length of a rectangle is 2 m more than the width.The area of the rectangle is Find the length andthe width.

27. The base of a triangle is 6 cm greater than twice theheight. The area is Find the height and the base.

2h � 6

h

28 cm2.48 m2.

28. Masonry Corner. A mason wants to be sure he has aright corner in a building’s foundation. He marks apoint 3 ft from the corner along one wall and anotherpoint 4 ft from the corner along the other wall. If thecorner is a right angle, what should the distance bebetween the two marked points?

3 ft

x

4 ft

Find the x-intercepts for the graph of the equation.

29. 30.

x

yy � 3x 2 � 5x � 2

x

y y � x 2 � 2x � 35

31. The length of a rectangle is five times its width. Whenthe length is decreased by 3 and the width is increasedby 2, the area of the new rectangle is 60. Find theoriginal length and width.

32. Factor: �a � 3�2 � 2�a � 3� � 35.

33. Solve: 34. If and then

a) 2 b) 10c) 34 d) 24

x2 � y2 � ?x � y � 6,x � y � 420x�x � 2� �x � 1� � 5x3 � 24x � 14x2.

SYNTHESIS

Ch05pgs388-392 1/19/06 9:59 AM Page 392

ISB

N:0-536-47742-6


myresource.phoenix.edumyresource.phoenix.edu/secure/resource/MAT117R7/mat117_week1… · EXAMPLE 5...

Documents

Transcript of myresource.phoenix.edumyresource.phoenix.edu/secure/resource/MAT117R7/mat117_week1… · EXAMPLE 5...