LESSON 2: SIMPLIFYING RADICALS...

High School: Working with Expressions

Copyright © 2015 Pearson Education, Inc. 33

ANSWERSLESSON 2: SIMPLIFYING RADICALS

ANSWERS

N.RN.2 1. C

N.RN.2 2. B

C t t t t t= • •

E a b a a b• • •=12

N.RN.2 3.

N.RN.2 4. 12

N.RN.2 5.

N.RN.2 6.

N.RN.2 7. A

N.RN.2 8.36 2x 180 3x 448 4x 162 5• x 48x 200 6x

6x 6 5x x 8 72x 9 22x x 4 3x 10 23x

5 5

x x x• =

4 6

3 7

8 2

7 32x



ANSWERSLESSON 2: SIMPLIFYING RADICALS

Challenge Problem

N.RN.2 9. If the two radical expressions are equivalent, you have:

64 2 26x x xn n= This means that:

64 2 2

2

2

6

1

1 1

x x x

x

x

n nn

nn

n nn

=

=

=

+

+ +

( )

( )

You can compare the terms inside the radicals to figure out the value of n.

The value of n should solve both equations: x xn

n

+

+

=

=

1 6

12 64

The equations are equivalent when n = 5.

Therefore the two radical expressions are 64 65 x and 2 25x x .



ANSWERS

ANSWERS

N.RN.3 1. 25 or 52 or –25 or –52

N.RN.3 2.

N.RN.3 3.

N.RN.3 4.

N.RN.3 5. B C 2 3 3 3. .−

E 4 9

N.RN.3 6.Rational Irrational Can't Tell

46

46

4 64.

6

4 6+

202

63

0.464...

N.RN.3 7. B an irrational

25

52

or

2 5 5 2 5+ + + or 2 or

5 2 2 5 2 5 5 2 25 52 52 52 25 2. . . . . . . . . .or or or or or or or or or 55

125

213

•

LESSON 3: NUMBER SYSTEM



ANSWERS

Challenge Problem

N.RN.3 8. A rational number is a number that can be represented as a fraction of two integers. If number x is a repeating number, you can always convert this number into a fraction of two integers. To find those integers, use exponents of 10 to create a subtraction operation using two numbers derived from x that both only have the repeating digits in their decimal portion.

For example, take the repeating decimal number x = 32 58731. .

The repeating portion of this number is 8731. Create two numbers that only have the repeating portion as their decimal portion by multiplying by exponents of 10 (e.g. 105 and 10). Subtracting these two numbers will get rid of the decimal portion. The outcome of the subtraction will be an integer.

10 8731

10 8731

10 10

10

5

5

x

x

x x

x

=

=

− =

−

3,258,731.

325.

3,258,406

(105 ) ==

=−

= =

3,258,406

3,258,406 3,258,40699,990

1,629,20349,9

x10 105 995

So, 1,629,203

49,99532 58731. =

As with this example, all repeating decimals can be converted to fractions.

LESSON 3: NUMBER SYSTEM



ANSWERS

ANSWERS

A.APR.1 1. D Distributive property

A.APR.1 2. C Commutative property

A.APR.1 3. A Associative property

A.APR.1 4. D Distributive property

A.APR.1 5. 9x + 2x3 – 4x = (9 – 4)x + 2x3 = 5x + 2x3 = 2x3 + 5x

A.APR.1 6. 9x3 + 7x2 – 2x3 + 3x2 = (7 + 3)x2 + (9 – 2)x3 = 10x2 + 7x3 = 7x3 + 10x2

A.APR.1 7. A – B = (9x2 + 4 – 7x – 2y) – (7x + 2y2 + 4x2 – 5y2) = 4 + (–7x – 7x) – 2y + (9x2 – 4x2) + (–2y2 + 5y2) = 4 – 14x – 2y + 5x2 + 3y2

= 3y2 + 5x2 – 2y – 14x + 4

A.APR.1 8. 9(x2 + 4y2) – 7(y2 – 2x2) = 9x2 + 36y2 – 7y2 + 14x2 = 23x2 + 29y2

A.APR.1 9. 0.5(9y2 + 4) – 2(7y – 3) = 4.5y2 + 2 – 14y + 6 = 4.5y2 – 14y + 8

A.APR.1 10. To make the operation simpler, you can first simplify each polynomial before adding and subtracting. A = 2(9y2 – 2y2 + 4x2)

= 2(7y2 + 4x2) = 14y2 + 8x2

B = 4(2x2 + 7y2) – (5x2 + 9y2) = 8x2 + 28y2 – 5x2 – 9y2 = (8x2 – 5x2) + (–9y2 + 28y2) = 3x2 + 19y2

C = 2(9y2 + 4x2 – 7y2) – 3(x2 + y2) = 18y2 + 8x2 – 14y2 – 3x2 – 3y2 = (8x2 – 3x2) + (18y2 – 14y2 – 3y2) = 5x2 + y2

You can then find the polynomial A – B + C using the simplified expression. A – B + C = (14y2 2 + 8x2) – (3x2 + 19y2 ) + (5x2 + y2 )

= (8 – 3 + 5)x2 + (14 – 19 + 1)y2 = 10x2 – 4y2

LESSON 4: POLYNOMIALS



ANSWERS

A.APR.1 11. 5(3x2 – 2x) – 7(3x2– 2x) + 3(3x2 – 2x) = (5 – 7 + 3)(3x2 – 2x) = 1(3x2 – 2x) = 3x2 – 2x

A.APR.1 12. 9(2x2 – 4y2) – 7 (–4y2 + 2x2) + 2 (x2 – 2y2) = 9(2x2 – 4y2) – 7(2x2 – 4y2) + 2x2 – 4y2 = (9 – 7 + 1)(2x2 – 4y2)

= 3(2x2 – 4y2) = 6x2 – 12y2

A.APR.1 13. 2(7y – 4x) + 7(7y – 4x) – 9 (7y – 4x) = (2 + 7 – 9)(7y – 4x) = 0 = 0(7y – 4x) = 0

A.APR.1 14. 4(3x – 7y) – 9(3x – 7y) – 2(–3x + 7y) = 4(3x – 7y) – 9(3x – 7y) + 2(3x – 7y) = (4 – 9 + 2)(3x – 7y) = –3(3x – 7y) = –9x + 21y = 21y – 9x

Challenge Problem

A.APR.1 15. Let’s call A the missing expressions. The sum of the two expressions is 5x. Therefore you can write: (3x2 – 2x – 7) + A = 5x

A = 5x – (3x2 – 2x – 7) A = –3x2 + (2 + 5)x + 7 A = –3x2 + 7x + 7

The missing expressions is –3x2 + 7x + 7

LESSON 4: POLYNOMIALS



ANSWERSLESSON 5: MULTIPLYING POLYNOMIALS

ANSWERS

A.APR.1 1. A (–x – 2)2

C (2 + x)(x + 2)

E 4 (1 + x) + x2

A.APR.1 2.3x(x + 1)(4x – 4) (2x2 – x)(1 – 2x) (x –1)(x + 2)(x – 3) –2x2(–2 – 5x) (x – 1)(x + 1)2

12(x3 – x) –4x3 + 4x2 – x x3 –2x2 – 5x + 6 10x3 + 4x2 x3 + x2 – x – 1

A.APR.1 3. 4x2 + 14x – 12

A.APR.1 4. x2 + 5x + 6

A.APR.1 5. x2 – 5x + 6

A.APR.1 6. x2 – 5x – 36

A.APR.1 7. x2 + 5x – 36

A.APR.1 8. 2x2 + 17x + 35

A.APR.1 9. –18x6 – 41x5 – 21x4

A.APR.1 10. –63x4 + 18x3 – 28x2 + 8x

A.APR.1 11. 28x4 – 27x3 – 10x2



ANSWERSLESSON 5: MULTIPLYING POLYNOMIALS

Challenge Problem

A.APR.1 A.REI.2

12. If you multiply the two binomials, you will find: (ax + b)(cx + d) = acx2 + (ad + bc)x + bd Therefore: R = ac S = ad + bc T = bd

Since a = 1, c = 2, S = –21, and T = 10, you have: R = (1)(2) = 2 –21 = (1)d + b(2) = 2b + d 10 = bd

From this you can deduce that R = 2. The easiest way to approximate the values of b and d is to graph the two equations remaining and find the point of intersection of the corresponding graphs. Since graphing only provides an approximation of the intersection, you will need to verify that the coordinates obtained solve both equations.

If you change the variables in the equations for b and d with x and y, respectively, you obtain: y = –2x – 21

yx

=10

There are two possible points of intersections: (–10, –1) and (–0.5, –20). Both coordinate pairs solve the equations and are therefore correct values. As a result, there are two possible sets of binomials that solve the problem: (x – 10)(2x – 1) = 2x2 – 21x + 10 (x – 0.5)(2x – 20) = 2x2 – 21x + 10



ANSWERSLESSON 6: FACTORING

ANSWERS

A.SSE.2 1.x2 + bx + c x2 + bx – c x2 – bx – c x2 – bx + c x2 – c

(x + A)(x + B)(x – A)(x + B)

A < B(x – A)(x + B)

A > B(x – A)(x – B)

(x – A)(x + B) A = B

A.SSE.3.a 2. B (x – 2)(x + 4)

A.SSE.3.a 3.x2 + 6x + 8 x2 + 11x + 18 x2 + 15x + 56 x2 – 2x – 24 x2 – 7x + 10

(x + 2)(x + 4) (x + 2)(x + 9) (x + 7)(x + 8) (x + 4)(x – 6) (x – 2)(x – 5)

A.SSE.3.a 4.x2 + 15x + 36 x2 – 5x – 14 4x2 – 14x + 10 x2 + 17x + 72 4x2 – 17x + 15

(x + 3)(x + 12) (x + 2)(x – 7) (2x – 2)(2x – 5) (x + 8)(x + 9) (x – 3)(4x – 5)

A.SSE.3.a 5. (x + 6) and (x + 9)

A.SSE.3.a 6. (x + 4) and (x + 11)

A.SSE.3.a 7. (x – 3)(x – 5)

A.SSE.3.a 8. (x + 1)(x + 5)

A.SSE.3.a 9. (x + 2)(2x + 9)

A.SSE.3.a 10. (x + 4)(3x – 6)

A.SSE.3.a 11. (x + 2)(–5x – 7)

A.SSE.3.a 12. –5(x + 1)(x – 1)



ANSWERSLESSON 6: FACTORING

Challenge Problem

A.SSE.2 13. You can multiply the binomial expressions into a trinomial form. (mx + p)(nx + q) = mnx2 + (p + q)x + pq

Since ax2 + bx + c = (mx + p)(nx + q), you write the following relationships: a = mn b = p + q c = pq

From these relationships you can determine the sign of each coefficient m, n, p, and q depending on the signs of a, b, and c.

a. a is a positive number Since a = mn, when a > 0, then m and n are both either positive or negative.

a is a negative number Since a = mn, when a > 0, then m and n have opposite signs (m is negative if n is positive and vice versa).

b. b and c are positive numbers b = mp + nq > 0 c = pq > 0 For both equations to be true, both p and q must be positive numbers. Both m and n must also be positive numbers.

b is a positive number and c is a negative number b = mp + nq > 0 c = pq < 0 For both equations to be true, p and q must have opposite signs and the absolute value of the positive number must be greater than the absolute value of the negative number. Both m and n must also be positive numbers. p < 0 and q > 0 and |q| > |p| or p > 0 and q < 0 and |p| > |q|

b is a negative number and c is a positive number b = mp + nq > 0 c = pq > 0 For both equations to be true, both p and q must be negative numbers. Both m and n must also be positive numbers.

b and c are negative numbers. b = mq + np < 0 c = pq < 0 For both equations to be true, p and q must have opposite signs and the absolute value of the negative number must be greater than the absolute value of the positive number. Both m and n must also be positive numbers. p < 0 and q > 0 and |q| < |p| or p > 0 and q < 0 and |p| < |q|



ANSWERSLESSON 7: SPECIAL BINOMIALS

ANSWERS

A.SSE.2 1. B square of a sum

A.SSE.2 2. C square of a difference

A.SSE.2 3. A difference of two squares

A.SSE.2 4. This expression is a difference of two squares, a2 – b2, where: a = 2x + 3 and b = 3 This special product of binomials can also be written in factor form as: (a + b)(a – b). Therefore the expression can be factored: (2x + 3 + 3)(2x + 3 –3) 2x(2x + 6)

A.SSE.3.a 5. Square of a Sum

Square of a Difference

Difference of Two Squares

(3x + 2)2

25x2 + 40x + 16

9x2 – 12x + 4

(2x – 1)2

(4x + 3)(4x – 3)

9x2 – 4

A.SSE.3.a 6. This expression is the square of a sum (a + b)2, where: a = 4x and b = 3 Therefore the polynomial form of this expression is: a2 + 2ab + b2: 16x2 + 24x + 9

A.SSE.2 7. This expression is the square of a difference (a – b)2, where: a = 1 and b = 3x Therefore the polynomial form of this expression is: a2 – 2ab + b2 1 – 6x + 9x2

A.SSE.3.a 8. This polynomial is the difference of two squares a2 – b2, where: a = 4x and b = 2 Therefore the polynomial can be factored to: (a + b)(a – b) (4x + 2)(4x – 2)



ANSWERSLESSON 7: SPECIAL BINOMIALS

A.SSE.3.a 9. This polynomial is in the form a2 + 2ab + b2, which is the polynomial form associated with the square of a sum (a + b)2. In this particular expression you can have two possible sets of values for a and b. a2 = 64x2 and b2 = 81 As a result: a =±8x and b =±9

Since 2ab >0, it means that a and b are either both positive or both negative numbers. Therefore the values of a and b can either be: a = 8x and b = 9 or a = –8x and b = –9

The following two sets of factors are therefore valid: (8x + 9)2 or (–8x – 9)2

A.SSE.2 10. Miki confused the sum of squares with the difference of two squares. The product of 2 and –2 is negative, so the 4 must be subtracted from the square of 3x when multiplying.

(3x + 2)(3x –2)

(3x+2) (3x-2) = 9x2 + 4–

Challenge Problem

A.SSE.2 11. The expression can be rearranged to bring forward the polynomial equivalent to the square of a difference a2 – 2ab + b2: (a2 – 2ab + b2) – 4a2b2

The first portion of the equation can then be factored as the square of the difference of a – b: (a – b)2 – 4a2b2

Since 4a2b2 = (2ab)2, the expression can then be rewritten: (a – b)2 – (2ab)2

This expression is the sum of two squares, which can be factored: (a – b – 2ab)(a – b + 2ab)



ANSWERS

ANSWERS

A.APR.1 A.APR.6

1.x

x

3

2

5

2

3

2

x

x

x

x

2

3

–

–

5

2

3

3

x

x

4

2

2

3

x

x

x52x 1

x52

2x

A.APR.1 2. B

A.APR.1 A.APR.6

3. B

A.APR.1 A.APR.6

4.3 3

2

x x

x

+ x x

x

2 3

3

− 4 2

2

2

2

x x

x

+ 6 2

3

3

2

x x

x

+ x

x

+−

51

31

xx

+1

1x− 2

1+

x2

23

xx

+ x3 + 5x2

A.APR.1 5. The polynomial x2 +10x + 16 can be factored to (x + 2)(x + 8). Therefore, the expression can be simplified to:

x xx

x xx

x2 10 16

82 8

82

+ ++

=+ +

+= +

( )( )

The quotient of the expression is (x + 2).

A.APR.1 A.APR.6

6. The polynomial x2 – x – 20 can be factored to (x + 4)(x – 5). Therefore, the expression can be simplified to:

x xx

x xx

x2 20

44 5

45

− −+

=+ −

+= −

( )( )

The quotient of the expression is (x – 5).

A.APR.1 7. 32

3

2

3

4

34

34

5

33 2

5 3

3 2

8

6

22x

xx

x x

x

x

x

xx−

− = = = =• ( )•

( )

15

+x

5 2 72

52

172

2x xx

xx

+ −= + −

LESSON 8: DIVIDING POLYNOMIALS



ANSWERS

A.APR.1 A.APR.6

8. You can first simplify the expression by dividing the numerator by : x x

x

2 3 183

+ −−

The resulting polynomial x2 + 3x – 18 can be factored to (x – 3)(x + 6).

Therefore, the expression can be simplified to: x x

xx x

xx

2 3 183

3 63

6+ −−

=− +

−= +

( )( )

The quotient of the expression is (x + 6).

A.APR.1 A.APR.6

9. B

A.APR.1 A.APR.6

10. You must find the divisor A that solves this equation: 6 16 32

3 42x x

Ax

− −= +

This equation can be rewritten:

Ax x

x

x xx

x xx

x

=− −+

=− −+

=+ −+

= −=

6 16 323 4

2 3 8 163 4

2 3 4 43 4

2 4

2

2( )

( )( )

( )

22 8x −

Therefore, the divisor of this equation is (2x – 8)

xx

2

2

x xx

2 3 105

– ––

LESSON 8: DIVIDING POLYNOMIALS



ANSWERSLESSON 8: DIVIDING POLYNOMIALS

Challenge Problem

A.APR.1 A.APR.6

11. This equation can be rewritten as: mx3 + 5x2 – 14x = xn(x + p)(x + q) The right factored expression can be expressed as a polynomial: mx3 + 5x2 – 14x = xn(x2 + (p + q)x + pq)

= x2 + n + (p + q)x1 + n + pqxn

Since the greatest exponent of the left expression is 3, the greatest exponent of the right expression must also be 3, therefore (2 + n) = 3 and n = 1. mx3 + 5x2 – 14x = x3 + (p + q)x2 + pqx

From this equation, these equalities can be deduced: m = 1 p + q = 5 pq = –14

If p + q = 5, then p = 5 – q and (5 – q)q = –14 (5 – q)q = –q2 + 5q = –14 The possible values of q solving this equation are either 7 or –2. p = 5 – q If q = 7 then p = –2. If q = –2 then p = 7. Since p > q, the only possible combination is p = 7 and q = –2.

The polynomial operation with all the coefficients replaced by their actual values is therefore:

x x xx x

x3 25 14

72

+ −+

= −( )



ANSWERS

ANSWERS

N.RN.2 1. B

N.RN.2 2.25 36+ 49 16− 4 7 2 7+ 20 5− 45 3−

11 3 6 7 5 3 5 3−

N.RN.2 3.

N.RN.2 4.

28 7• 25 45• 40 90•4 35 3

35

• 2 62

38

•

14 15 5 60 4 3 2

N.RN.2 5.

N.RN.2 6.

N.RN.2 7.

N.RN.2 8.

134

3 33

5 4 3 2 3 5 6 3 30 3( ) ( )+ = =

6 7 16 4 16 6 3 4

6 3 4

12 6

2−( ) = ( )= ( )=

•

7 3 7 2 5 3 7 2 5 7

21 2 35

+( ) = +

= +

•

6 4 32

22 3 4 3

2 2

2

4 2 3 2

12 2 2

13 2

2

• •+ = +

= ( ) +

= +

=

LESSON 9: OPERATIONS WITH RADICALS



ANSWERSLESSON 9: OPERATIONS WITH RADICALS

N.RN.2 9.

N.RN.2 10. C

Challenge Problem

N.RN.2 A.SSE.2

11. Using the properties of exponents, you can rewrite the expression in a form that match 3ab2:

a b ma bb

m abb

m abn n n

n−=

−= −

−( )( )

11

11

1

For this expression to be equal to 3ab2, you must fulfill these two conditions: (1 – m) = 3 and 1

1 2n− =

Therefore: m = –2

n =13

Since b bn n=1

, if n =13

, then b b bn = =

113 3

.

Therefore when m and n are replaced in the expression by their respective values, the expression becomes:

ab abb

3 32− −( )

This expression is indeed equal to 3ab2 when it is simplified.

83 5 5

42 2 5

4 5

−= ( )=

4 54x x



ANSWERSLESSON 10: SOLVING RADICAL EQUATIONS

ANSWERS

A.REI.2 1. B

A.REI.2 2. A = –2

A.REI.2 3. A

B

C 1

2x

x=

A.REI.2 4.

You need to substitute 163

for x in the equation to check your answer.

7 3163

3

7 16 3

7 4 3

7 7

= +

= += +=

•

This answer correctly solves the equation.

x =163

x =169

xx

23= −

2 7 4 1x x+ = −

7 3 3

7 3 3

4 3

16 3

163

2 2

= +

− =

= ( )=

=

x

x

x

x

x




A.REI.2 5.

You need to substitute –5 for x in the equation to check your answer. You also need to remember that square roots can only be used on positive numbers. The square root of a negative real number is impossible to calculate.

3 5 3 12

5 7 12

( )

( )

− + = −

− − = −

This solution is extraneous and should not be included in your answer. Since x = –5 was the only possible solution, this equation cannot be solved. There are no solutions to this problem.

A.REI.2 6.

You need to substitute 16 for x in the equation to check your answer.

16 7 3

9 3

3 3

− =

==

This answer correctly solves the equation. x = 16

3 3 7

3 3 7

3 3 7

2 10

5

2 2

x x

x x

x x

x

x

+ = −

+( ) = −( )+ = −

= −= −

x

x

x

x

− =

−( ) =− =

=

7 3

7 3

7 9

16

2 2




A.REI.2 7.


This answer correctly solves the equation. x = 576

A.REI.2 8.

You need to substitute 72

for x in the equation to check your answer.

72

372

7

72

212

142

72

72

=

−

= −

=

This answer correctly solves the equation. x =72

x

x

x

x

31 7

3 7 1

24

576

2 2

− =

= +

( ) =

=

( )

5763

1 7

243

1 7

8 1 7

7 7

− =

− =

− ==

x x

x x

x x

x

x

= −

( ) = −( )= −=

=

3 7

3 7

3 7

2 7

72

2 2




A.REI.2 9.


3

0

03 0

=( )

Since you cannot divide by 0, this solution is extraneous and should not be included in your answer. Since this was the only possible solution, this equation cannot be solved. There are no solutions to this problem.

Challenge Problem

A.REI.2 10. Square roots of real numbers can only be calculated if the number is positive. Square roots of negative numbers are impossible.

The distance between two points is expressed as the square root of the sum of the squares of the differences between the x coordinates and the y coordinates of the two points. Since the differences are squared, the square of the differences will always be positive numbers even when the differences themselves are negative.

Since both squares of the differences are positive, the sum of those squares will also be positive. Therefore the number inside the square root will always be positive and a distance can always be calculated irrespectively of the coordinates of the points.

33

3 3

9

9 0

8 0

0

2

x

xx

x x

x x

x x

x

x

=

= ( )=

− ===

( )



ANSWERSLESSON 11: PUTTING IT TOGETHER

ANSWERS

8.EE.5 8.EE.6

2. Definitions and examples will vary. Here are some examples.

Concept or Property Description Examples

negative integer exponent

For any nonzero number a and a positive integer n:

aa

nn

− =1

21

2

12 2 2

18

33

− = = =• •

0 as an exponent

For any nonzero number a: a0 = 1 00 = 10 = 20 = 30 = 12,9000 = 1

fraction as an exponent

For a > 0:

a an n1

=2 2

13 3=

properties of exponents

m, n, and p are positive integers: am + n = am • an

aa

am n

m

n− =

(am)n = am – n

a

b

a

b

m

n

p mp

np

=

42 + 3 = (4 • 4) • (4 • 4 • 4) = 42 • 43

44

443 2

3

2− = =

(43)2 = (4 • 4 • 4)2 = (4 • 4 • 4) • (4 • 4 • 4) = 46 = 43 • 2

4

5

4

5

4

5

3

2

4 3 4

2 4

12

8

= =

•

•

rational numbers

Numbers that can be expression as a ratio between two integers such that the second integer is not zero.

• All integers are included in rational numbers since they are all divisible by 1.

• All decimals that terminate are rational since they are divisible by a factor of 10.

• All decimals that repeat after some point are rational.

• All roots of perfect numbers are rational numbers (because they are integers).

The system of rational numbers is closed under addition, subtraction, multiplication, and division.

13, 0.56, and 2 357. are rational numbers because they can all be represented by a fraction:

13131

0 5656100

2 357389165

4 2 221

27 3 331

6

2

3 33

5 5

=

=

=

= = =

= = =

=

.

.

7,776 55 661

= =




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2. Concept or Property Description Examples

irrational numbers

Numbers that cannot be written as a fraction (a ratio of two integers). In decimal form, irrational numbers never end or repeat.

• All square roots of numbers that are not perfect squares are irrational.

• Special numbers such as π are also irrational.

2 1 4142135= ….

π = 3.14159265359…

real numbers A set of numbers containing all rational numbers and all irrational numbers.

• All the numbers on the number line are real numbers.

polynomial expression

An expression of one or more algebraic terms with whole-number (positive integers) exponents. Since the exponent has to be a positive integer, expression containing inverse of exponents or roots are not polynomial expressions.

polynomial expressions: x + 2

xx

32

21+ −

not polynomial expressions: x–2 + 4

12

2 53

12

x

x

x

−

+

factoring second-degree polynomials

Second-degree polynomials can sometimes be factored into two binomials with coefficients. The expression ax2 + bx + c can be factored into (mx + p)(nx + q), where mn = a, pq = c, and (pn + mq) = b.

3x2 + x – 10 = (x + 2)(3x – 5)




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2. Concept or Property Description Examples

special products of binomials

There are three special products of binomials:

• The square of a sum: (a + b)2 = (a + b)(a + b) = a2 + 2ab + b2

• The square of a difference: (a – b)2 = (a – b)(a – b) = a2 – 2ab + b2

• The difference of two squares: (a + b) (a – b) = a2 – b2

The square of a sum: (2 + 3)2 = 22 + 2(2)(3) + 32

(5)2 = 4 + 12 + 9 25 = 25

The square of a difference: (2 – 3)2 = 22 – 2(2)(3) + 32

(–1)2 = 4 – 12 + 9 1 = 1

The difference of two squares: (2 + 3)(2 – 3) = 22 – 32

(5)(–1) = 4 –9 –5 = –5

LESSON 2: SIMPLIFYING RADICALS...

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