2.5 Zeros of Polynomial Functions - cbsd.org€¦ · Because the factor produces no real zeros, you...

Section 2.5 Zeros of Polynomial Functions 169

The Fundamental Theorem of AlgebraYou know that an th-degree polynomial can have at most real zeros. In thecomplex number system, this statement can be improved. That is, in the complexnumber system, every th-degree polynomial function has precisely zeros. Thisimportant result is derived from the Fundamental Theorem of Algebra, firstproved by the German mathematician Carl Friedrich Gauss (1777–1855).

Using the Fundamental Theorem of Algebra and the equivalence of zerosand factors, you obtain the Linear Factorization Theorem.

For a proof of the Linear Factorization Theorem, see Proofs in Mathematics onpage 214.

Note that the Fundamental Theorem of Algebra and the Linear FactorizationTheorem tell you only that the zeros or factors of a polynomial exist, not how tofind them. Such theorems are called existence theorems.

Zeros of Polynomial Functions

a. The first-degree polynomial has exactly one zero:

b. Counting multiplicity, the second-degree polynomial function

has exactly two zeros: and (This is called a repeated zero.)

c. The third-degree polynomial function

has exactly three zeros: and

d. The fourth-degree polynomial function

has exactly four zeros: and

Now try Exercise 1.

x � �i.x � i,x � �1,x � 1,

f �x� � x4 � 1 � �x � 1��x � 1��x � i��x � i�

x � �2i.x � 0, x � 2i,

f �x� � x3 � 4x � x�x2 � 4� � x�x � 2i��x � 2i�

x � 3.x � 3

f �x� � x2 � 6x � 9 � �x � 3��x � 3�

x � 2.f �x� � x � 2

nn

nn

What you should learn• Use the Fundamental Theorem

of Algebra to determine thenumber of zeros of polynomialfunctions.

• Find rational zeros of polyno-mial functions.

• Find conjugate pairs of com-plex zeros.

• Find zeros of polynomials byfactoring.

• Use Descartes’s Rule of Signsand the Upper and LowerBound Rules to find zeros ofpolynomials.

Why you should learn itFinding zeros of polynomialfunctions is an important part ofsolving real-life problems. Forinstance, in Exercise 112 on page182, the zeros of a polynomialfunction can help you analyzethe attendance at women’scollege basketball games.

Zeros of Polynomial Functions2.5

The Fundamental Theorem of AlgebraIf is a polynomial of degree where then has at least onezero in the complex number system.

fn > 0,n,f �x�

Linear Factorization TheoremIf is a polynomial of degree where then has precisely linear factors

where are complex numbers.c1, c2, . . . , cn

f �x� � an�x � c1��x � c2� . . . �x � cn�

nfn > 0,n,f �x�

Recall that in order to find thezeros of a function set equal to 0 and solve the resultingequation for For instance, thefunction in Example 1(a) has azero at because

x � 2.

x � 2 � 0

x � 2

x.

f �x�f �x�,

Example 1

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170 Chapter 2 Polynomial and Rational Functions

Finding zeros of polynomial functions isa very important concept in algebra.This is a good place to discuss the factthat polynomials do not necessarilyhave rational zeros but may have zerosthat are irrational or complex.

−3 −2 21 3x

−3

−2

−1

1

2

3

f(x) = x3 + x + 1y

FIGURE 2.30

The Rational Zero TestThe Rational Zero Test relates the possible rational zeros of a polynomial(having integer coefficients) to the leading coefficient and to the constant term ofthe polynomial.

To use the Rational Zero Test, you should first list all rational numbers whosenumerators are factors of the constant term and whose denominators are factorsof the leading coefficient.

Having formed this list of possible rational zeros, use a trial-and-error method todetermine which, if any, are actual zeros of the polynomial. Note that when theleading coefficient is 1, the possible rational zeros are simply the factors of theconstant term.

Rational Zero Test with Leading Coefficient of 1

Find the rational zeros of

SolutionBecause the leading coefficient is 1, the possible rational zeros are the fac-tors of the constant term. By testing these possible zeros, you can see that neitherworks.

So, you can conclude that the given polynomial has no rational zeros. Note fromthe graph of in Figure 2.30 that does have one real zero between and 0.However, by the Rational Zero Test, you know that this real zero is not a rationalnumber.

Now try Exercise 7.

�1ff

� �1

f ��1� � ��1�3 � ��1� � 1

� 3

f �1� � �1�3 � 1 � 1

±1,

f �x� � x3 � x � 1.

Possible rational zeros �factors of constant term

factors of leading coefficient

The Rational Zero TestIf the polynomial has integer coefficients, every rational zero of has the form

where and have no common factors other than 1, and

q � a factor of the leading coefficient an.

p � a factor of the constant term a0

qp

Rational zero �p

q

fa1x � a0f �x� � anxn � an�1xn�1 � . . . � a2 x 2 �

Example 2

Historical NoteAlthough they were notcontemporaries, Jean Le Rondd’Alembert (1717–1783)worked independently of Carl Gauss in trying to provethe Fundamental Theorem of Algebra. His efforts weresuch that, in France, theFundamental Theorem ofAlgebra is frequently known as the Theorem of d’Alembert.

Fog

g A

rt M

use

um

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Rational Zero Test with Leading Coefficient of 1


SolutionBecause the leading coefficient is 1, the possible rational zeros are the factors ofthe constant term.

Possible rational zeros:

By applying synthetic division successively, you can determine that andare the only two rational zeros.

So, factors as

Because the factor produces no real zeros, you can conclude thatand are the only real zeros of which is verified in Figure 2.31.

FIGURE 2.31

Now try Exercise 11.

If the leading coefficient of a polynomial is not 1, the list of possible rationalzeros can increase dramatically. In such cases, the search can be shortened inseveral ways: (1) a programmable calculator can be used to speed up the calcula-tions; (2) a graph, drawn either by hand or with a graphing utility, can give a goodestimate of the locations of the zeros; (3) the Intermediate Value Theorem alongwith a table generated by a graphing utility can give approximations of zeros; and(4) synthetic division can be used to test the possible rational zeros.

Finding the first zero is often the most difficult part. After that, the search issimplified by working with the lower-degree polynomial obtained in syntheticdivision, as shown in Example 3.

−4 −2 6x

6

4

−6

8

−8

( 1, 0)− (2, 0)

8−6−8

f x x x x x( ) = + 3 64 3 2− − −

y

f,x � 2x � �1�x2 � 3�

f �x� � �x � 1��x � 2��x2 � 3�.

f �x�

2 1

1

�2

2

0

3

0

3

�6

6

0

�1 1

1

�1

�1

�2

1

2

3

�3

�3

�6

�6

6

0

x � 2x � �1

±1, ±2, ±3, ±6

f �x� � x4 � x3 � x2 � 3x � 6.

Additional Example

List the possible rational zeros of

Solution

The leading coefficient is 1, so the possi-ble rational zeros are

and To decide which possiblerational zeros should be tested usingsynthetic division, graph the function.From the graph, you can see that thezero is positive and less than 10, so theonly values of that should be testedare and

x−8

−1600

800

8−16

y

x � 7.x � 3, x � 5,x � 1,x

±525.±15, ±21, ±25, ±35, ±75, ±105, ±175,

±1, ±3, ±5, ±7,

f �x� � x 3 � 8x2 � 40x � 525.

Example 3

0 remainder, so is a zero.x � �1

0 remainder, so is a zero.x � 2

When the list of possible rationalzeros is small, as in Example 2,it may be quicker to test thezeros by evaluating the function.When the list of possible rationalzeros is large, as in Example 3, itmay be quicker to use a differentapproach to test the zeros, suchas using synthetic division orsketching a graph.

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Using the Rational Zero Test


SolutionThe leading coefficient is 2 and the constant term is 3.

Possible rational zeros:

By synthetic division, you can determine that is a rational zero.

So, factors as

and you can conclude that the rational zeros of are and


Recall from Section 2.2 that if is a zero of the polynomial function then is a solution of the polynomial equation

Solving a Polynomial Equation

Find all the real solutions of

SolutionThe leading coefficient is and the constant term is

Possible rational solutions:

With so many possibilities (32, in fact), it is worth your time to stop and sketch agraph. From Figure 2.32, it looks like three reasonable solutions would be

and Testing these by synthetic division shows that is the only rational solution. So, you have

Using the Quadratic Formula for the second factor, you find that the two addi-tional solutions are irrational numbers.

and


x ��5 � �265

20� 0.5639

x ��5 � �265

20� �1.0639

�x � 2��10x2 � 5x � 6� � 0.

x � 2x � 2.x �12,x � �

65,

Factors of �12

Factors of �10�

±1, ±2, ±3, ±4, ±6, ±12

±1, ±2, ±5, ±10

�12.�10

�10x3 � 15x2 � 16x � 12 � 0.

f �x� � 0.x � af,x � a

x � �3.x �12, x � 1,f

� �x � 1��2x � 1��x � 3�

f �x� � �x � 1��2x2 � 5x � 3�

f �x�

1 2

2

3

2

5

�8

5

�3

3

�3

0

x � 1

Factors of 3

Factors of 2�

±1, ±3

±1, ±2� ±1, ±3, ±

1

2, ±

3

2

f �x� � 2x3 � 3x2 � 8x � 3.

x

f x x x x( ) = 10 + 15 + 16 12− −3 2

1

15

5

−10

10

−5

y

FIGURE 2.32

Example 4

Example 5

Remember that when you try to find the rational zeros of apolynomial function with manypossible rational zeros, as inExample 4, you must use trialand error. There is no quickalgebraic method to determinewhich of the possibilities is anactual zero; however, sketchinga graph may be helpful.

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Conjugate PairsIn Example 1(c) and (d), note that the pairs of complex zeros are conjugates.That is, they are of the form and

Be sure you see that this result is true only if the polynomial function has realcoefficients. For instance, the result applies to the function given by

but not to the function given by

Finding a Polynomial with Given Zeros

Find a fourth-degree polynomial function with real coefficients that has and as zeros.

SolutionBecause is a zero and the polynomial is stated to have real coefficients, youknow that the conjugate must also be a zero. So, from the Linear FactorizationTheorem, can be written as

For simplicity, let to obtain


Factoring a PolynomialThe Linear Factorization Theorem shows that you can write any th-degreepolynomial as the product of linear factors.

However, this result includes the possibility that some of the values of arecomplex. The following theorem says that even if you do not want to get involvedwith “complex factors,” you can still write as the product of linear and/orquadratic factors. For a proof of this theorem, see Proofs in Mathematics onpage 214.

f �x�

ci

f �x� � an�x � c1��x � c2��x � c3� . . . �x � cn�

nn

� x4 � 2x3 � 10x2 � 18x � 9.

f �x� � �x2 � 2x � 1��x2 � 9�

a � 1

f �x� � a�x � 1��x � 1��x � 3i��x � 3i�.

f �x��3i

3i

3i�1,�1,

g�x� � x � i.x2 � 1f �x� �

a � bi.a � bi

Complex Zeros Occur in Conjugate PairsLet be a polynomial function that has real coefficients. If where

is a zero of the function, the conjugate is also a zero of thefunction.

a � bib � 0,a � bi,f �x�

Factors of a PolynomialEvery polynomial of degree with real coefficients can be written asthe product of linear and quadratic factors with real coefficients, where thequadratic factors have no real zeros.

n > 0

Example 6

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A quadratic factor with no real zeros is said to be prime or irreducible overthe reals. Be sure you see that this is not the same as being irreducible over therationals. For example, the quadratic is irreducible overthe reals (and therefore over the rationals). On the other hand, the quadratic

is irreducible over the rationals but reducibleover the reals.x2 � 2 � �x � �2 ��x � �2 �

x2 � 1 � �x � i��x � i�

In Example 7, if you were not told that is a zero of you could stillfind all zeros of the function by using synthetic division to find the real zeros and 3. Then you could factor the polynomial as Finally, by using the Quadratic Formula, you could determine that the zeros are

and x � 1 � 3i.x � 1 � 3i,x � 3,x � �2,

�x � 2��x � 3��x2 � 2x � 10�.�2

f,1 � 3i

You may want to remind studentsthat a graphing calculator is helpful indetermining real zeros, which in turnare useful in finding the complex zeros.

Finding the Zeros of a Polynomial Function

Find all the zeros of given that is azero of f.

1 � 3if �x� � x4 � 3x3 � 6x2 � 2x � 60

Example 7

Algebraic SolutionBecause complex zeros occur in conjugate pairs, you know that

is also a zero of This means that both

and

are factors of Multiplying these two factors produces

Using long division, you can divide into to obtainthe following.

So, you have

and you can conclude that the zeros of are and


x � �2.x � 3,x � 1 � 3i,x � 1 � 3i,f

� �x2 � 2x � 10��x � 3��x � 2�

f �x� � �x2 � 2x � 10��x2 � x � 6�

0

�6x2 � 12x � 60

�6x2 � 12x � 60

�x3 � 2x2 � 10x

�x3 � 4x2 � 2x

x4 � 2x3 � 10x2

x2 � 2x � 10 ) x4 � 3x3 � 6x2 � 2x � 60

x2 � x � 6

fx2 � 2x � 10

� x2 � 2x � 10.

� �x � 1�2 � 9i2

�x � �1 � 3i��x � �1 � 3i�� x � 1� � 3i��x � 1� � 3i�

f.

�x � �1 � 3i��x � �1 � 3i��

f.1 � 3i

Graphical SolutionBecause complex zeros always occur in conju-gate pairs, you know that is also a zero of

Because the polynomial is a fourth-degreepolynomial, you know that there are at most twoother zeros of the function. Use a graphing utilityto graph

as shown in Figure 2.33.

FIGURE 2.33

You can see that and 3 appear to be zeros ofthe graph of the function. Use the zero or rootfeature or the zoom and trace features of thegraphing utility to confirm that and

are zeros of the graph. So, you canconclude that the zeros of are

and x � �2.x � 3,x � 1 � 3i,x � 1 � 3i,f

x � 3x � �2

�2

−80

−4 5

80

y = x4 − 3x3 + 6x2 + 2x − 60

y � x4 � 3x3 � 6x2 � 2x � 60

f.1 � 3i

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−4 2 4x

y

5

10

(−2, 0) (1, 0)

f(x) = x5 + x3 + 2x2 −12x + 8

FIGURE 2.34

Example 8 shows how to find all the zeros of a polynomial function,including complex zeros.


Write as the product of linear factors, and listall of its zeros.

SolutionThe possible rational zeros are and Synthetic division producesthe following.

So, you have

You can factor as and by factoring as

you obtain

which gives the following five zeros of

and

From the graph of shown in Figure 2.34, you can see that the real zeros are theonly ones that appear as -intercepts. Note that is a repeated zero.


x � 1xf

x � �2ix � 2i,x � �2,x � 1,x � 1,

f.

f �x� � �x � 1��x � 1��x � 2��x � 2i��x � 2i�

� �x � 2i��x � 2i�

x2 � ��4� � �x � ��4 ��x � ��4 �x2 � 4�x � 1��x2 � 4�,x3 � x2 � 4x � 4

� �x � 1��x � 2��x3 � x2 � 4x � 4�.

f �x� � x5 � x3 � 2x2 � 12x � 8

�2 1

1

1

�2

�1

2

2

4

4

�8

�4

�8

8

0

1 1

1

0

1

1

1

1

2

2

2

4

�12

4

�8

8

�8

0

±8.±1, ±2, ±4,

f �x� � x5 � x3 � 2x2 � 12x � 8

1 is a zero.

is a zero.�2

In Example 8, the fifth-degreepolynomial function has threereal zeros. In such cases, youcan use the zoom and tracefeatures or the zero or rootfeature of a graphing utility toapproximate the real zeros. Youcan then use these real zeros todetermine the complex zerosalgebraically.

You can use the table feature of a graphing utility to helpyou determine which of the possible rational zeros arezeros of the polynomial in Example 8. The table should beset to ask mode. Then enter each of the possible rationalzeros in the table. When you do this, you will see thatthere are two rational zeros, and 1, as shown at theright.

�2

Techno logy

Example 8

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Other Tests for Zeros of PolynomialsYou know that an th-degree polynomial function can have at most real zeros.Of course, many th-degree polynomials do not have that many real zeros. Forinstance, has no real zeros, and has only one realzero. The following theorem, called Descartes’s Rule of Signs, sheds more lighton the number of real zeros of a polynomial.

A variation in sign means that two consecutive coefficients have oppositesigns.

When using Descartes’s Rule of Signs, a zero of multiplicity should becounted as zeros. For instance, the polynomial has two variationsin sign, and so has either two positive or no positive real zeros. Because

you can see that the two positive real zeros are of multiplicity 2.

Using Descartes’s Rule of Signs

Describe the possible real zeros of

SolutionThe original polynomial has three variations in sign.

to to

to

The polynomial

has no variations in sign. So, from Descartes’s Rule of Signs, the polynomialhas either three positive real zeros or one positive

real zero, and has no negative real zeros. From the graph in Figure 2.35, you cansee that the function has only one real zero (it is a positive number, near ).


x � 1

f �x� � 3x3 � 5x2 � 6x � 4

� �3x3 � 5x2 � 6x � 4

f ��x� � 3��x�3 � 5��x�2 � 6��x� � 4

��

f �x� � 3x3 � 5x2 � 6x � 4

��

f �x� � 3x3 � 5x2 � 6x � 4.

x � 1

�x � 1��x � 1��x � 2�x3 � 3x � 2 �

x3 � 3x � 2kk

f �x� � x3 � 1f �x� � x2 � 1n

nn

Descartes’s Rule of SignsLet be a polynomial withreal coefficients and

1. The number of positive real zeros of is either equal to the number ofvariations in sign of or less than that number by an even integer.

2. The number of negative real zeros of is either equal to the number ofvariations in sign of or less than that number by an even integer.f ��x�

f

f �x�f

a0 � 0.f (x) � anxn � an�1x

n�1 � . . . � a2x2 � a1x � a0

−3 −2 −1 2 3x

y

−3

−2

−1

1

2

3

f(x) = 3x3 − 5x2 + 6x − 4

FIGURE 2.35

Example 9

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Another test for zeros of a polynomial function is related to the sign patternin the last row of the synthetic division array. This test can give you an upper orlower bound of the real zeros of A real number is an upper bound for thereal zeros of if no zeros are greater than Similarly, is a lower bound if noreal zeros of are less than


Find the real zeros of

SolutionThe possible real zeros are as follows.

The original polynomial has three variations in sign. The polynomial

has no variations in sign. As a result of these two findings, you can applyDescartes’s Rule of Signs to conclude that there are three positive real zeros or onepositive real zero, and no negative zeros. Trying produces the following.

So, is not a zero, but because the last row has all positive entries, you knowthat is an upper bound for the real zeros. So, you can restrict the search tozeros between 0 and 1. By trial and error, you can determine that is a zero.So,

Because has no real zeros, it follows that is the only real zero.


x �236x2 � 3

f �x� � �x �2

3��6x2 � 3�.

x �23

x � 1x � 1

1 6

6

�4

6

2

3

2

5

�2

5

3

x � 1

� �6x3 � 4x2 � 3x � 2

f ��x� � 6��x�3 � 4��x�2 � 3��x� � 2

f �x�

� ±1, ±1

2, ±

1

3, ±

1

6, ±

2

3, ±2

Factors of 2

Factors of 6�

±1, ±2

±1, ±2, ±3, ±6

f �x� � 6x3 � 4x2 � 3x � 2.

b.fbb.f

bf.

In Example 10, notice how the RationalZero Test, Descartes’s Rule of Signs, andthe Upper and Lower Bound Rules maybe used together in a search for all realzeros of a polynomial function.

Upper and Lower Bound RulesLet be a polynomial with real coefficients and a positive leading coeffi-cient. Suppose is divided by using synthetic division.

1. If and each number in the last row is either positive or zero, is anupper bound for the real zeros of

2. If and the numbers in the last row are alternately positive and negative (zero entries count as positive or negative), is a lower boundfor the real zeros of f.

cc < 0

f.cc > 0

x � c,f�x�f �x�

Example 10

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Before concluding this section, here are two additional hints that can helpyou find the real zeros of a polynomial.

1. If the terms of have a common monomial factor, it should be factored outbefore applying the tests in this section. For instance, by writing

you can see that is a zero of and that the remaining zeros can beobtained by analyzing the cubic factor.

2. If you are able to find all but two zeros of you can always use theQuadratic Formula on the remaining quadratic factor. For instance, if yousucceeded in writing

you can apply the Quadratic Formula to to conclude that the tworemaining zeros are and

Using a Polynomial Model

You are designing candle-making kits. Each kit contains 25 cubic inches ofcandle wax and a mold for making a pyramid-shaped candle. You want the heightof the candle to be 2 inches less than the length of each side of the candle’ssquare base. What should the dimensions of your candle mold be?

SolutionThe volume of a pyramid is where is the area of the base and is theheight. The area of the base is and the height is So, the volume of thepyramid is Substituting 25 for the volume yields the following.

Substitute 25 for

Multiply each side by 3.

Write in general form.

The possible rational solutions are Use synthetic division to test some of the possible solutions. Note that in this case,it makes sense to test only positive -values. Using synthetic division, you candetermine that is a solution.

The other two solutions, which satisfy are imaginary and can be discarded. You can conclude that the base of the candle mold should be 5 inches by 5 inches and the height of the mold should be inches.


5 � 2 � 3

x2 � 3x � 15 � 0,

5 1

1

�253

01515

�75750

x � 5x

±75.±25,±15,±5,±3,x � ±1,

0 � x3 � 2x2 � 75

75 � x3 � 2x2

V. 25 �13

x2�x � 2�

V �13 x2�x � 2�.

�x � 2�.x2hBV �

13 Bh,

x � 2 � �5.x � 2 � �5x2 � 4x � 1

� x�x � 1��x2 � 4x � 1�

f �x� � x4 � 5x3 � 3x2 � x

f �x�,

fx � 0

� x�x3 � 5x2 � 3x � 1�

f �x� � x4 � 5x3 � 3x2 � x

f �x�

Activities

1. Write as a product of linear factors:

Answer:

2. Find a third-degree polynomial withinteger coefficients that has 2,and as zeros.

Answer:

3. Use the zero to find all the zerosof

Answer: �2, 3, 2i, �2i

f �x� � x4 � x3 � 2x2 � 4x � 24.x � 2i

x3 � 8x2 � 22x � 20

3 � i3 � i,

�x � 2��x � 2��x � 2i��x � 2i�f �x� � x 4 � 16.

Example 11

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Exercises 2.5

In Exercises 1–6, find all the zeros of the function.

1.

2.

3.

4.

5.

6.

In Exercises 7–10, use the Rational Zero Test to listall possible rational zeros of Verify that the zeros of shown on the graph are contained in the list.

7.

8.

9.

10.

In Exercises 11–20, find all the rational zeros of the function.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20. f �x� � 2x4 � 15x3 � 23x 2 � 15x � 25

f �x� � 9x4 � 9x3 � 58x 2 � 4x � 24

f �x� � 3x3 � 19x 2 � 33x � 9

C�x� � 2x3 � 3x 2 � 1

p�x� � x3 � 9x 2 � 27x � 27

h�t� � t 3 � 12t 2 � 21t � 10

h�x� � x3 � 9x 2 � 20x � 12

g�x� � x3 � 4x 2 � x � 4

f �x� � x3 � 7x � 6

f �x� � x3 � 6x 2 � 11x � 6

x−4 −2 4

2

yf �x� � 4x5 � 8x4 � 5x3 � 10x 2 � x � 2

x−8 −4 8

−20

−30

−40

yf �x� � 2x4 � 17x3 � 35x 2 � 9x � 45

x−6 6 12

18

y

f �x� � x3 � 4x 2 � 4x � 16

4

2

−4

−4 −2 2x

y

f �x� � x3 � 3x 2 � x � 3

ff.

h�t� � �t � 3��t � 2��t � 3i ��t � 3i �f �x� � �x � 6��x � i��x � i�f �x� � �x � 5��x � 8�2

g�x) � �x � 2��x � 4�3

f �x� � x2�x � 3��x2 � 1�f �x� � x�x � 6�2

VOCABULARY CHECK: Fill in the blanks.

1. The ________ ________ of ________ states that if is a polynomial of degree then has at least one zero in the complex number system.

2. The ________ ________ ________ states that if is a polynomial of degree then has precisely linear factors where are complex numbers.

3. The test that gives a list of the possible rational zeros of a polynomial function is called the ________ ________ Test.

4. If is a complex zero of a polynomial with real coefficients, then so is its ________,

5. A quadratic factor that cannot be factored further as a product of linear factors containing real numbers is said to be ________ over the ________.

6. The theorem that can be used to determine the possible numbers of positive real zeros and negative real zeros of a function is called ________ ________ of ________.

7. A real number is a(n) ________ bound for the real zeros of if no real zeros are less than and is a(n) ________ bound if no real zeros are greater than

PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com.

b.b,fb

a � bi.a � bi

c1, c2, . . . , cnf �x� � an�x � c1��x � c2� . . . �x � cn�nfn �n > 0�,f �x�

fn �n > 0�,f �x�

333202_0205.qxd 12/7/05 9:36 AM Page 179


In Exercises 21–24, find all real solutions of the polynomialequation.

21.

22.

23.

24.

In Exercises 25–28, (a) list the possible rational zeros of (b) sketch the graph of so that some of the possible zerosin part (a) can be disregarded, and then (c) determine allreal zeros of

25.

26.

27.

28.

In Exercises 29–32, (a) list the possible rational zeros of (b) use a graphing utility to graph so that some of thepossible zeros in part (a) can be disregarded, and then (c) determine all real zeros of

29.

30.

31.

32.

Graphical Analysis In Exercises 33–36, (a) use the zero or root feature of a graphing utility to approximate the zerosof the function accurate to three decimal places, (b) deter-mine one of the exact zeros (use synthetic division to verifyyour result), and (c) factor the polynomial completely.

33. 34.

35.

36.

In Exercises 37– 42, find a polynomial function with realcoefficients that has the given zeros. (There are manycorrect answers.)

37. 38.

39. 40.

41. 42.

In Exercises 43– 46, write the polynomial (a) as the productof factors that are irreducible over the rationals, (b) as theproduct of linear and quadratic factors that are irreducibleover the reals, and (c) in completely factored form.

43.

44.(Hint: One factor is )



In Exercises 47–54, use the given zero to find all the zeros ofthe function.

Function Zero

47.

48.

49.

50.

51.

52.

53.

54.

In Exercises 55–72, find all the zeros of the function andwrite the polynomial as a product of linear factors.

55. 56.

57. 58.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69.

70.

71. 72.

In Exercises 73–78, find all the zeros of the function. Whenthere is an extended list of possible rational zeros, use agraphing utility to graph the function in order to discardany rational zeros that are obviously not zeros of thefunction.

73.

74.

75.

76.

77.

78. g�x� � x 5 � 8x 4 � 28x 3 � 56x 2 � 64x � 32

f �x� � 2x4 � 5x 3 � 4x 2 � 5x � 2

f �x� � 9x 3 � 15x 2 � 11x � 5

f �x� � 16x 3 � 20x 2 � 4x � 15

f �s� � 2s3 � 5s2 � 12s � 5

f �x� � x3 � 24x 2 � 214x � 740

f �x� � x 4 � 29x 2 � 100f �x� � x 4 � 10x 2 � 9

h �x� � x 4 � 6x3 � 10x2 � 6x � 9

g�x� � x 4 � 4x3 � 8x2 � 16x � 16

g�x� � 3x3 � 4x2 � 8x � 8

f �x� � 5x 3 � 9x 2 � 28x � 6

h �x� � x3 � 9x2 � 27x � 35

h �x� � x3 � x � 6

f �x� � x 3 � 2x 2 � 11x � 52

g�x� � x3 � 6x2 � 13x � 10

h(x) � x 3 � 3x 2 � 4x � 2

f �z� � z 2 � 2z � 2

f �y� � y4 � 625

f �x� � x 4 � 81

g�x� � x 2 � 10x � 23h�x� � x 2 � 4x � 1

f �x� � x 2 � x � 56f �x� � x 2 � 25

�1 � 3if �x� � x 3 � 4x 2 � 14x � 20

�3 ��2 if �x� � x 4 � 3x 3 � 5x 2 � 21x � 22

1 � �3 ih �x� � 3x 3 � 4x 2 � 8x � 8

�3 � ig �x� � 4x 3 � 23x 2 � 34x � 10

5 � 2ig �x� � x 3 � 7x 2 � x � 87

2if �x� � 2x 4 � x 3 � 7x 2 � 4x � 4

3if �x� � x3 � x 2 � 9x � 9

5if �x� � 2x 3 � 3x 2 � 50x � 75

x 2 � 4.f �x� � x 4 � 3x 3 � x 2 � 12x � 20

x 2 � 2x � 2.f �x� � x 4 � 4x 3 � 5x 2 � 2x � 6

x 2 � 6.f �x� � x 4 � 2x 3 � 3x 2 � 12x � 18

f �x� � x 4 � 6x 2 � 27

�5, �5, 1 ��3 i23, �1, 3 � �2 i

2, 4 � i, 4 � i6, �5 � 2i, �5 � 2i

4, 3i, �3i1, 5i, �5i

g�x� � 6x4 � 11x3 � 51x 2 � 99x � 27

h�x� � x5 � 7x4 � 10x3 � 14x 2 � 24x

P�t� � t 4 � 7t 2 � 12f �x� � x 4 � 3x 2 � 2

f �x� � 4x3 � 7x 2 � 11x � 18

f �x� � 32x3 � 52x 2 � 17x � 3

f �x� � 4x4 � 17x 2 � 4

f �x� � �2x4 � 13x 3 � 21x 2 � 2x � 8

f.

ff,

f �x� � 4x3 � 12x 2 � x � 15

f �x� � �4x3 � 15x 2 � 8x � 3

f �x� � �3x3 � 20x 2 � 36x � 16

f �x� � x3 � x 2 � 4x � 4

f.

ff,

x5 � x4 � 3x3 � 5x 2 � 2x � 0

2y4 � 7y 3 � 26y 2 � 23y � 6 � 0

x 4 � 13x 2 � 12x � 0

z4 � z3 � 2z � 4 � 0

333202_0205.qxd 12/7/05 9:36 AM Page 180


In Exercises 79– 86, use Descartes’s Rule of Signs to deter-mine the possible numbers of positive and negative zerosof the function.

79. 80.

81. 82.

83.

84.

85.

86.

In Exercises 87– 90, use synthetic division to verify theupper and lower bounds of the real zeros of

87.

(a) Upper: (b) Lower:

88.


89.


90.


In Exercises 91–94, find all the real zeros of the function.

91.

92.

93.

94.

In Exercises 95–98, find all the rational zeros of the polyno-mial function.

95.

96.

97.

98.

In Exercises 99–102, match the cubic function with thenumbers of rational and irrational zeros.

(a) Rational zeros: 0; irrational zeros: 1(b) Rational zeros: 3; irrational zeros: 0(c) Rational zeros: 1; irrational zeros: 2(d) Rational zeros: 1; irrational zeros: 0

99. 100.

101. 102.

103. Geometry An open box is to be made from a rectangularpiece of material, 15 centimeters by 9 centimeters, by cutting equal squares from the corners and turning up the sides.

(a) Let represent the length of the sides of the squaresremoved. Draw a diagram showing the squaresremoved from the original piece of material and theresulting dimensions of the open box.

(b) Use the diagram to write the volume of the box asa function of Determine the domain of the function.

(c) Sketch the graph of the function and approximate thedimensions of the box that will yield a maximumvolume.

(d) Find values of such that Which of thesevalues is a physical impossibility in the constructionof the box? Explain.

104. Geometry A rectangular package to be sent by adelivery service (see figure) can have a maximumcombined length and girth (perimeter of a cross section)of 120 inches.

(a) Show that the volume of the package is

(b) Use a graphing utility to graph the function andapproximate the dimensions of the package that willyield a maximum volume.

(c) Find values of such that Which ofthese values is a physical impossibility in theconstruction of the package? Explain.

105. Advertising Cost A company that produces MP3players estimates that the profit (in dollars) for selling aparticular model is given by

where is the advertising expense (in tens of thousands of dollars). Using this model, find the smaller of twoadvertising amounts that will yield a profit of $2,500,000.

106. Advertising Cost A company that manufactures bicy-cles estimates that the profit (in dollars) for selling aparticular model is given by

where is the advertising expense (in tens of thousands of dollars). Using this model, find the smaller of twoadvertising amounts that will yield a profit of $800,000.

x

0 ≤ x ≤ 50P � �45x 3 � 2500x 2 � 275,000,

P

x

0 ≤ x ≤ 60P � �76x 3 � 4830x 2 � 320,000,

P

V � 13,500.x

V�x� � 4x 2�30 � x�.

y

x

x

V � 56.x

x.V

x

f �x� � x3 � 2xf �x� � x3 � x

f �x� � x3 � 2f �x� � x3 � 1

f �z� � z3 �116 z2 �

12z �

13 �

16�6z3�11z2 �3z � 2�

f �x� � x3 �14 x 2 � x �

14 �

14�4x3 � x 2 � 4x � 1�

f �x�� x3�32 x2 �

232 x � 6 �

12�2x3�3x 2 �23x �12�

P�x� � x 4 �254 x 2 � 9 �

14�4x4 � 25x 2 � 36�

g�x� � 3x3 � 2x 2 � 15x � 10

f �y� � 4y3 � 3y 2 � 8y � 6

f �z� � 12z3 � 4z 2 � 27z � 9

f �x� � 4x3 � 3x � 1

x � �4x � 3

f �x� � 2x4 � 8x � 3

x � �3x � 5

f �x� � x4 � 4x3 � 16x � 16

x � �3x � 4

f �x� � 2x3 � 3x 2 � 12x � 8

x � �1x � 4

f �x� � x4 � 4x3 � 15

f.

f �x� � 3x3 � 2x 2 � x � 3

f �x� � �5x3 � x 2 � x � 5

f �x� � 4x3 � 3x 2 � 2x � 1

g�x� � 2x3 � 3x 2 � 3

h�x� � 2x4 � 3x � 2h�x� � 3x4 � 2x 2 � 1

h�x� � 4x2 � 8x � 3g�x� � 5x5 � 10x

333202_0205.qxd 12/7/05 9:36 AM Page 181


107. Geometry A bulk food storage bin with dimensions 2 feet by 3 feet by 4 feet needs to be increased in size tohold five times as much food as the current bin. (Assumeeach dimension is increased by the same amount.)

(a) Write a function that represents the volume of thenew bin.

(b) Find the dimensions of the new bin.

108. Geometry A rancher wants to enlarge an existingrectangular corral such that the total area of the newcorral is 1.5 times that of the original corral. The currentcorral’s dimensions are 250 feet by 160 feet. The rancherwants to increase each dimension by the same amount.

(a) Write a function that represents the area of the newcorral.

(b) Find the dimensions of the new corral.

(c) A rancher wants to add a length to the sides of thecorral that are 160 feet, and twice the length to thesides that are 250 feet, such that the total area of thenew corral is 1.5 times that of the original corral.Repeat parts (a) and (b). Explain your results.

109. Cost The ordering and transportation cost (inthousands of dollars) for the components used in manu-facturing a product is given by

where is the order size (in hundreds). In calculus, it canbe shown that the cost is a minimum when

Use a calculator to approximate the optimal order size tothe nearest hundred units.

110. Height of a Baseball A baseball is thrown upward froma height of 6 feet with an initial velocity of 48 feet persecond, and its height (in feet) is

where is the time (in seconds). You are told the ballreaches a height of 64 feet. Is this possible?

111. Profit The demand equation for a certain product iswhere is the unit price (in dollars)

of the product and is the number of units produced andsold. The cost equation for the product is

where is the total cost (in dollars)and is the number of units produced. The total profitobtained by producing and selling units is

You are working in the marketing department of the com-pany that produces this product, and you are asked todetermine a price that will yield a profit of 9 milliondollars. Is this possible? Explain.

Synthesis

True or False? In Exercises 113 and 114, decide whetherthe statement is true or false. Justify your answer.

113. It is possible for a third-degree polynomial function withinteger coefficients to have no real zeros.

114. If is a zero of the function given by

then must also be a zero of

Think About It In Exercises 115–120, determine (if possi-ble) the zeros of the function if the function has zeros at

and

115. 116. g�x� � 3f �x�g�x� � �f �x�

x � r3.x � r1, x � r2,fg

f.x � i

f �x� � x 3 �ix2 � ix � 1

x � �i

p

P � R � C � xp � C.

xx

CC � 80x � 150,000,

xpp � 140 � 0.0001x,

t

h�t� � �16t 2 � 48t � 6, 0 ≤ t ≤ 3

h

3x3 � 40x 2 � 2400x � 36,000 � 0.

x

x ≥ 1C � 100�200

x 2�

x

x � 30�,

C

A

V

112. Athletics The attendance (in millions) at NCAAwomen’s college basketball games for the years 1997through 2003 is shown in the table, where representsthe year, with corresponding to 1997.(Source: National Collegiate Athletic Association)

(a) Use the regression feature of a graphing utility tofind a cubic model for the data.

(b) Use the graphing utility to create a scatter plot ofthe data. Then graph the model and the scatterplot in the same viewing window. How do theycompare?

(c) According to the model found in part (a), in whatyear did attendance reach 8.5 million?

(d) According to the model found in part (a), in whatyear did attendance reach 9 million?

(e) According to the right-hand behavior of themodel, will the attendance continue to increase?Explain.

t � 7t

A

Model It

Year, t Attendance, A

7 6.7

8 7.4

9 8.0

10 8.7

11 8.8

12 9.5

13 10.2

333202_0205.qxd 12/7/05 9:36 AM Page 182


117. 118.

119. 120.

121. Exploration Use a graphing utility to graph the func-tion given by for different values of

Find values of such that the zeros of satisfy thespecified characteristics. (Some parts do not have uniqueanswers.)

(a) Four real zeros

(b) Two real zeros, each of multiplicity 2

(c) Two real zeros and two complex zeros

(d) Four complex zeros

122. Think About It Will the answers to Exercise 121change for the function

(a) (b)

123. Think About It A third-degree polynomial function has real zeros and 3, and its leading coefficient isnegative. Write an equation for Sketch the graph of How many different polynomial functions are possible for

124. Think About It Sketch the graph of a fifth-degreepolynomial function whose leading coefficient is positiveand that has one zero at of multiplicity 2.

125. Writing Compile a list of all the various techniques forfactoring a polynomial that have been covered so far inthe text. Give an example illustrating each technique, andwrite a paragraph discussing when the use of eachtechnique is appropriate.

126. Use the information in the table to answer each question.

(a) What are the three real zeros of the polynomial func-tion

(b) What can be said about the behavior of the graph of at

(c) What is the least possible degree of Explain. Canthe degree of ever be odd? Explain.

(d) Is the leading coefficient of positive or negative?Explain.

(e) Write an equation for (There are many correctanswers.)

(f) Sketch a graph of the equation you wrote inpart (e).

127. (a) Find a quadratic function (with integer coefficients)that has as zeros. Assume that is a positiveinteger.

(b) Find a quadratic function (with integer coefficients)that has as zeros. Assume that is a positiveinteger.

128. Graphical Reasoning The graph of one of the following functions is shown below. Identify the functionshown in the graph. Explain why each of the others is notthe correct function. Use a graphing utility to verify yourresult.

(a)

(b)

(c)

(d)

Skills Review

In Exercises 129–132, perform the operation and simplify.

129.

130.

131.

132.

In Exercises 133–138, use the graph of f to sketch the graphof g. To print an enlarged copy of the graph, go to thewebsite www.mathgraphs.com.

133.

134.

135.

136.

137.

138. g�x� � f � 12x�g�x� � f �2x�g�x� � f ��x�g�x� � 2 f �x�g�x� � f �x� � 2

x(−2, 0)

(4, 4)

(2, 2)

(0, 2)

21 43

4

5

f

yg�x� � f �x � 2�

�9 � 5i��9 � 5i��6 � 2i��1 � 7i��12 � 5i� � 16i

��3 � 6i� � �8 � 3i�

2 4

–40

–30

–20

10

x

y

k �x� � �x � 1)�x � 2��x � 3.5�h �x� � �x � 2)�x � 3.5��x 2 � 1�g �x� � �x � 2)�x � 3.5�f �x� � x 2�x � 2)�x � 3.5�

ba ± bif

b±�bif

f.

f

ff ?

x � 1?f

f ?

x � 3

f ?

f.f.

12,�2,

f

g�x� � f �2x�g�x� � f �x � 2�g?

fkk.f �x� � x 4 � 4x 2 � k

g�x� � f ��x�g�x� � 3 � f �x�g�x� � f �2x�g�x� � f �x � 5�

Interval Value of

Positive

Negative

Negative

Positive�4, ��

�1, 4�

��2, 1�

��, �2�

f �x�

333202_0205.qxd 12/7/05 9:36 AM Page 183

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