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D. Descartes’ Rule of Signs and Upper/Lower Bounds

# D. Descartes’ Rule of Signs and Upper/Lower Bounds

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Section 4.2 The Zeroes of Polynomial Functions

Descartes’ Rule of Signs

Given the real polynomial equation P1x2 ϭ 0,

1. The number of positive real zeroes is equal to the number of variations in

sign for P(x), or an even number less.

2. The number of negative real zeroes is equal to the number of variations in

sign for P1Ϫx2 , or an even number less.

EXAMPLE 10

Finding the Zeroes of a Polynomial

For P1x2 ϭ 2x5 Ϫ 5x4 ϩ x3 ϩ x2 Ϫ x ϩ 6,

a. Use the rational zeroes theorem to list all possible rational zeroes.

b. Apply Descartes’ rule to count the number of possible positive, negative, and

complex zeroes.

c. Use this information and the tools of this section to find all zeroes of P.

Solution

a. The factors of 2 are 5Ϯ1, Ϯ26 and the factors of 6 are 5Ϯ1, Ϯ6, Ϯ2, Ϯ36 . The

possible rational zeroes for P are 5Ϯ1, Ϯ6, Ϯ2, Ϯ3, Ϯ12, Ϯ32 6 .

b. For Descartes’ rule, we organize our work in a table. Since P has degree 5,

there must be a total of five zeroes. For this illustration, positive terms are in

blue and negative terms in red: P1x2 ϭ 2x5 Ϫ 5x4 ϩ x3 ϩ x2 Ϫ x ϩ 6. The

terms change sign a total of four times, meaning there are four, two, or zero

positive roots. For the negative roots, recall that P1Ϫx2 will change the sign of

all odd-degree terms, giving P1Ϫx2 ϭ Ϫ2x5 Ϫ 5x4 Ϫ x3 ϩ x2 ϩ x ϩ 6. This

time there is only one sign change (from negative to positive) showing there

will be exactly one negative root, a fact that is highlighted in the following

table. Since there must be 5 zeroes, the number of possible complex zeroes is:

none, two, or four, as shown.

possible

positive zeroes

known

negative zeroes

possibilities for

complex roots

total number

must be 5

4

1

0

5

2

1

2

5

0

1

4

5

c. Testing 1 and Ϫ1 shows x ϭ 1 is not a root, but x ϭ Ϫ1 is, and using Ϫ1 in

synthetic division gives:

use Ϫ1 as a “divisor”

Ϫ1

2

WORTHY OF NOTE

As you recall from our study of

quadratics, it’s entirely possible for

a polynomial function to have no

real zeroes. Also, if the zeroes are

irrational, complex, or a combination

of these, they cannot be found

using the rational zeroes theorem.

For a look at ways to determine

these zeroes, see the Reinforcing

Basic Skills feature that follows

Section 4.3.

Ϫ5

Ϫ2

Ϫ7

2

1

7

8

1

Ϫ8

Ϫ7

Ϫ1

7

6

6

Ϫ6

0

coefficients of P (x )

q1(x ) is not easily factored

Since there is only one negative root, we need only check the remaining

positive zeroes. The quotient q1(x) is not easily factored, so we continue with

synthetic division using the next larger positive root, x ϭ 2.

use 2 as a “divisor”

2

2

2

Ϫ7

4

Ϫ3

8

Ϫ6

2

Ϫ7

4

Ϫ3

6

Ϫ6

0

coefficients of q1(x )

q2(x )

The partially factored form is P1x2 ϭ 1x ϩ 121x Ϫ 2212x3 Ϫ 3x2 ϩ 2x Ϫ 32 .

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CHAPTER 4 Polynomial and Rational Functions

10

Graphing P(x) at this point (see the figure)

verifies that Ϫ1 and 2 are zeroes, but also

indicates there is an additional zero between

1 and 2. If it’s a rational zero, it

4

Ϫ3

3

must be x ϭ from our list of possible zeroes.

2

3

Checking using synthetic division and

2

Ϫ5

q2(x) we have

3

2 Ϫ3 2 Ϫ3

2

T

3 0

3

2

0 2

0

3

3

Since the remainder is zero, P a b ϭ 0 (remainder theorem) and ax Ϫ b is a

2

2

factor (factor theorem). Since the zero is a fraction, we’ll use the ideas discussed in

Section 4.1 to help write P(x) in completely factored form:

3

P1x2 ϭ 1x ϩ 12 1x Ϫ 22ax Ϫ b12x2 ϩ 22

2

3

ϭ 1x ϩ 12 1x Ϫ 22ax Ϫ b122 1x2 ϩ 12

2

ϭ 1x ϩ 12 1x Ϫ 22 12x Ϫ 32 1x2 ϩ 12

ϭ 1x ϩ 12 1x Ϫ 22 12x Ϫ 32 1x ϩ i2 1x Ϫ i2

partially factored form

factor out 2

3

multiply 2 ax Ϫ b

2

completely factored form

The zeroes of P are Ϫ1, 2, 32, Ϫi and i, with two positive, one negative, and two

complex zeroes (row two of the table).

Now try Exercises 83 through 96 ᮣ

One final idea that helps reduce the number of possible zeroes is the upper and

lower bounds property. A number b is an upper bound on the positive zeroes of a

function if no positive zero is greater than b. In the same way, a number a is a lower

bound on the negative zeroes if no negative zero is less than a.

Upper and Lower Bounds Property

Given P(x) is a polynomial with real coefficients.

1. If P(x) is divided by x Ϫ b 1b 7 02 using synthetic division and all

coefficients in the quotient row are either positive or zero, then b is

an upper bound on the zeroes of P.

2. If P(x) is divided by x Ϫ a 1a 6 02 using synthetic division and all

coefficients in the quotient row alternate in sign, then a is a lower

bound on the zeroes of P.

For both 1 and 2, zero coefficients can be either positive or negative as needed.

D. You just seen how we

the zeroes of real polynomials

using Descartes’ rule of signs

and upper/lower bounds

theorem

While this test certainly helps narrow the possibilities, we gain the additional

benefit of knowing the property actually places boundaries on all real zeroes of the

polynomial, both rational and irrational. In Part (c) of Example 10, the quotient row

of the first division alternates in sign, showing x ϭ Ϫ1 is both a zero and a lower

bound on the real zeroes of P. For more on the upper and lower bounds property, see

Exercise 111.

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Section 4.2 The Zeroes of Polynomial Functions

E. Applications of Polynomial Functions

Polynomial functions can be very accurate models of real-world phenomena, though

we often must restrict their domain, as illustrated in Example 11.

EXAMPLE 11

Using the Remainder Theorem to Solve an Oceanography Application

As part of an environmental study, scientists use radar to map the ocean floor from

the coastline to a distance 12 mi from shore. In this study, ocean trenches appear as

negative values and underwater mountains as positive values, as measured from the

surrounding ocean floor. The terrain due west of a particular island can be modeled

by h1x2 ϭ x4 Ϫ 25x3 ϩ 200x2 Ϫ 560x ϩ 384, where h(x) represents the height in

feet, x mi from shore 10 6 x Յ 122.

a. Use the remainder theorem to find the “height of the ocean floor” 10 mi out.

b. Use the tools developed in this section to find the number of times the ocean

floor has height h1x2 ϭ 0 in this interval, given this occurs 12 mi out.

Solution

a. For part (a) we simply evaluate h(10) using the remainder theorem.

use 10 as a “divisor”

10

1

1

Ϫ25

10

Ϫ15

200

Ϫ150

50

Ϫ560

500

Ϫ60

384

Ϫ600

Ϫ216

coefficients of h(x)

remainder is Ϫ216

Ten miles from shore, there is an ocean trench 216 ft deep.

b. For part (b), we're given 12 is zero, so we again use the remainder theorem and

work with the quotient polynomial.

use 12 as a “divisor”

12

1

Ϫ25

12

Ϫ13

3

2

1

Ϫ560

528

Ϫ32

200

Ϫ156

44

384

Ϫ384

0

coefficients of h(x)

q1(x)

The quotient is q1 1x2 ϭ x Ϫ 13x ϩ 44x Ϫ 32. Since a ϭ 1, we know the

remaining rational zeroes must be factors of Ϫ32: 5Ϯ1, Ϯ32, Ϯ2, Ϯ16, Ϯ4, Ϯ86.

Using x ϭ 1 gives

use 1 as a “divisor”

1

1

1

Ϫ13

1

Ϫ12

44

Ϫ12

32

Ϫ32

32

0

coefficients of q1(x)

q2(x)

The function can now be written as h1x2 ϭ 1x Ϫ 122 1x Ϫ 121x2 Ϫ 12x ϩ 322

and in completely factored form h1x2 ϭ 1x Ϫ 1221x Ϫ 121x Ϫ 421x Ϫ 82 . The

ocean floor has height zero at distances of

450

1, 4, 8, and 12 mi from shore.

The graph of h(x) is shown in the figure. The

graph shows a great deal of variation in the

ocean floor, but the zeroes occurring at 1, 4, 8,

and 12 mi out are clearly evident.

E. You’ve just seen how

we can solve an application

of polynomial functions

0

13

Ϫ450

Now try Exercises 99 through 110 ᮣ

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CHAPTER 4 Polynomial and Rational Functions

4.2 EXERCISES

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A complex polynomial is one where one or more

are complex numbers.

2. A polynomial function of degree n will have

exactly

zeroes, real or

, where

zeroes of multiplicity m are counted m times.

3. If a ϩ bi is a complex zero of polynomial P with

real coefficients, then

is also a zero.

4. According to Descartes’ rule of signs, there are as

many

real roots as changes in sign from

term to term, or an

number less.

5. Which of the following values is not a possible root

of f 1x2 ϭ 6x3 Ϫ 2x2 ϩ 5x Ϫ 12:

a. x ϭ 43

b. x ϭ 34

c. x ϭ 12

6. Discuss/Explain each of the following:

(a) irreducible quadratic factors, (b) factors that are

complex conjugates, (c) zeroes of multiplicity m,

and (d) upper bounds on the zeroes of a

polynomial.

Discuss/Explain why.

Rewrite each polynomial as a product of linear factors,

and find the zeroes of the polynomial.

7. P1x2 ϭ x4 ϩ 5x2 Ϫ 36

8. Q1x2 ϭ x4 ϩ 21x2 Ϫ 100

9. Q1x2 ϭ x4 Ϫ 16

19. degree 3, x ϭ 3, x ϭ 2i

20. degree 3, x ϭ Ϫ5, x ϭ Ϫ3i

10. P1x2 ϭ x4 Ϫ 81

21. degree 4, x ϭ Ϫ1, x ϭ 2, x ϭ i

11. P1x2 ϭ x3 ϩ x2 Ϫ x Ϫ 1

12. Q1x2 ϭ x3 Ϫ 3x2 Ϫ 9x ϩ 27

13. Q1x2 ϭ x3 Ϫ 5x2 Ϫ 25x ϩ 125

14. P1x2 ϭ x3 ϩ 4x2 Ϫ 16x Ϫ 64

Factor each polynomial completely. Write any repeated

factors in exponential form, then name all zeroes and

their multiplicity.

15. p1x2 ϭ 1x2 Ϫ 10x ϩ 252 1x2 ϩ 4x Ϫ 452 1x ϩ 92

16. q1x2 ϭ 1x2 ϩ 12x ϩ 362 1x2 ϩ 2x Ϫ 242 1x Ϫ 42

17. P1x2 ϭ 1x Ϫ 5x Ϫ 1421x Ϫ 492 1x ϩ 22

2

Find a polynomial P(x) having real coefficients,

with the degree and zeroes indicated. All real

zeroes are given. Assume the lead coefficient is 1.

Recall 1a ؉ bi21a ؊ bi2 ‫ ؍‬a2 ؉ b2.

2

18. Q1x2 ϭ 1x2 Ϫ 9x ϩ 182 1x2 Ϫ 3621x Ϫ 32

22. degree 4, x ϭ Ϫ1, x ϭ 3, x ϭ Ϫ2i

23. degree 4, x ϭ 3, x ϭ 2i

24. degree 4, x ϭ Ϫ2, x ϭ Ϫ3i

25. degree 4, x ϭ Ϫ1, x ϭ 1 ϩ 2i

26. degree 4, x ϭ Ϫ1, x ϭ 1 Ϫ 3i

27. degree 4, x ϭ Ϫ3, x ϭ 1 ϩ i12

28. degree 4, x ϭ Ϫ2, x ϭ 1 ϩ i 13

Use the intermediate value theorem to verify the given

polynomial has at least one zero “ci” in the intervals

specified. Do not find the zeroes.

29. f 1x2 ϭ x3 ϩ 2x2 Ϫ 8x Ϫ 5

a. 3 Ϫ4, Ϫ34

b. [2, 3]

30. g1x2 ϭ x4 Ϫ 2x2 ϩ 6x Ϫ 3

a. 3Ϫ3, Ϫ24

b. [0, 1]

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D. Descartes’ Rule of Signs and Upper/Lower Bounds

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