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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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
can obtain more information on
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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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.
ᮣ
DEVELOPING YOUR SKILLS
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]