SULLMC05_013157759X.QXD 11/21/06 12:45 PM Page 378
378
CHAPTER 5
Polynomial and Rational Functions
The next theorem concerns the number of real zeros that a polynomial function
may have. In counting the zeros of a polynomial, we count each zero as many times
as its multiplicity.
THEOREM
Number of Real Zeros
A polynomial function cannot have more real zeros than its degree.
Proof The proof is based on the Factor Theorem. If r is a real zero of a polynomial
function f, then f1r2 = 0 and, hence, x - r is a factor of f1x2. Each real zero corresponds to a factor of degree 1. Because f cannot have more first-degree factors
than its degree, the result follows.
2 Use Descartes’ Rule of Signs to Determine the Number
of Positive and the Number of Negative Real Zeros
of a Polynomial Function
Descartes’ Rule of Signs provides information about the number and location of
the real zeros of a polynomial function written in standard form (descending powers
of x). It requires that we count the number of variations in the sign of the coefficients of f1x2 and f1-x2.
For example, the following polynomial function has two variations in the signs
of the coefficients.
f(x) 3x7 4x4 3x2 2x 1
3x7 0x6 0x5 4x4 0x3 3x2 2x 1
to to Notice that we ignored the zero coefficients in 0x6, 0x5, and 0x3 in counting the
number of variations in the sign of f1x2. Replacing x by -x, we get
f (x ) 3(x )7 4(x )4 3(x )2 2(x ) 1
3x7 4x4 3x2 2x 1
to which has one variation in sign.
THEOREM
Descartes’ Rule of Signs
Let f denote a polynomial function written in standard form.
The number of positive real zeros of f either equals the number of variations
in the sign of the nonzero coefficients of f1x2 or else equals that number less
an even integer.
The number of negative real zeros of f either equals the number of variations
in the sign of the nonzero coefficients of f1-x2 or else equals that number less
an even integer.
We shall not prove Descartes’ Rule of Signs. Let’s see how it is used.
EXAMPLE 3
Using the Number of Real Zeros Theorem
and Descartes’ Rule of Signs
Discuss the real zeros of f1x2 = 3x6 - 4x4 + 3x3 + 2x2 - x - 3.
Solution
Because the polynomial is of degree 6, by the Number of Real Zeros Theorem there
are at most six real zeros. Since there are three variations in the sign of the nonzero
SULLMC05_013157759X.QXD 11/21/06 12:45 PM Page 379
SECTION 5.5
The Real Zeros of a Polynomial Function
379
coefficients of f1x2, by Descartes’ Rule of Signs we expect either three or one positive real zeros. To continue, we look at f1-x2.
f1-x2 = 3x6 - 4x4 - 3x3 + 2x2 + x - 3
There are three variations in sign, so we expect either three or one negative real
zeros. Equivalently, we now know that the graph of f has either three or one positive x-intercepts and three or one negative x-intercepts.
Now Work
PROBLEM
21
3 Use the Rational Zeros Theorem to List the Potential Rational
Zeros of a Polynomial Function
The next result, called the Rational Zeros Theorem, provides information about the
rational zeros of a polynomial with integer coefficients.
THEOREM
Rational Zeros Theorem
Let f be a polynomial function of degree 1 or higher of the form
f1x2 = anxn + an - 1xn - 1 + Á + a1x + a0
an Z 0, a0 Z 0
p
, in lowest terms, is a rational zero of
q
f, then p must be a factor of a0 , and q must be a factor of an .
where each coefficient is an integer. If
EXAMPLE 4
Listing Potential Rational Zeros
List the potential rational zeros of
f1x2 = 2x3 + 11x2 - 7x - 6
Solution
Because f has integer coefficients, we may use the Rational Zeros Theorem. First,
we list all the integers p that are factors of the constant term a0 = -6 and all the
integers q that are factors of the leading coefficient a3 = 2.
p:
q:
;1, ;2, ;3, ;6
;1, ;2
Now we form all possible ratios
In Words
For the polynomial function
f(x) = 2x3 + 11x2 - 7x - 6, we
know 5 is not a zero, because 5
is not in the list of potential
rational zeros. However, -1 may
or may not be a zero.
p
:
q
Factors of -6
Factors of 2
p
.
q
1
3
;1, ;2, ;3, ;6, ; , ;
2
2
If f has a rational zero, it will be found in this list, which contains 12 possibilities.
Now Work
PROBLEM
33
Be sure that you understand what the Rational Zeros Theorem says: For a polynomial with integer coefficients, if there is a rational zero, it is one of those listed. It
may be the case that the function does not have any rational zeros.
Long division, synthetic division, or substitution can be used to test each potential rational zero to determine whether it is indeed a zero. To make the work easier,
integers are usually tested first. Let’s continue this example.
SULLMC05_013157759X.QXD 11/21/06 12:45 PM Page 387
SECTION 5.5
The Real Zeros of a Polynomial Function
13. f1x2 = 3x4 - 6x3 - 5x + 10; x - 2
14. f1x2 = 4x4 - 15x2 - 4; x - 2
15. f1x2 = 3x6 + 82x3 + 27; x + 3
16. f1x2 = 2x6 - 18x4 + x2 - 9; x + 3
17. f1x2 = 4x6 - 64x4 + x2 - 15; x + 4
18. f1x2 = x6 - 16x4 + x2 - 16; x + 4
19. f1x2 = 2x4 - x3 + 2x - 1; x -
1
2
20. f1x2 = 3x4 + x3 - 3x + 1; x +
387
1
3
In Problems 21–32, tell the maximum number of real zeros that each polynomial function may have. Then use Descartes’ Rule of Signs
to determine how many positive and how many negative zeros each polynomial function may have. Do not attempt to find the zeros.
21. f1x2 = -4x7 + x3 - x2 + 2
22. f1x2 = 5x4 + 2x2 - 6x - 5
23. f1x2 = 2x6 - 3x2 - x + 1
24. f1x2 = -3x5 + 4x4 + 2
25. f1x2 = 3x3 - 2x2 + x + 2
26. f1x2 = -x3 - x2 + x + 1
27. f1x2 = -x4 + x2 - 1
28. f1x2 = x4 + 5x3 - 2
29. f1x2 = x5 + x4 + x2 + x + 1
30. f1x2 = x5 - x4 + x3 - x2 + x - 1
31. f1x2 = x6 - 1
32. f1x2 = x6 + 1
In Problems 33–44, list the potential rational zeros of each polynomial function. Do not attempt to find the zeros.
33. f1x2 = 3x4 - 3x3 + x2 - x + 1
34. f1x2 = x5 - x4 + 2x2 + 3
35. f1x2 = x5 - 6x2 + 9x - 3
36. f1x2 = 2x5 - x4 - x2 + 1
37. f1x2 = -4x3 - x2 + x + 2
38. f1x2 = 6x4 - x2 + 2
39. f1x2 = 6x4 - x2 + 9
40. f1x2 = -4x3 + x2 + x + 6
41. f1x2 = 2x5 - x3 + 2x2 + 12
42. f1x2 = 3x5 - x2 + 2x + 18
43. f1x2 = 6x4 + 2x3 - x2 + 20
44. f1x2 = -6x3 - x2 + x + 10
In Problems 45–56, use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor f over
the real numbers.
45. f1x2 = x3 + 2x2 - 5x - 6
46. f1x2 = x3 + 8x2 + 11x - 20
47. f1x2 = 2x3 - x2 + 2x - 1
48. f1x2 = 2x3 + x2 + 2x + 1
49. f1x2 = 2x3 - 4x2 - 10x + 20
50. f1x2 = 3x3 + 6x2 - 15x - 30
51. f1x2 = 2x4 + x3 - 7x2 - 3x + 3
52. f1x2 = 2x4 - x3 - 5x2 + 2x + 2
53. f1x2 = x4 + x3 - 3x2 - x + 2
54. f1x2 = x4 - x3 - 6x2 + 4x + 8
55. f1x2 = 4x4 + 5x3 + 9x2 + 10x + 2
56. f1x2 = 3x4 + 4x3 + 7x2 + 8x + 2
In Problems 57–68, solve each equation in the real number system.
57. x4 - x3 + 2x2 - 4x - 8 = 0
58. 2x3 + 3x2 + 2x + 3 = 0
59. 3x3 + 4x2 - 7x + 2 = 0
60. 2x3 - 3x2 - 3x - 5 = 0
61. 3x3 - x2 - 15x + 5 = 0
62. 2x3 - 11x2 + 10x + 8 = 0
63. x4 + 4x3 + 2x2 - x + 6 = 0
64. x4 - 2x3 + 10x2 - 18x + 9 = 0
65. x3 -
2 2
8
x + x + 1 = 0
3
3
67. 2x4 - 19x3 + 57x2 - 64x + 20 = 0
66. x3 +
3 2
x + 3x - 2 = 0
2
68. 2x4 + x3 - 24x2 + 20x + 16 = 0