2-2 Polynomial Functions
For each function, (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of
any repeated zeros, (c) find a few additional points, and then (d) graph the function.
37. f (x) = −x(x – 3)(x + 2)3
SOLUTION: a. The degree is 5, and the leading coefficient is −1. Because the degree is odd and the leading coefficient is negative,
b. The zeros are 0, 3, and −2. The zero −2 has multiplicity 3 since (x + 2) is a factor of the polynomial three times.
c. Sample answer: Evaluate the function for a few x-values in its domain.
Choose x-values that fall in the intervals determined by the zeros of the function. Interval
(–∞, –2)
(–2, 0)
(0, 3)
(3, ∞)
x
−3
−1
1
4
f(x)
18
−4
54
−864
d. Evaluate the function for several x-values in its domain.
x
−4
−2
0
2
3
f(x)
224
0
0
128
0
Use these points to construct a graph.
39. f (x) = 3x3 – 3x2 – 36x
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SOLUTION: Page 1
a. The degree is 3, and the leading coefficient is 3. Because the degree is odd and the leading coefficient is positive,
2-2 Polynomial Functions
39. f (x) = 3x3 – 3x2 – 36x
SOLUTION: a. The degree is 3, and the leading coefficient is 3. Because the degree is odd and the leading coefficient is positive,
b.
The zeros are 0, 4, and −3.
c. Sample answer: Evaluate the function for a few x-values in its domain.
Choose x-values that fall in the intervals determined by the zeros of the function. Interval
(–∞, –3)
(–3, 0)
(0, 4)
(4, ∞)
x
−4
−2
2
5
f(x)
−96
36
−60
120
d. Evaluate the function for several x-values in its domain.
x
−3
−1
0
1
3
4
f(x)
0
30
0
−36
−54
0
Use these points to construct a graph.
41. f (x) = x4 + x3 – 20x2
SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,
b.
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2-2 Polynomial Functions
41. f (x) = x4 + x3 – 20x2
SOLUTION: a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive,
b.
The zeros are 0, 4, and −5. The zero 0 has multiplicity 2 since x is a factor of the polynomial twice.
c. Sample answer: Evaluate the function for a few x-values in its domain.
Choose x-values that fall in the intervals determined by the zeros of the function. Interval
(–∞, –5)
(–5, 0)
(0, ∞)
(0, ∞)
x
−6
−2
2
5
f(x)
360
−72
−56
250
d.
Evaluate the function for several x-values in its domain.
x
−5
−4
−3
−1
0
1
3
4
f(x)
0
−128
−126
−20
0
−18
−72
0
Use these points to construct a graph.
Find a polynomial function of degree n with only the following real zeros. More than one answer is
possible.
55. 3; n = 3
SOLUTION: Sample answer: If 3 is a zero of the polynomial function, then (x − 3) is a factor. For the function to have a degree
of 3, raise (x − 3) to the third power.
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2-2 Polynomial Functions
Find a polynomial function of degree n with only the following real zeros. More than one answer is
possible.
55. 3; n = 3
SOLUTION: Sample answer: If 3 is a zero of the polynomial function, then (x − 3) is a factor. For the function to have a degree
of 3, raise (x − 3) to the third power.
57. –5, 4; n = 4
SOLUTION: Sample answer: If −5 and 4 are zeros of the polynomial function, then (x + 5) and (x − 4) are factors. For the
function to have a degree of 4, raise (x + 5) to the third power.
59. 0, –4; n = 5
SOLUTION: Sample answer: If 0 and −4 are zeros of the polynomial function, then x and (x + 4) are factors. For the function to
have a degree of 5, raise x to the fourth power.
61. 0, 3, –2; n = 5
SOLUTION: Sample answer: If 0, 3, and −2 are zeros of the polynomial function, then x, (x −3), and (x + 2) are factors. For the
function to have a degree of 5, raise x to the third power.
63. no real zeros; n = 6
SOLUTION: 2
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2
Sample answer: If there are no real zeros of the polynomial function, then (x + 1) can be a factor because x + Page
1= 4
2
0 has no real solutions. For the function to have a degree of 6, raise (x + 1) to the third power.
2-2 Polynomial Functions
63. no real zeros; n = 6
SOLUTION: 2
2
Sample answer: If there are no real zeros of the polynomial function, then (x + 1) can be a factor because x + 1 =
2
0 has no real solutions. For the function to have a degree of 6, raise (x + 1) to the third power.
Determine a polynomial function that has each set of zeros. More than one answer is possible.
71. 3, 0, 4, –1, 3
SOLUTION: 2
Sample answer: If 3, 0, 4, −1, and 3 are zeros of the polynomial function, then (x − 3) , x, (x − 4), and (x + 1) are
factors.
State the number of possible real zeros and turning points of each function. Then find all of the real zeros
by factoring.
85. f (x) = 16x4 + 72x2 + 80
SOLUTION: The degree of the function is 4, so f has at most 4 distinct real zeros and at most 3 turning points. To find the real
zeros, solve the related equation f (x) = 0 by factoring.
2
Let u = x .
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2-2 Polynomial Functions
State the number of possible real zeros and turning points of each function. Then find all of the real zeros
by factoring.
85. f (x) = 16x4 + 72x2 + 80
SOLUTION: The degree of the function is 4, so f has at most 4 distinct real zeros and at most 3 turning points. To find the real
zeros, solve the related equation f (x) = 0 by factoring.
2
Let u = x .
Because ±
and ±
are not real zeros, f has no real zeros.
87. f (x) = −24x4 + 24x3 – 6x2
SOLUTION: The degree of the function is 4, so f has at most 4 distinct real zeros and at most 3 turning points. To find the real
zeros, solve the related equation f (x) = 0 by factoring.
The expression above has 4 factors, but solving for x yields only 2 distinct real zeros, x = 0 and x =
. Both of the
zeros have multiplicity 2.
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