2-4 Zeros of Polynomial Functions
Write a polynomial function of least degree with
real coefficients in standard form that has the
given zeros.
34. –5, 3, 4 + i
SOLUTION: Because 4 + i is a zero and the polynomial is to have
real coefficients, you know that 4 − i must also be a
zero. Using the Linear Factorization Theorem and the
zeros –5, 3, 4 + i, and 4 − i, write f (x) as follows.
f(x) = a[x − (−5)][x − (3)][x − (4 + i)][x − (4 − i)]
Therefore, a function of least degree that has –1, 8,
4
3
2
6 − i, and 6 + i as zeros is f (x) = x – 19x + 113x –
163x − 296 or any nonzero multiple of f (x).
Write each function as (a) the product of linear
and irreducible quadratic factors and (b) the
product of linear factors. Then (c) list all of its
zeros.
43. g(x) = x4 – 3x3 – 12x2 + 8
SOLUTION: a. g(x) has possible rational zeros of ±1, ±2, ±4, ±8.
By using synthetic division, it can be determined that
−2 is a rational zero.
Let a = 1. Then write the function in standard form.
By using synthetic division on the depressed
polynomial, it can be determined that −1 is a rational
zero.
Therefore, a function of least degree that has –5, 3,
4
3
2
4 + i, and 4 − i as zeros is f (x) = x – 6x – 14x +
154x − 255 or any nonzero multiple of f (x).
2
The remaining quadratic factor (x − 6x + 4) yields
no rational zeros. Use the quadratic formula to find
the zeros.
35. –1, 8, 6 − i
SOLUTION: Because 6 − i is a zero and the polynomial is to have
real coefficients, you know that 6 + i must also be a
zero. Using the Linear Factorization Theorem and
the zeros –1, 8, 6 − i, and 6 + i, write f (x) as follows.
f(x) = a[x − (−1)][x − (8)][x − (6 − i)][x − (6 + i)]
Let a = 1. Then write the function in standard form.
So, g(x) written as a product of linear and irreducible
quadratic factors is g(x) = (x + 2)(x + 1)(x – 3 +
)(x – 3 −
).
b. g(x) written as a product of linear factors is g(x)
= (x + 2)(x + 1)(x – 3 +
c. The zeros are –2, –1, 3 −
)(x – 3 −
,3+
).
.
45. f (x) = 4x4 – 35x3 + 140x2 – 295x + 156
Therefore, a function of least degree that has –1, 8,
4
3
2
6 − i, and 6 + i as zeros is f (x) = x – 19x + 113x –
163x − 296 or any nonzero multiple of f (x).
SOLUTION: a. f (x) has possible rational zeros of ±1, ±2, ±3, ±4,
±12, ±13, ±26, ±39, ±52, ±78,
±156,
Write each function as (a) the product of linear
and irreducible quadratic factors and (b) the
product of linear factors. Then (c) list all of its
zeros.
By
using synthetic division, it can be determined that 4 is
a rational zero.
43. g(x) = x4 – 3x3 – 12x2 + 8
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SOLUTION: a. g(x) has possible rational zeros of ±1, ±2, ±4, ±8.
By using synthetic division, it can be determined that
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By using synthetic division on the depressed
polynomial, it can be determined that
is a rational )(x – 3 −
).
b. g(x) written as a product of linear factors is g(x)
= (x + 2)(x + 1)(x – 3 +
)(x – 3 −
).
2-4 c.
Zeros
of Polynomial
The zeros
are –2, –1, 3 − Functions
,3+
.
c. The zeros are
45. f (x) = 4x4 – 35x3 + 140x2 – 295x + 156
SOLUTION: a. f (x) has possible rational zeros of ±1, ±2, ±3, ±4,
±12, ±13, ±26, ±39, ±52, ±78,
±156,
By
using synthetic division, it can be determined that 4 is
a rational zero.
By using synthetic division on the depressed
polynomial, it can be determined that
− 3i)]. Thus, f (x) written as a product of linear
factors is f (x) = (4x – 3)(x – 4)(x – 2 + 3i)(x – 2 –
3i).
, 4, 2 – 3i, and 2 + 3i.
Use the given zero to find all complex zeros of
each function. Then write the linear
factorization of the function.
49. h(x) = 2x5 + x4 – 7x3 + 21x2 – 225x + 108; 3i
SOLUTION: Use synthetic substitution to verify that 3i is a zero
of h(x).
Because x = 3i is a zero of h, x = −3i is also a zero
of h. Divide the depressed polynomial by −3i.
is a rational zero.
2
The remaining quadratic factor (4x −16x + 52) can
Using these two zeros and the depressed polynomial
from the last division, write h(x) = (x + 3i)(x − 3i)
3
2
3
2
(2x + x − 25x + 12). 2x + x − 25x + 12 has
possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±12,
2
be written as 4(x − 4x + 13) and yields no real zeros
and is therefore, irreducible over the reals. So, f (x)
written as a product of linear and irreducible
quadratic factors is
. By using synthetic division, it can be
determined that 3 is a rational zero.
. This can be
2
rewritten as (x – 4x + 13)(4x – 3)(x – 4).
2
b. Use the quadratic formula to find the zeros of x
− 4x + 13.
2
The remaining depressed polynomial 2x +7x − 4 can
be written as (x + 4)(2x − 1). The zeros of the
depressed polynomial are −4 and
zeros of h are 3, –4,
. Therefore, the
, 3i, and –3i. The linear factorization of h is h(x) = (x – 3)(x + 4)
(2x – 1)(x + 3i)(x – 3i).
2
x − 16x + 52 can be written as [x − (2 + 3i)][x − (2
− 3i)]. Thus, f (x) written as a product of linear
factors is f (x) = (4x – 3)(x – 4)(x – 2 + 3i)(x – 2 –
3i).
c. The zeros are
Use each graph to write g as the product of
linear factors. Then list all of its zeros.
61. g(x) = 3x4 – 15x3 + 87x2 – 375x + 300
, 4, 2 – 3i, and 2 + 3i.
Use the given zero to find all complex zeros of
linear
factorization of the function.
49. h(x) = 2x5 + x4 – 7x3 + 21x2 – 225x + 108; 3i
eachManual
function.
Then
write the
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SOLUTION: The graph suggests 1 and 4 are zeros of g(x). Use
zeros of h are 3, –4,
2-4
, 3i, and –3i. The linear factorization of h is h(x) = (x – 3)(x + 4)
Zeros
Functions
(2x
3i)(x – 3i).
– 1)(xof+ Polynomial
Use each graph to write g as the product of
linear factors. Then list all of its zeros.
61. g(x) = 3x4 – 15x3 + 87x2 – 375x + 300
2
written as 3(x + 25) or 3(x + 5i)(x − 5i). So, g
written as the product of linear factors is g(x) = 3(x
– 4)(x – 1)(x + 5i)(x – 5i). The zeros of g are 4, 1,
5i.
62. g(x) = 2x5 + 2x4 + 28x3 + 32x2 – 64x
SOLUTION: The graph suggests −2 and 1 are zeros of g(x). Use
synthetic substitution to test this possibility.
SOLUTION: The graph suggests 1 and 4 are zeros of g(x). Use
synthetic substitution to test this possibility.
Use synthetic division on the depressed polynomial to
test 4.
Use synthetic division on the depressed polynomial to
test −2.
3
2
The remaining quadratic factor (3x + 75) can be
2
written as 3(x + 25) or 3(x + 5i)(x − 5i). So, g
written as the product of linear factors is g(x) = 3(x
– 4)(x – 1)(x + 5i)(x – 5i). The zeros of g are 4, 1,
5i.
The remaining factor (2x + 32x) can be written as
2
2x(x + 16) or 2x(x + 4i)(x − 4i). So, g written as the
product of linear factors is g(x) = 2x(x – 1)(x + 2)(x
+ 4 i)(x – 4 i). The zeros of g are 0, 1, –2, 4i.
62. g(x) = 2x5 + 2x4 + 28x3 + 32x2 – 64x
SOLUTION: The graph suggests −2 and 1 are zeros of g(x). Use
synthetic substitution to test this possibility.
Use synthetic division on the depressed polynomial to
test −2.
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The remaining factor (2x + 32x) can be written as
2
2x(x + 16) or 2x(x + 4i)(x − 4i). So, g written as the
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