2-4 Zeros of Polynomial Functions
Write a polynomial function of least degree with real coefficients in standard form that has the given
zeros.
33. –2, –4, –3, 5
SOLUTION: Using the Linear Factorization Theorem and the zeros −2, −4, −3, and 5, write f (x) as follows.
f(x) = a[x − (−2)][x − (−4)][x − (−3)][x − (5)]
Let a = 1. Then write the function in standard form.
4
3
2
Therefore, a function of least degree that has −2, −4, −3, and 5 as zeros is f (x) = x + 4x – 19x – 106x – 120 or
any nonzero multiple of f (x).
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.
4
3
2
Therefore, a function of least degree that has –1, 8, 6 − i, and 6 + i as zeros is f (x) = x – 19x + 113x – 163x −
296 or any nonzero multiple of f (x).
37. −5, 2, 4 −
,4+
SOLUTION: Using the Linear Factorization Theorem and the zeros −5, 2, 4 −
f(x) = a[x − (−5)][x − (2)][x − (4 −
)][x − (4 +
Let a = 1. Then write the function in standard form.
,–
, write f (x) as follows.
)]
Therefore, a function of least degree that has −5, 2, 4 −
− 130 or any nonzero multiple of f (x).
39. , and 4 +
, and 4 +
4
3
2
as zeros is f (x) = x – 5x – 21x + 119x
, 3 − 4i
SOLUTION: Because 3 − 4i is a zero and the polynomial is to have real coefficients, you know that 3 + 4i must also be a zero.
Using the Linear Factorization Theorem and the zeros
,–
, 3 − 4i, and 3 + 4i, write f (x) as follows.
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f(x) = a[x − (
)][x − (−
)][x − (3 − 4i)][x − (3 + 4i)]
Let a = 1. Then write the function in standard form.
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function of leastFunctions
degree that has −5, 2, 4 −
2-4 Therefore,
Zeros ofaPolynomial
, and 4 +
4
3
2
as zeros is f (x) = x – 5x – 21x + 119x
− 130 or any nonzero multiple of f (x).
39. ,–
, 3 − 4i
SOLUTION: Because 3 − 4i is a zero and the polynomial is to have real coefficients, you know that 3 + 4i must also be a zero.
Using the Linear Factorization Theorem and the zeros
,–
, 3 − 4i, and 3 + 4i, write f (x) as follows.
f(x) = a[x − (
)][x − (−
)][x − (3 − 4i)][x − (3 + 4i)]
Let a = 1. Then write the function in standard form.
Therefore, a function of least degree that has
− 150 or any nonzero multiple of f (x).
41. 6 −
,6+
,–
4
3
2
, 3 − 4i, and 3 + 4i as zeros is f (x) = x – 6x + 19x + 36x
, 8 – 3i
SOLUTION: Because 8 – 3i is a zero and the polynomial is to have real coefficients, you know that 8 + 3i must also be a zero.
Using the Linear Factorization Theorem and the zeros 6 −
,6+
, 8 – 3i, and 8 + 3i, write f (x) as follows.
f(x) = a[x − (6 −
)][x − (6 +
)][x − (8 – 3i)][x − (8 + 3i)]
Let a = 1. Then write the function in standard form.
Therefore, a function of least degree that has 6 −
,6+
4
3
, 8 – 3i, and 8 + 3i as zeros is f (x) = x – 28x +
2
296x – 1372x + 2263 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 x = −2 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = −1 is a rational zero.
2
The remaining quadratic factor (x − 6x + 4) yields no rational zeros. Use the quadratic formula to find the zeros.
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Therefore, a function of least degree that has 6 −
2-4 Zeros
2 of Polynomial Functions
,6+
4
3
, 8 – 3i, and 8 + 3i as zeros is f (x) = x – 28x +
296x – 1372x + 2263 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 x = −2 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = −1 is a rational zero.
2
The remaining quadratic factor (x − 6x + 4) yields no rational zeros. Use the quadratic formula to find the zeros.
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 −
,3+
)(x – 3 −
).
.
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 x = 4 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x =
2
is a rational zero.
2
The remaining quadratic factor (4x −16x + 52) can be written 4(x − 4x + 13) and yields no real zeros and is
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therefore,
over the reals. So, f (x) written as a product of linear and irreducible quadratic factors is
2
or (x – 4x + 13)(4x – 3)(x – 4)
Page 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+
.
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 x = 4 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x =
2
is a rational zero.
2
The remaining quadratic factor (4x −16x + 52) can be written 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
2
or (x – 4x + 13)(4x – 3)(x – 4)
b. Use the quadratic formula to find the zeros of x2 −4x + 13.
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
, 4, 2 – 3i, and 2 + 3i.
47. h(x) = x4 − 2x3 − 17x2 + 4x + 30
SOLUTION: a. h(x) has possible rational zeros of ±1, ±3, ±5, ±6, ±10, and ±30. By using synthetic division, it can be determined
that x = −3 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero.
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2
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).
2-4 c.
Zeros
of Polynomial
The zeros
are , 4, 2 – 3iFunctions
, and 2 + 3i.
47. h(x) = x4 − 2x3 − 17x2 + 4x + 30
SOLUTION: a. h(x) has possible rational zeros of ±1, ±3, ±5, ±6, ±10, and ±30. By using synthetic division, it can be determined
that x = −3 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero.
2
The remaining quadratic factor (x − 2) yields no rational zeros and can be written as (x +
)(x −
). So, h(x)
written as a product of linear and irreducible quadratic factors is h(x) = (x + 3)(x − 5)(x +
)(x −
).
b. h(x) written as a product of linear factors is h(x) = (x + 3)(x − 5)(x +
c. The zeros are –3, 5,
, and –
)(x −
).
.
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.
3
2
Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 3i)(x − 3i)(2x + x −
3
2
25x + 12). 2x + x − 25x + 12 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±12,
. By using
synthetic division, it can be determined that x = 3 is a rational zero.
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
. Therefore, the zeros of h are 3, –4,
, 3i, and –3i. The linear factorization of h is h(x) =
(x – 3)(x + 4)(2x – 1)(x + 3i)(x – 3i).
51. g(x) = x5 – 2x4 – 13x3 + 28x2 + 46x – 60; 3 – i
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SOLUTION: Use synthetic substitution to verify that 3 − i is a zero of g(x).
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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
. Therefore, the zeros of h are 3, –4,
, 3i, and –3i. The linear factorization of h is h(x) =
2-4 (x
Zeros
Polynomial
Functions
+ 4)(2x
– 3)(xof
– 1)(x + 3i)(x
– 3i).
51. g(x) = x5 – 2x4 – 13x3 + 28x2 + 46x – 60; 3 – i
SOLUTION: Use synthetic substitution to verify that 3 − i is a zero of g(x).
Because x = 3 − i is a zero of g, x = 3 + i is also a zero of g. Divide the depressed polynomial by 3 + i.
3
Using these two zeros and the depressed polynomial from the last division, write h(x) = [x − (3 − i)][x − (3 + i)](x +
2
3
2
4x + x − 6). x + 4x + x − 6 has possible rational zeros of ±1, ±2, ±3, and ±6. By using synthetic division, it can be
determined that x = 1 is a rational zero.
2
The remaining depressed polynomial x + 5x + 6 can be written as (x + 3)(x + 2). The zeros of the depressed
polynomial are −3 and −2. Therefore, the zeros of g are –3, –2, 1, 3 + i, and 3 – i. The linear factorization of g is g
(x) = (x + 3)(x + 2)(x – 1)(x – 3 + i)(x – 3 – i).
53. f (x) = x5 – 3x4 – 4x3 + 12x2 – 32x + 96; –2i
SOLUTION: Use synthetic substitution to verify that −2i is a zero of f (x).
Because x = −2i is a zero of f , x = 2i is also a zero of f . Divide the depressed polynomial by 2i.
3
2
Using these two zeros and the depressed polynomial from the last division, write f (x) = (x + 2i)(x − 2i)(x − 3x − 8x
3
2
+ 24). x − 3x − 8x + 24 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24. By using synthetic
division, it can be determined that x = 3 is a rational zero.
2
The remaining depressed polynomial x − 8 can be written as (x −
zeros of the depressed polynomial are 2
and −2
The linear factorization of f is f (x) = (x – 3)(x + 2
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)(x +
) or (x − 2
. Therefore, the zeros of f are 3, 2
)(x − 2
)(x + 2
, –2
). The
, 2i, and –2i.
)(x + 2i)(x – 2i).
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