2-4 Zeros of Polynomial Functions
Solve each equation.
11. x4 + 9x3 + 23x2 + 3x − 36 = 0
SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is
1, the possible rational zeros are the integer factors of the constant term −36. Therefore, the possible rational zeros
are
.
By using synthetic division, it can be determined that 1 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that −4 is a rational zero.
Because (x − 1) and (x + 4) are factors of the equation, we can use the final quotient to write a factored form as (x
2
2
− 1)(x + 4)(x + 6x + 9) = 0. Factoring the quadratic expression yields (x − 1)(x + 4)(x + 3) = 0. Thus, the solutions
are 1, −4, and −3 (multiplicity: 2).
13. x4 – 3x3 – 20x2 + 84x – 80 = 0
SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. Because the leading coefficient is
1, the possible rational zeros are the integer factors of the constant term −80. Therefore, the possible rational zeros
are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±40, and ±80. 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 −5 is a rational zero.
Because (x − 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as (x
2
2
− 4)(x + 5)(x − 4x + 4) = 0. Factoring the quadratic expression yields (x − 4)(x + 5)(x − 2) = 0. Thus, the solutions
are 4, −5, and 2 (multiplicity: 2).
15. 6x4 + 41x3 + 42x2 – 96x + 6 = –26
SOLUTION: 4
3
2
The equation can be written as 6x + 41x + 42x − 96x + 32 = 0. Apply the Rational Zeros Theorem to find
possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational
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zeros are
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= ±1, ±2, ±4, ±8, ±16, ±32, ± , ± , ± , ± ,± , ±
,±
, and ± .
2-4
Because (x − 4) and (x + 5) are factors of the equation, we can use the final quotient to write a factored form as (x
2
2
4)(x + 5)(x
Factoring the quadratic expression yields (x − 4)(x + 5)(x − 2) = 0. Thus, the solutions
−
− 4x + 4) = 0. Functions
Zeros
of Polynomial
are 4, −5, and 2 (multiplicity: 2).
15. 6x4 + 41x3 + 42x2 – 96x + 6 = –26
SOLUTION: 4
3
2
The equation can be written as 6x + 41x + 42x − 96x + 32 = 0. Apply the Rational Zeros Theorem to find
possible rational zeros of the equation. The leading coefficient is 6 and the constant term is 32. The possible rational
= ±1, ±2, ±4, ±8, ±16, ±32, ± , ± , ± , ± ,± , ±
zeros are
,±
, and ± .
By using synthetic division, it can be determined that
is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that Because
and is a rational zero.
are factors of the equation, we can use the final quotient to write a factored form as
. Factoring the quadratic expression yields
.Thus, the solutions are
,
, and −4
(multiplicity: 2).
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
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nonzero
multiple
f (x).
35. –1, 8, 6 − i
Factoring the quadratic expression yields
.Thus, the solutions are
2-4 Zeros of Polynomial Functions
,
, and −4
(multiplicity: 2).
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.
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, and 4 +
, write f (x) as follows.
)]
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2-4
4
3
2
Therefore,
function of leastFunctions
degree that has –1, 8, 6 − i, and 6 + i as zeros is f (x) = x – 19x + 113x – 163x −
Zeros ofaPolynomial
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.
, and 4 +
, 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 +
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.
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).
,–
4
3
2
, 3 − 4i, and 3 + 4i as zeros is f (x) = x – 6x + 19x + 36x
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.
By using synthetic division on the depressed polynomial, it can be determined that −1 is a rational zero.
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function of leastFunctions
degree that has
2-4 Therefore,
Zeros ofaPolynomial
,–
4
3
2
, 3 − 4i, and 3 + 4i as zeros is f (x) = x – 6x + 19x + 36x
− 150 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.
By using synthetic division on the depressed polynomial, it can be determined that −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 4 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that
2
is a rational zero.
2
The remaining quadratic factor (4x −16x + 52) can be written as 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 Page 5
2
. This can be rewritten as (x – 4x + 13)(4x – 3)(x – 4).
).
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 4 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that
2
is a rational zero.
2
The remaining quadratic factor (4x −16x + 52) can 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
2
. This can be rewritten as (x – 4x + 13)(4x – 3)(x – 4).
2
b. Use the quadratic formula to find the zeros of x − 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 −3 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that 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 −3 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that 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 –
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)(x −
).
.
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