Download 31.  33.  35. 

5
4
3
2
So, (2x + 5x − 5x + x − 18x + 10) ÷ (2x – 1) =
Study Guide and Review - Chapter 2
Divide using long division.
3
4
3
2
x + 3x − x − 9 +
.
Divide using synthetic division.
2
35. (x4 – x3 + 7x2 – 9x – 18) ÷ (x – 2)
31. (x + 8x – 5) ÷ (x – 2)
SOLUTION: SOLUTION: Because x − 2, c = 2. Set up the synthetic division as
follows. Then follow the synthetic division
procedure.
3
2
The quotient is x + x + 9x + 9.
3
2
2
So, (x + 8x – 5) ÷ (x – 2) = x + 10x + 20 +
.
Use the Factor Theorem to determine if the
binomials given are factors of f (x). Use the
binomials that are factors to write a factored
form of f (x).
37. f (x) = x3 + 3x2 – 8x – 24; (x + 3)
SOLUTION: 33. (2x5 + 5x4 − 5x3 + x2 − 18x + 10) ÷ (2x – 1)
Use synthetic division to test the factor (x + 3).
SOLUTION: Because the remainder when the polynomial is
divided by (x + 3) is 0, (x + 3) is a factor of f (x).
Because (x + 3) is a factor of f (x), we can use the
final quotient to write a factored form of f (x) as f (x)
2
= (x + 3)(x – 8).
39. f (x) = x4 – 2x3 – 3x2 + 4x + 4; (x + 1), (x – 2)
SOLUTION: Use synthetic division to test each factor, (x + 1) and
(x − 2).
5
4
3
2
So, (2x + 5x − 5x + x − 18x + 10) ÷ (2x – 1) =
4
3
2
x + 3x − x − 9 +
.
Divide using synthetic division.
Because the remainder when f (x) is divided by (x +
1) is 0, (x + 1) is a factor. Test the second factor, (x
3
2
− 2), with the depressed polynomial x − 3x + 4.
35. (x4 – x3 + 7x2 – 9x – 18) ÷ (x – 2)
SOLUTION: Because x − 2, c = 2. Set up the synthetic division as
follows. Then follow the synthetic division
procedure.
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Because the remainder when the depressed
polynomial is divided by (x − 2) is 0, (x − 2) is a
factor of f (x).
Because (x + 1) and (x − 2) are factors of f (x), we
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can use the final quotient to write a factored form
of
2
3
2
The quotient is x + x + 9x + 9.
f(x) as f (x) = (x + 1)(x − 2)(x − x − 2). Factoring
2
divided by (x + 3) is 0, (x + 3) is a factor of f (x).
Because (x + 3) is a factor of f (x), we can use the
final quotient to write a factored form of f (x) as f (x)
2
Study
= (xGuide
+ 3)(x –and
8). Review - Chapter 2
39. f (x) = x4 – 2x3 – 3x2 + 4x + 4; (x + 1), (x – 2)
Because (x + 3) is a factor of f (x), we can use the
final quotient to write a factored form of f (x) as f (x)
2
2
= (x + 3)(x −3x − 5). Because the factor (x − 3x −
5) yields no rational zeros, the rational zero of f is −3.
43. f (x) = 3x4 – 14x3 – 2x2 + 31x + 10
SOLUTION: SOLUTION: Use synthetic division to test each factor, (x + 1) and
(x − 2).
The leading coefficient is 3 and the constant term is
10. The possible rational zeros are 1, 2, 5,
10,
Because the remainder when f (x) is divided by (x +
1) is 0, (x + 1) is a factor. Test the second factor, (x
3
2
− 2), with the depressed polynomial x − 3x + 4.
.
By using synthetic division, it can be determined that
2 is a rational zero.
By using synthetic division on the depressed
Because the remainder when the depressed
polynomial is divided by (x − 2) is 0, (x − 2) is a
factor of f (x).
Because (x + 1) and (x − 2) are factors of f (x), we
can use the final quotient to write a factored form of
polynomial, it can be determined that
rational zero.
is a 2
f(x) as f (x) = (x + 1)(x − 2)(x − x − 2). Factoring
2
the quadratic expression yields f (x) = (x – 2) (x + 1)
2
.
Because (x − 2) and
List all possible rational zeros of each function.
Then determine which, if any, are zeros.
41. f (x) = x3 – 14x − 15
SOLUTION: are factors of f (x), we
can use the final quotient to write a factored form of
Since f(x) as
2
Because the leading coefficient is 1, the possible
rational zeros are the integer factors of the constant
term −15. Therefore, the possible rational zeros of f
are 1, 3, 5, and 15.
By using synthetic division, it can be determined that
–3 is a rational zero.
Because (x + 3) is a factor of f (x), we can use the
final quotient to write a factored form of f (x) as f (x)
2
2
= (x + 3)(x −3x − 5). Because the factor (x − 3x −
5) yields no rational zeros, the rational zero of f is −3.
(3x − 9x − 15) yields no rational zeros, the rational
zeros of f are
Solve each equation.
45. 6x3 – 23x2 + 26x – 8 = 0
SOLUTION: Apply the Rational Zeros Theorem to find possible
rational zeros of the equation. The leading coefficient
is 6 and the constant term is −8. The possible rational
zeros are 1, 2, 4, 8,
.
By using synthetic division, it can be determined that
2 is a rational zero.
43. f (x) = 3x4 – 14x3 – 2x2 + 31x + 10
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TheManual
leading
coefficient
is 3 and
the constant term is
10. The possible rational zeros are 1, 2, 5,
10,
.
Because (x − 2) is a factor of the equation, we can
use the final quotient to write a factored form asPage
(x −2
2
2)(6x − 11x + 4) = 0. Factoring the quadratic
expression yields (x − 2)(3x − 4)(2x − 1) = 0.Thus,
Since f(x) as
2
(3x − 9x − 15) yields no rational zeros, the rational
the solutions are 2,
Study
Guide
zeros
of f areand Review - Chapter 2
, and
.
47. 2x4 – 11x3 + 44x = –4x2 + 48
Solve each equation.
45. 6x3 – 23x2 + 26x – 8 = 0
SOLUTION: SOLUTION: 4
Apply the Rational Zeros Theorem to find possible
rational zeros of the equation. The leading coefficient
is 6 and the constant term is −8. The possible rational
zeros are 1, 2, 4, 8,
.
Because (x − 2) is a factor of the equation, we can
use the final quotient to write a factored form as (x −
2
2)(6x − 11x + 4) = 0. Factoring the quadratic
expression yields (x − 2)(3x − 4)(2x − 1) = 0.Thus,
, and
.
3
2
The equation can be written as 2x – 11x + 4x +
44x − 48 = 0. Apply the Rational Zeros Theorem to
find possible rational zeros of the equation. The
leading coefficient is 2 and the constant term is −48.
The possible rational zeros are ±1, ±2, ±3, ±6, ±8,
±16, ±24, ±48, ±
By using synthetic division, it can be determined that
2 is a rational zero.
the solutions are 2,
2)(6x − 11x + 4) = 0. Factoring the quadratic
expression yields (x − 2)(3x − 4)(2x − 1) = 0.Thus,
, and ±
.
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 4 is a rational
zero.
47. 2x4 – 11x3 + 44x = –4x2 + 48
SOLUTION: 4
3
2
The equation can be written as 2x – 11x + 4x +
44x − 48 = 0. Apply the Rational Zeros Theorem to
find possible rational zeros of the equation. The
leading coefficient is 2 and the constant term is −48.
The possible rational zeros are ±1, ±2, ±3, ±6, ±8,
±16, ±24, ±48, ±
, and ±
.
By using synthetic division, it can be determined that
2 is a rational zero.
Because (x − 2) and (x − 4) are factors of the
equation, we can use the final quotient to write a
2
factored form as (x − 2)(x − 4)(2x + x − 6) = 0.
Factoring the quadratic expression yields (x − 2)(x −
4)(2x − 3)(x + 2) = 0. Thus, the solutions are
, –2,
2, and 4.
Use the given zero to find all complex zeros of
each function. Then write the linear
factorization of the function.
49. f (x) = x4 – 4x3 + 7x2 − 16x + 12; −2i
SOLUTION: Use synthetic substitution to verify that −2i is a zero
of f (x).
By using synthetic division on the depressed
polynomial, it can be determined that 4 is a rational
zero.
Because −2i is a zero of f , 2i is also a zero of f .
Divide the depressed polynomial by 2i.
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Because
− 2) and
− 4) are
factors of the
equation, we can use the final quotient to write a
2
factored form as (x − 2)(x − 4)(2x + x − 6) = 0.
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factored form as (x − 2)(x − 4)(2x + x − 6) = 0.
Factoring the quadratic expression yields (x − 2)(x −
4)(2x − 3)(x + 2) = 0. Thus, the solutions are
Study Guide and Review - Chapter 2
, –2,
2, and 4.
Use the given zero to find all complex zeros of
each function. Then write the linear
factorization of the function.
49. f (x) = x4 – 4x3 + 7x2 − 16x + 12; −2i
SOLUTION: Use synthetic substitution to verify that −2i is a zero
of f (x).
Because −2i is a zero of f , 2i is also a zero of f .
Divide the depressed polynomial by 2i.
Using these two zeros and the depressed polynomial
2
from the last division, write f (x) = (x + 2i)(x − 2i)(x
− 4x + 3). Factoring the quadratic expression yields f
(x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the
zeros of f are 1, 3, 2i, and −2i.
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