2-3 The Remainder and Factor Theorems
Factor each polynomial completely using the given factor and long division.
1. x3 + 2x2 – 23x – 60; x + 4
SOLUTION: 3
2
2
So, x + 2x – 23x – 60 = (x + 4)(x − 2x − 15).
3
2
Factoring the quadratic expression yields x + 2x – 23x – 60 = (x + 4)(x − 5)(x + 3).
3. x3 + 3x2 – 18x – 40; x – 4
SOLUTION: 3
2
2
So, x + 3x – 18x – 40 = (x − 4)(x + 7x + 10).
3
2
Factoring the quadratic expression yields x + 3x – 18x – 40 = (x − 4)(x + 2)(x + 5).
5. –3x3 + 15x2 + 108x – 540; x – 6
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3
2-3
2
2
So, x + 3x – 18x – 40 = (x − 4)(x + 7x + 10).
The Remainder and Factor Theorems
3
2
Factoring the quadratic expression yields x + 3x – 18x – 40 = (x − 4)(x + 2)(x + 5).
5. –3x3 + 15x2 + 108x – 540; x – 6
SOLUTION: 3
2
2
So, –3x + 15x + 108x – 540 = (x − 6)(−3x − 3x + 90).
3
2
Factoring the quadratic expression yields –3x + 15x + 108x – 540 = −3(x − 6)(x + 6)(x − 5).
7. x4 + 12x3 + 38x2 + 12x – 63; x2 + 6x + 9
SOLUTION: 4
3
2
2
2
So, x + 12x + 38x + 12x – 63 = (x + 6x + 9)(x + 6x − 7).
4
3
2
2
Factoring both quadratic expressions yield x + 12x + 38x + 12x – 63 = (x + 3) (x + 7)(x − 1).
Divide using long division.
9. (5x4 – 3x3 + 6x2 – x + 12) ÷ (x – 4)
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4
2-3
3
2
2
2
So, x + 12x + 38x + 12x – 63 = (x + 6x + 9)(x + 6x − 7).
The Remainder and Factor Theorems
4
3
2
2
Factoring both quadratic expressions yield x + 12x + 38x + 12x – 63 = (x + 3) (x + 7)(x − 1).
Divide using long division.
9. (5x4 – 3x3 + 6x2 – x + 12) ÷ (x – 4)
SOLUTION: 3
2
5x + 17x + 74x + 295 +
11. (4x4 – 8x3 + 12x2 – 6x + 12) ÷ (2x + 4)
SOLUTION: The remainder
can be written as .
3
2
2x – 8x + 22x – 47 +
13. (6x6 − 3x5 + 6x4 − 15x3 + 2x2 + 10x − 6) ÷ (2x – 1)
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The remainder
can be written as 2-3
3
2
The
Remainder
2x
+ 22x – 47 +and
– 8x
.
Factor Theorems
13. (6x6 − 3x5 + 6x4 − 15x3 + 2x2 + 10x − 6) ÷ (2x – 1)
SOLUTION: 15. (x4 + x3 + 6x2 + 18x – 216) ÷ (x3 – 3x2 + 18x – 54)
SOLUTION: x+4
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2-3 The Remainder and Factor Theorems
x+4
17. SOLUTION: 2
2x − 4x + 2 +
Divide using synthetic division.
19. (x4 – x3 + 3x2 – 6x – 6) ÷ (x – 2)
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 + 5x + 4 +
.
21. (3x4 – 9x3 – 24x – 48) ÷ (x – 4)
SOLUTION: 2
Because x − 4, c = 4. Set up the synthetic division as follows, using a zero placeholder for the missing x -term in the
dividend. Then follow the synthetic division procedure.
3
2
The quotient is 3x + 3x + 12x + 24 +
.
23. (12x5 + 10x4 – 18x3 – 12x2 – 8) ÷ (2x – 3)
SOLUTION: Rewrite the division expression so that the divisor is of the form x − c.
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3
2
2-3 The
Thequotient
Remainder
Factor
is 3x + 3xand
+ 12x
+ 24 + Theorems
.
23. (12x5 + 10x4 – 18x3 – 12x2 – 8) ÷ (2x – 3)
SOLUTION: Rewrite the division expression so that the divisor is of the form x − c.
c =
Because
. Set up the synthetic division as follows, using a zero placeholder for the missing x-term in
the dividend. Then follow the synthetic division procedure.
can be written as The remainder
4
3
2
. So, the quotient is 6x + 14x + 12x + 12x + 18 +
.
25. (45x5 + 6x4 + 3x3 + 8x + 12) ÷ (3x – 2)
SOLUTION: Rewrite the division expression so that the divisor is of the form x − c.
c =
Because
2
. Set up the synthetic division as follows, using a zero placeholder for the missing x -term in
the dividend. Then follow the synthetic division procedure.
The remainder
can be written as eSolutions Manual - Powered by Cognero
.
4
3
2
. So, the quotient is 15x + 12x + 9x + 6x +
+ Page 6
can be written as The remainder
4
3
2
. So, the quotient is 6x + 14x + 12x + 12x + 18 +
.
2-3 The Remainder and Factor Theorems
25. (45x5 + 6x4 + 3x3 + 8x + 12) ÷ (3x – 2)
SOLUTION: Rewrite the division expression so that the divisor is of the form x − c.
c =
Because
2
. Set up the synthetic division as follows, using a zero placeholder for the missing x -term in
the dividend. Then follow the synthetic division procedure.
The remainder
can be written as 4
3
2
. So, the quotient is 15x + 12x + 9x + 6x +
+ .
Find each f (c) using synthetic substitution.
31. f (x) = 4x5 – 3x4 + x3 – 6x2 + 8x – 15; c = 3
SOLUTION: The remainder is 711. Therefore, f (3) = 711.
33. f (x) = 2x6 + 5x5 – 3x4 + 6x3 – 9x2 + 3x – 4; c = 5
SOLUTION: The remainder is 45,536. Therefore, f (5) = 45,536.
35. f (x) = 10x5 + 6x4 – 8x3 + 7x2 – 3x + 8; c = –6
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2-3 The Remainder and Factor Theorems
The remainder is 45,536. Therefore, f (5) = 45,536.
35. f (x) = 10x5 + 6x4 – 8x3 + 7x2 – 3x + 8; c = –6
SOLUTION: The remainder is −67,978. Therefore, f (−6) = −67,978.
37. f (x) = −2x8 + 6x5 – 4x4 + 12x3 – 6x + 24; c = 4
SOLUTION: The remainder is −125,184. Therefore, f (4) = −125,184.
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).
39. f (x) = x4 + 2x3 – 5x2 + 8x + 12; (x – 1), (x + 3)
SOLUTION: Use synthetic division to test each factor, (x − 1) and (x + 3).
Because the remainder when f (x) is divided by (x − 1) is 18, (x − 1) is not a factor.
Because the remainder when f (x) is divided by (x + 3) is −30, (x + 3) is not a factor.
41. f (x) = 3x4 – 22x3 + 13x2 + 118x – 40; (3x – 1), (x – 5)
SOLUTION: Use synthetic division to test each factor, (3x − 1) and (x − 5).
For (3x − 1), rewrite the division expression so that the divisor is of the form x − c.
Because
Set up the synthetic division as follows. Then follow the synthetic division procedure.
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2-3 Because
The Remainder
Theorems
the remainderand
whenFactor
f (x) is divided
by (x + 3) is −30, (x + 3) is not a factor.
41. f (x) = 3x4 – 22x3 + 13x2 + 118x – 40; (3x – 1), (x – 5)
SOLUTION: Use synthetic division to test each factor, (3x − 1) and (x − 5).
For (3x − 1), rewrite the division expression so that the divisor is of the form x − c.
Because
Set up the synthetic division as follows. Then follow the synthetic division procedure.
Because the remainder when f (x) is divided by (3x − 1) is 0, (3x − 1) is a factor. Test the second factor, (x − 5),
3
2
with the depressed polynomial x − 7x + 2x + 40.
Because the remainder when the depressed polynomial is divided by (x − 5) is 0, (x − 5) is a factor of f (x).
Because (3x − 1) and (x − 5) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f
2
(x) = (3x − 1)(x − 5)(x − 2x − 8). Factoring the quadratic expression yields f (x) = (3x – 1)(x – 5)(x – 4)(x + 2).
43. f (x) = 3x4 – 35x3 + 38x2 + 56x + 64; (3x – 2), (x + 2)
SOLUTION: Use synthetic division to test each factor, (3x − 2) and (x + 2).
For (3x − 2), rewrite the division expression so that the divisor is of the form x − c.
Because
Set up the synthetic division as follows. Then follow the synthetic division procedure.
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2-3
Because the remainder when the depressed polynomial is divided by (x − 5) is 0, (x − 5) is a factor of f (x).
Because (3x − 1) and (x − 5) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f
The Remainder 2and Factor Theorems
(x) = (3x − 1)(x − 5)(x − 2x − 8). Factoring the quadratic expression yields f (x) = (3x – 1)(x – 5)(x – 4)(x + 2).
43. f (x) = 3x4 – 35x3 + 38x2 + 56x + 64; (3x – 2), (x + 2)
SOLUTION: Use synthetic division to test each factor, (3x − 2) and (x + 2).
For (3x − 2), rewrite the division expression so that the divisor is of the form x − c.
Because
Set up the synthetic division as follows. Then follow the synthetic division procedure.
Because the remainder when f (x) is divided by (3x − 2) is
, (3x − 2) is not a factor.
Test (x + 2).
Because the remainder when f (x) is divided by (x + 2) is 432, (x + 2) is not a factor.
45. f (x) = 4x5 – 9x4 + 39x3 + 24x2 + 75x + 63; (4x + 3), (x – 1)
SOLUTION: Use synthetic division to test each factor, (4x + 3) and (x − 1).
For (4x + 3), rewrite the division expression so that the divisor is of the form x − c.
Because
Set up the synthetic division as follows. Then follow the synthetic division procedure.
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2-3 The Remainder and Factor Theorems
Because the remainder when f (x) is divided by (x + 2) is 432, (x + 2) is not a factor.
45. f (x) = 4x5 – 9x4 + 39x3 + 24x2 + 75x + 63; (4x + 3), (x – 1)
SOLUTION: Use synthetic division to test each factor, (4x + 3) and (x − 1).
For (4x + 3), rewrite the division expression so that the divisor is of the form x − c.
Because
Set up the synthetic division as follows. Then follow the synthetic division procedure.
Because the remainder when f (x) is divided by (4x + 3) is 0, (4x + 3) is a factor. Test the second factor, (x − 1),
4
3
2
with the depressed polynomial x − 3x + 12x − 3x + 21.
Because the remainder when the depressed polynomial is divided by (x − 1) is 28, (x − 1) is not a factor of f (x).
Because (4x + 3) is a factor of f (x), we can use the quotient of f (x) ÷ (4x + 3) to write a factored form of f (x) as f
4
3
2
(x) = (4x + 3) (x – 3x + 12x – 3x + 21).
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