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Checkpoint: Assess Your Understanding, pages 28–31
1.1
1. Multiple Choice When synthetic division is used to divide
x4 ⫺ 7x2 ⫹ 2x ⫹ 3 by x ⫹ 1, the result is: 1 ⫺1 ⫺6 8 ⫺5
Which is the correct division statement?
A. x4 ⫺ 7x2 ⫹ 2x ⫹ 3 ⫽ (x ⫹ 1)(x4 ⫺ x3 ⫺ 6x2 ⫹ 8x ⫺ 5) ⫹ 0
B. x4 ⫺ 7x2 ⫹ 2x ⫹ 3 ⫽ (x ⫹ 1)(x3 ⫺ x2 ⫺ 6x ⫹ 8) ⫺ 5
C. x4 ⫺ 7x2 ⫹ 2x ⫹ 3 ⫽ (x ⫹ 1)(x3 ⫺ x2 ⫺ 6x ⫹ 8) ⫹ 5
D. x4 ⫺ 7x2 ⫹ 2x ⫺ 2 ⫽ (x ⫹ 1)(x3 ⫺ x2 ⫺ 6x ⫹ 8)
2. a) Use long division to divide:
i) 2x2 - 11x - 19 by x ⫺ 7 ii) 3x3 + 4x2 - 15x + 30 by x ⫹ 4
2x ⴙ 3
x ⴚ 7 冄 2x ⴚ 11x ⴚ 19
2x2 ⴚ 14x
3x ⴚ 19
3x ⴚ 21
2
3x2 ⴚ 8x ⴙ 17
x ⴙ 4 冄 3x ⴙ 4x 2 ⴚ 15x ⴙ 30
3x 3 ⴙ 12x2
ⴚ8x2 ⴚ 15x
ⴚ 8x2 ⴚ 32x
17x ⴙ 30
17x ⴙ 68
ⴚ38
The quotient is 2x ⴙ 3 and
the remainder is 2.
The quotient is 3x2 ⴚ 8x ⴙ 17 and
the remainder is ⴚ38.
2
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b) Verify the answer to one division in part a.
Sample response:
In part ii, multiply the quotient by the divisor, then add the remainder.
(x ⴙ 4)(3x2 ⴚ 8x ⴙ 17) ⴙ (ⴚ38)
ⴝ 3x3 ⴚ 8x2 ⴙ 17x ⴙ 12x2 ⴚ 32x ⴙ 68 ⴚ 38
ⴝ 3x3 ⴙ 4x2 ⴚ 15x ⴙ 30
Since this is the original polynomial, the answer is correct.
3. Use synthetic division to divide 2x4 - 7x3 - 29x2 - 8x + 12 by
each binomial.
a) x ⫺ 1
1
2
ⴚ7
ⴚ29
ⴚ8
12
2
ⴚ5
ⴚ34
ⴚ42
ⴚ42
ⴚ30
ⴚ5
2
ⴚ34
The quotient is 2x ⴚ 5x ⴚ 34x ⴚ 42 and the remainder is ⴚ30.
3
2
b) x ⫺ 6
6
2
2
ⴚ7
ⴚ29
ⴚ8
12
12
30
6
ⴚ12
ⴚ2
0
5
1
The quotient is 2x ⴙ 5x ⴙ x ⴚ 2 and the remainder is 0.
3
2
c) x ⫹ 2
ⴚ2
2
2
ⴚ7
ⴚ29
ⴚ8
12
ⴚ4
22
14
ⴚ12
6
0
ⴚ11
ⴚ7
The quotient is 2x ⴚ 11x ⴚ 7x ⴙ 6 and the remainder is 0.
3
2
d) x ⫹ 3
ⴚ3
2
2
ⴚ7
ⴚ29
ⴚ8
12
ⴚ6
39
ⴚ30
114
ⴚ38
126
ⴚ13
10
The quotient is 2x ⴚ 13x ⴙ 10x ⴚ 38 and the remainder is 126.
3
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1.2
4. Multiple Choice When a polynomial P(x) is divided by x ⫹ 3, the
remainder is ⫺4. Which statement is true?
A. P(⫺4) ⫽ ⫺3
B. P(3) ⫽ ⫺4
C. P(⫺3) ⫽ ⫺4
D. P(⫺3) ⫽ 0
5. a) Determine the remainder when 3x4 + 8x3 - 15x2 - 32x + 12 is
divided by x ⫹ 1.
Let P(x) ⴝ 3x4 ⴙ 8x3 ⴚ 15x2 ⴚ 32x ⴙ 12
P(ⴚ1) ⴝ 3(ⴚ1)4 ⴙ 8(ⴚ1)3 ⴚ 15(ⴚ1)2 ⴚ 32(ⴚ1) ⴙ 12
ⴝ 3 ⴚ 8 ⴚ 15 ⴙ 32 ⴙ 12
ⴝ 24
The remainder is 24.
b) Is x ⫹ 1 a factor of the polynomial in part a? If your answer is
yes, explain how you know. If your answer is no, determine a
binomial of the form x ⫺ a, a H ⺪, that is a factor.
x ⴙ 1 is not a factor of the polynomial because the remainder after
dividing by x ⴙ 1 is not 0.
The factors of the constant term, 12, are: 1, ⴚ1, 2, ⴚ2, 3, ⴚ3, 4, ⴚ4, 6,
ⴚ6, 12, ⴚ12. Use mental math to substitute x ⴝ 1 to determine that
x ⴚ 1 is not a factor.
Try x ⴝ 2: P(2) ⴝ 3(2)4 ⴙ 8(2)3 ⴚ 15(2)2 ⴚ 32(2) ⴙ 12
ⴝ 48 ⴙ 64 ⴚ 60 ⴚ 64 ⴙ 12
ⴝ0
So, x ⴚ 2 is a factor of 3x4 ⴙ 8x3 ⴚ 15x2 ⴚ 32x ⴙ 12.
6. For each polynomial, determine one factor of the form x ⫺ a, a H ⺪.
a) x3 - 5x2 - 17x + 21
Sample response:
Let P(x) ⴝ x3 ⴚ 5x2 ⴚ 17x ⴙ 21
The factors of 21 are: 1, ⴚ1, 3, ⴚ3, 7, ⴚ7, 21, ⴚ21
Try x ⴝ 1: P(1) ⴝ (1)3 ⴚ 5(1)2 ⴚ 17(1) ⴙ 21
ⴝ0
So, x ⴚ 1 is a factor of x3 ⴚ 5x2 ⴚ 17x ⴙ 21.
b) 4x4 - 15x3 - 32x2 + 33x + 10
Sample response:
Let P(x) ⴝ 4x4 ⴚ 15x3 ⴚ 32x2 ⴙ 33x ⴙ 10
The factors of 10 are: 1, ⴚ1, 2, ⴚ2, 5, ⴚ5, 10, ⴚ10
Try x ⴝ 1: P(1) ⴝ 4(1)4 ⴚ 15(1)3 ⴚ 32(1)2 ⴙ 33(1) ⴙ 10
ⴝ0
So, x ⴚ 1 is a factor of 4x4 ⴚ 15x3 ⴚ 32x2 ⴙ 33x ⴙ 10.
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7. Factor this polynomial.
4x4 - 12x3 + 3x2 + 13x - 6
Let P(x) ⴝ 4x4 ⴚ 12x3 ⴙ 3x2 ⴙ 13x ⴚ 6
The factors of ⴚ6 are: 1, ⴚ1, 2, ⴚ2, 3, ⴚ3, 6, ⴚ6
Try x ⴝ 1: P(1) ⴝ 4(1)4 ⴚ 12(1)3 ⴙ 3(1)2 ⴙ 13(1) ⫺ 6
ⴝ2
So, x ⴚ 1 is not a factor.
Try x ⴝ ⴚ1: P(ⴚ1) ⴝ 4(ⴚ1)4 ⴚ 12(ⴚ1)3 ⴙ 3(ⴚ1)2 ⴙ 13(ⴚ1) ⫺ 6
ⴝ0
So, x ⴙ 1 is a factor.
Divide to determine the other factor.
ⴚ1
4
4
ⴚ12
3
13
ⴚ6
ⴚ4
16
ⴚ19
6
19
ⴚ6
0
ⴚ16
So, 4x ⴚ 12x ⴙ 3x ⴙ 13x ⴚ 6 ⴝ (x ⴙ 1)(4x3 ⴚ 16x2 ⴙ 19x ⴚ 6)
Let P1(x) ⴝ 4x3 ⴚ 16x2 ⴙ 19x ⴚ 6
Try x ⴝ 2: P1(2) ⴝ 4(2)3 ⴚ 16(2)2 ⴙ 19(2) ⴚ 6
ⴝ0
So, x ⴚ 2 is a factor.
Divide to determine the other factor.
2
4
3
4
ⴚ16
19
ⴚ6
8
ⴚ16
6
3
0
4
2
ⴚ8
So, 4x ⴚ 12x ⴙ 3x ⴙ 13x ⴚ 6 ⴝ (x ⴙ 1)(x ⴚ 2)(4x2 ⴚ 8x ⴙ 3)
Factor the trinomial: 4x2 ⴚ 8x ⴙ 3 ⴝ (2x ⴚ 1)(2x ⴚ 3)
So, 4x4 ⴚ 12x3 ⴙ 3x2 ⴙ 13x ⴚ 6 ⴝ (x ⴙ 1)(x ⴚ 2)(2x ⴚ 1)(2x ⴚ 3)
4
16
3
2
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