1. Simplify each expression by combining like terms.
Polynomial Functions
ALGEBRA 2 LESSON 6-1
(For help, go to Lesson 1-2.)
Simplify each expression by combining like terms.
1. 3x + 5x – 7x    2. –8xy2 – 2x2y + 5x2y    3. –4x + 7x2 + x
Find the number of terms in each expression.
4. bh             5. 1 – x                   6. 4x3 – x2 – 9 1 2
6-1

2. 2. –8xy2 – 2x2y + 5x2y = (–2 + 5)x2y – 8xy2 = 3x2y – 8xy2
Polynomial Functions
ALGEBRA 2 LESSON 6-1
Solutions
1. 3x + 5x – 7x = (3 + 5 – 7)x = 1x = x
2. –8xy2 – 2x2y + 5x2y = (–2 + 5)x2y – 8xy2 = 3x2y – 8xy2
3. –4x + 7x2 + x = 7x2 + (–4 + 1)x = 7x2 – 3x
4. bh; 1 term
5. 1 – x; 2 terms
6. 4x3 – x2 – 9; 3 terms 1 2
6-1

3. The term with the largest degree is x3,so the polynomial is degree 3.
Polynomial Functions
ALGEBRA 2 LESSON 6-1
Write each polynomial in standard form. Then classify it by degree and by number of terms.
a. 9 + x3 b. x3 – 2x2 – 3x4
x3 + 9
–3x4 + x3 – 2x2
The term with the largest
degree is x3,so the polynomial
is degree 3.
The term with the largest
degree is –3x4, so the
polynomial is degree 4.
It has two terms.
The polynomial is a
cubic binomial.
It has three terms.
The polynomial is
a quartic trinomial.
6-1

4. Enter the data. Use the LinReg, QuadReg, and CubicReg
Polynomial Functions
ALGEBRA 2 LESSON 6-1
Using a graphing calculator, determine whether a linear, quadratic, or cubic model best fits the values in the table.
x y
0 2.8
2 5
4 6
6 5.5
8 4
Enter the data. Use the LinReg, QuadReg, and CubicReg
options of a graphing calculator to find the best-fitting model
for each polynomial classification.
Graph each model and compare.
Linear model
Quadratic model
Cubic model
The quadratic model appears to best fit the given values.
6-1

5. To find a cubic model, use the CubicReg
Polynomial Functions
ALGEBRA 2 LESSON 6-1
The table shows data on the number of employees that a small company had from 1975 to Find a cubic function to model the data. Use it to estimate the number of employees in Let 0 represent 1975.
Number of
Employees
Year
Enter the data.
To find a cubic model, use the CubicReg
option of a graphing calculator.
Graph the model.
The function ƒ(x) = x3 – 0.375x x is
an approximate model for the cubic function.
To estimate the number of employees for 1988, you can use the Table function
option of a graphing calculator to find that ƒ(13) According to the model,
there were about 62 employees in 1988.
6-1

6. 3. 2m2 + 7m – 3; quadratic binomial 4. x4 – x3 + x; quartic trinomial
Polynomial Functions
ALGEBRA 2 LESSON 6-1
pages 303–305  Exercises
1. 10x + 5; linear binomial
2. –3x + 5; linear binomial
3. 2m2 + 7m – 3; quadratic binomial
4. x4 – x3 + x; quartic trinomial
5. 2p2 – p; quadratic binomial
6. 3a3 + 5a2 + 1; cubic trinomial
7. –x5; quintic monomial
8. 12x4 + 3; quartic binomial
9. 5x3; cubic monomial
10. –2x3; cubic monomial
11. 5x2 + 4x + 8; quadratic trinomial
12. –x4 + 3x3; quartic binomial
13. y = x3 + 1
14. y = 2x3 – 12
15. y = 1.5x3 + x2 – 2x + 1
16. y = –3x3 – 10x
17. a. males: y = – x2 +
0.2829x
females: y = – x2 +
0.2514x
b. males: y = x3 –
x x
females: y = x3 –
x x
c. The cubic model is a better fit.
6-1

7. 26. –4a4 + a3 + a2; quartic trinomial 27. 7; constant monomial
Polynomial Functions
ALGEBRA 2 LESSON 6-1
18. y = x3 – 2x2; 4335
19. y = x3 – 10x2; 2023
20. y = –0.5x3 + 10x2; 433.5
21. y = – x x2 –
17.93x ;
22. y = – x x2 –
6.009x ; 26.34
23. y = x3 – x2 +
5.002x ; 25.39
24. Check students’ work.
25. x3 + 4x; cubic binomial
26. –4a4 + a3 + a2; quartic trinomial
27. 7; constant monomial
28. 6x2; quadratic monomial
29. x4 + 2x3; quartic binomial
x x; quintic binomial
31. a. V = 10 r 2
b. V = r 3
c. V = r r 2
32. Answers may vary. Sample:
Cubic functions represent curvature
in the data. Because of their turning
points they can be unreliable for
extrapolation.
33. –c2 + 16; binomial
34. –9d3 – 13; binomial 1 2 2 3 2 3 2 3
6-1

8. Polynomial Functions 35. 16x2 – x – 5; trinomial
ALGEBRA 2 LESSON 6-1
35. 16x2 – x – 5; trinomial
36. 2x3 – 6x + 17; trinomial
37. a + 4b; binomial
38. –12y; monomial
39. 8x2 – 6y; binomial
40. –3a + 2; binomial
41. 2x3 + 9x2 + 5x + 27;
polynomial of 4 terms
42. –4x4 – 3x3 + 5x – 54;
43. 80x3 – 109x2 + 7x – 75;
44. 2x3 – 2x2 + 8x – 27;
45. 6a2 + 3ab – 8; trinomial
46. 8x3 + 2x2; binomial
47. 30x3 – 10x2; binomial
48. 2a3 – 5a2 – 2a + 5;
polynomial of 4 terms
49. b3 – 6b2 + 9b; trinomial
50. x3 – 6x2 + 12x – 8;
51. x4 + 2x2 + 1; trinomial
52. 8x3 + 60x x + 126;
53. a3 – a2b – b2a + b3;
6-1

9. The quartic model fits best. c. 72.2  1015
Polynomial Functions
ALGEBRA 2 LESSON 6-1
54. a4 – 4a3 + 6a2 – 4a + 1;
polynomial of 5 terms
55. 12s3 + 61s2 + 68s – 21;
polynomial of 4 terms
56. x3 + 2x2 – x – 2; trinomial
57. 8c3 – 26c + 12; trinomial
58. s4 – 2t 2s2 + t 4; trinomial
59. a. y = x
y = x3 – x2 +
2.2929x
y = – x x3 –
0.1647x x b.
The quartic model fits best.
c  1015
6-1

10. 65. [2] If it is in standard form, the degree is the exponent
Polynomial Functions
ALGEBRA 2 LESSON 6-1
 108 cm3
61. a. up 4 units
b. y = 4x3 is more narrow.
c. y = x3
62. B
63. A
64. A
65. [2] If it is in standard form,
the degree is the exponent
on the first variable.
[1] incorrect reason for why it is
easier to find in standard form
66. 2
67. 2
68. none
69.
(5, –3) –
–8 –3 0 –3
–3 –6
5 –2 0
72. –2 –1 –2 –1
–3 –3 –4 –4
6-1

11. Write each polynomial in standard form. Then classify it by degree
Polynomial Functions
ALGEBRA 2 LESSON 6-1
Write each polynomial in standard form. Then classify it by degree
and by number of terms.
1. –x2 + 2x + x2 2. 7x x3 3. 3x(4x) + x2(2x2) 4.
a. Find a cubic and a quadratic
model for the table of values.
b. Which model appears to give the better fit?
c. Using the model you selected in part b, estimate
the value of y when x = 12.
2x; degree 1,
linear monomial
4x3 + 7x2 + 10;
degree 3, cubic trinomial
2x4 + 12x2; degree 4,
quartic binomial x y
y = – x x2 – 5.29x ,
y = –0.1319x x
the cubic model
–49.3
6-1

12. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
(For help, go to Lessons 5-1 and 5-4.)
Factor each quadratic expression.
1. x2 + 7x x2 + 8x – x2 – 14x + 24
Find each product.
4. x(x + 4) 5. (x + 1)2 6. (x – 3)2(x + 2)
6-2

13. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Solutions
1. Factors of 12 with a sum of 7: 4 and 3
x2 + 7x + 12 = (x + 4)(x + 3)
2. Factors of –20 with a sum of 8: 10 and –2
x2 + 8x – 20 = (x + 10)(x – 2)
3. Factors of 24 with a sum of –14: –12 and –2
x2 – 14x + 24 = (x – 12)(x – 2)
4. x(x + 4) = x(x) + x(4) = x2 + 4x
5. (x + 1)2 = x2 + 2(1)x + 12 = x2 + 2x + 1
6. (x – 3)2(x + 2) = (x2 – 2(3)x + 32)(x + 2)
= (x2 – 6x + 9)(x + 2)
= (x2 – 6x + 9)(x) + (x2 – 6x + 9)(2)
= (x3 – 6x2 + 9x) + (2x2 – 12x + 18)
= x3 + (–6 + 2)x2 + (9 – 12)x + 18
= x3 – 4x2 – 3x + 18
6-2

14. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Write (x – 1)(x + 3)(x + 4) as a polynomial in standard form.
(x – 1)(x + 3)(x + 4) = (x – 1)(x2 + 4x + 3x + 12) Multiply (x + 3)
and (x + 4).
= (x – 1)(x2 + 7x + 12) Simplify.
= x(x2 + 7x + 12) – 1(x2 + 7x + 12) Distributive Property
= x3 + 7x2 + 12x – x2 – 7x – 12 Multiply.
= x3 + 6x2 + 5x – 12 Simplify.
The expression (x – 1)(x + 3)(x + 4) is the factored form
of x3 + 6x2 + 5x – 12.
6-2

15. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Write 3x3 – 18x2 + 24x in factored form.
3x3 – 18x2 + 24x = 3x(x2 – 6x + 8) Factor out the GCF, 3x.
= 3x(x – 4)(x – 2) Factor x2 – 6x + 8.
Check: 3x(x – 4)(x – 2) = 3x(x2 – 6x + 8) Multiply (x – 4)(x – 2).
= 3x3 – 18x2 + 24x Distributive Property
6-2

16. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Another airline has different carry-on luggage regulations. The sum of the length, width, and depth may not exceed 50 in.
a. Assume that the sum of the length, width, and depth is 50 in. and the length
is 10 in. greater than the depth. Graph the function relating the volume v to
depth x. Find the x-intercepts. What do they represent?
Relate: Volume = depth • length • width
Define: Let x = depth. Then x = length, and
50 – (depth + length) = width.
Write: V(x) = x ( x + 10 )( 50 – (x + x + 10) )
= x (x + 10)(40 – 2x)
The x-intercepts of
the function are x = 0, x = –10, x = 20. These values
of x produce a volume of zero.
Graph the function for volume.
6-2

17. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
(continued)
b. Describe a realistic domain for V(x).
The function has values over the set of all real numbers x.
Since x represents the depth of the luggage, x > 0.
Since the volume must be positive, x < 20.
A realistic domain is 0 < x < 20.
c. What is the maximum possible volume of the box?
What are the corresponding dimensions of the box?
Look for the greatest value of y that occurs within the domain 0 < x < 20.
Use the Maximum feature of a graphing calculator to find the maximum volume.
A volume of approximately 4225 in.3 occurs for a depth of about 12.2 in.
Then the length is about 22.2 in. and the width is about 15.6 in.
6-2

18. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Find the zeros of y = (x + 1)(x – 1)(x + 3). Then graph the function using a graphing calculator.
Using the Zero Product Property, find a zero for each linear factor.
x + 1 = 0    or     x – 1 = 0    or    x + 3 = 0
x = –1 x = x = –3
The zeros of the function are –1, 1, –3.
Now sketch and label the function.
6-2

19. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Write a polynomial in standard form with zeros at 2, –3, and 0.
2 –3 0 Zeros
ƒ(x) = (x – 2)(x + 3)(x) Write a linear factor for each zero.
= (x – 2)(x2 + 3x) Multiply (x + 3)(x).
= x(x2 + 3x) – 2(x2 + 3x) Distributive Property
= x3 + 3x2 – 2x2 – 6x Multiply.
= x3 + x2 – 6x Simplify.
The function ƒ(x) = x3 + x2 – 6x has zeros at 2, –3, and 0.
6-2

20. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
Find any multiple zeros of ƒ(x) = x5 – 6x4 + 9x3 and state the multiplicity.
ƒ(x) = x5 – 6x4 + 9x3
ƒ(x) = x3(x2 – 6x + 9) Factor out the GCF, x3.
ƒ(x) = x3(x – 3)(x – 3) Factor x2 – 6x + 9.
Since you can rewrite x3 as (x – 0)(x – 0)(x – 0), or
(x – 0)3, the number 0 is a multiple zero of the function, with multiplicity 3.
Since you can rewrite (x – 3)(x – 3) as (x – 3)2, the number 3 is a multiple
zero of the function with multiplicity 2.
6-2

21. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
pages 311–313  Exercises
1. x2 + x – 6
2. x3 + 12x2 + 47x + 60
3. x3 – 7x2 + 15x – 9
4. x3 + 4x2 + 4x
5. x3 + 10x2 + 25x
6. x3 – x
7. x(x – 6)(x + 6)
8. 3x(3x – 1)(x + 1)
9. 5x(2x2 – 2x + 3)
10. x(x + 5)(x + 2)
11. x(x + 4)2
12. x(x – 9)(x + 2)
, –1.4, 0, –5, 1
, –16.9, 2, 6, 8
15. a. h = x, = 16 – 2x,
w = 12 – 2x
b. V = x(16 – 2x)(12 – 2x) c.
194 in.3, 2.26 in.
16. 1, –2 
17. 2, –9
18. 0, –5, 8
19. –1, 2, 3 
6-2

22. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
30. 0, 1 (mult. 3)
31. –1, 0,
32. –1, 0, 1
33. 4 (mult. 2)
34. 1, 2 (mult. 2)
35. – , 1 (mult. 2)
36. –1 (mult. 2), 1, 2
37. 2 x3 blocks, 15 x2 blocks,
31 x blocks, 12 unit blocks
38. a. V = 2x3 + 15x2 + 31x + 12;
2x3 + 7x2 + 7x + 2
b. V = 8x2 + 24x + 10
39. V = 12x3 – 27x
20. –1, 1, 2 
21. y = x3 – 18x x – 210
22. y = x3 + x2 – 2x
23. y = x3 + 9x2 + 15x – 25
24. y = x3 – 9x2 + 27x – 27
25. y = x3 + 2x2 – x – 2
26. y = x3 + 6x2 + 11x + 6
27. y = x3 – 2x2
28. y = x3 – x2 – 2x
29. –3 (mult. 3) 1 2 3 2 7 2
6-2

23. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
46. y = x x – x +
, 5.1; , 4, 6
, –6.9, –1.4; 0, –3, –1, 1
, –6.17; 1.5
50. none, –1; –2, 0
51–53.  Answers may vary.
Samples are given.
51. y = x3 – 3x2 – 10x
52. y = x3 – 21x x – 343
53. y = x4 – 4x3 – 7x2 + 22x + 24
54. –4, 5 (mult. 3)
55. 0 (mult. 2), –1 (mult. 2)
40. a. h = x + 3; w = x b.
0 < –3 < 2; where the volume is zero
c. 0 < x < 2
d ft3
41. y = –2x3 + 9x2 – x – 12
42. y = 5x4 – 23x3 – 250x x + 504
43. y = 3x(x – 8)(x – 1)
44. y = –2x(x + 5)(x – 4)
45. y = x2(x + 4)(x – 1) 1 2 1 2 1 2 3 2
6-2

24. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
56. 0, 6, –6
57. Answers may vary. Sample:
Write the polynomial in standard
form. The constant term is the
value of the y-intercept.
58. 1 ft
59. Answers may vary. Sample:
y = x4 – x2, and zeros are 0, ±1.
60. Answers may vary. Sample:
The linear factors can be
determined by examining
the x-intercepts of the graph.
61. x + 2a
62. a. A = –x3 + 2x2 + 4x
b. 6 square units
63. Answers may vary. Sample:
y = (x – 1)(x + 1)(x – i )(x + i );
y = x4 – 1
64. a. Answers may vary. Sample:
translation to the right 4 units
b. No; the second graph is not the
result of a horizontal translation.
c. Answers may vary. Sample:
rotation of 180° about the origin
65. C
66. H
67. B
68. [2] ƒ(x) = x2(x + 9)(x – 1)
[1] incomplete factoring of x2 + 8x – 9
or other minor errors 7 8
6-2

25. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
69. [4] The student multiplies (x + 2),
(x – 5), (x – 6) and –2 to get
y = –2x3 + 18x2 – 16x – 120.
[3] appropriate methods but with
one computational error
[2] appropriate methods to find basic
polynomial with given roots, but no
multiplication of polynomial by –2
[1] correct function, without work shown
70. –3x5 + 3x2 – 1; quintic trinomial
71. –7x4 – x3; quartic binomial
72. x3 – 2; cubic binomial
73. (x + 4)(x + 1)
74. (x – 5)(x + 3)
75. (x – 6)2
76. 8
77. –11
78. 0
6-2

26. Polynomials and Linear Factors
ALGEBRA 2 LESSON 6-2
1. Find the zeros of y = (x + 7)(x – 3)(x – 2). Then write the polynomial
in standard form.
2. Factor x3 – 2x2 – 15x. Find the relative maximum and relative minimum.
3. Write a polynomial function in standard form with zeros at –5, –4, and 3.
4. Find any multiple zeros of ƒ(x) = x4 – 25x2 and state the multiplicity.
–7, 2, 3; x3 + 2x2 – 29x + 42
x(x + 3)(x – 5); , –36
400 27
Answers may vary. Sample: ƒ(x) = x3 + 6x2 – 7x – 60
The number 0 is a zero with multiplicity 2.
6-2

27. Simplify each expression. 1. (x + 3)(x – 4) + 2 2. (2x + 1)(x – 3)
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
(For help, go to Lessons 5-1 and 6-1.)
Simplify each expression.
1. (x + 3)(x – 4) (2x + 1)(x – 3)
3. (x + 2)(x + 1) – –3(2 – x)(x + 5)
Write each polynomial in standard form. Then list the coefficients.
5. 5x – 2x x x3 – 9x2
7. 3x + x2 – x + 7 – 2x2 8. –4x4 – 7x2 + x3 + x4
6-3

28. 2. (2x + 1)(x – 3) = 2x2 – 6x + x – 3 = 2x2 – 5x – 3
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
Solutions
1. (x + 3)(x – 4) + 2 = x2 – 4x + 3x –
= x2 – x – 10
2. (2x + 1)(x – 3) = 2x2 – 6x + x – 3 = 2x2 – 5x – 3
3. (x + 2)(x + 1) – 11 = x2 + x + 2x + 2 – 11
= x2 + 3x – 9
4. –3(2 – x)(x + 5) = –3(2x + 10 – x2 – 5x) = –3(–x2 – 3x + 10) = 3x2 + 9x – 30
5. 5x – 2x x3 = 4x3 – 2x2 + 5x + 9 coefficients: 4, –2, 5, 9
x3 – 9x2 = 5x3 – 9x coefficients: 5, –9, 0, 10
7. 3x + x2 – x + 7 – 2x2 = –x2 + 2x + 7 coefficients: –1, 2, 7
8. –4x4 – 7x2 + x3 + x4 = –3x4 + x3 – 7x2 coefficients: –3, 1, –7, 0, 0
6-3

29. x2 – 5x Multiply: x(x – 5) = x2 – 5x 7x
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
Divide x2 + 2x – 30 by x – 5. 
x Divide = x.
x – 5 x2 + 2x – 30 x2 x
x2 – 5x Multiply: x(x – 5) = x2 – 5x 7x
– 30 Subtract: (x2 + 2x) – (x2 – 5x) = 7x.
Bring down –30.
Repeat the process of dividing, multiplying, and subtracting.
x Divide = 7.
x – 5 x2 + 2x – 30
x2 – 5x
7x – 30 7x x
7x – 35 Multiply: 7(x – 5) = 7x – 35.
5 Subtract: (7x – 30) – (7x – 35) = 5.
The quotient is x + 7 with a remainder of 5, or simply x + 7, R 5.
6-3

30. Determine whether x + 2 is a factor of each polynomial.
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
Determine whether x + 2 is a factor of each polynomial.
a. x2 + 10x + 16 b. x3 + 7x2 – 5x – 6
x2 + 5x – 15
x + 2 x3 + 7x2 – 5x – 6
x3 + 2x2
5x2 – 5x
5x2 + 10x
–15x – 6
–15x – 30 24
x + 8
x + 2 x2 + 10x + 16
x2 + 2x
8x + 16
Since the remainder is zero,
x + 2 is a factor of x2 + 10x + 16.
Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6. /
6-3

31. Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.
Step 1: Reverse the sign of the constant term in the divisor.
Write the coefficients of the polynomial in standard form.
Write x – 3 5x3 – 6x2 + 4x – 1
as – –1
Step 2: Bring down the coefficient.
Bring down the 5.
3 5 –6 4 –1 This begins the quotient. 5
6-3

32. Step 3: Multiply the first coefficient by the new divisor.
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
(continued)
Step 3:  Multiply the first coefficient by the new divisor.
Write the result under the next coefficient. Add.
Multiply 3 by 5. Write
3 5 –6 4 –1 the result under –6.
x 15
Add –6 and 15.
Step 4:  Repeat the steps of multiplying and adding until
the remainder is found.
3 5 –6 4 –1
5x x   Remainder
The quotient is 5x2 + 9x + 31, R 92.
6-3

33. The volume in cubic feet of a shipping carton is
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
The volume in cubic feet of a shipping carton is
V(x) = x3 – 6x2 + 3x The height is x – 5 feet.
a. Find linear expressions for the other dimensions.
Assume that the length is greater than the width.
5 1 – Divide.
5 –5 –10
1 –1 –
x2 – x – 2  Remainder
x2 – x – 2 = (x – 2)(x + 1) Factor the quotient.
The length and the width are x + 1 and x – 2, respectively.
b. If the width of the carton is 4 feet, what are the other two dimensions?
x – 2 = 4  Substitute 4 into the expression for width. Find x.
x = 6
Since the height equals x – 5 and the length equals
x + 1, the height is 1 ft. and the length is 7 ft.
6-3

34. Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.
By the Remainder Theorem, P(3) equals the remainder when
P(x) is divided by x – 3.
3 1 – –9
The remainder is 21, so P(3) = 21.
6-3

35. Dividing Polynomials pages 318–320 Exercises 1. x – 8 2. 3x – 5
ALGEBRA 2 LESSON 6-3
pages 318–320  Exercises
1. x – 8
2. 3x – 5
3. x2 + 4x + 3, R 5
4. 2x2 + 5x + 2
5. 3x2 – 7x + 2
6. 9x – 12, R –32
7. x – 10, R 40
8. x2 + 4x + 3
9. no
10. yes
11. yes
12. no
13. x2 + 4x + 3
14. x2 – 2x + 2
15. x2 – 11x + 37, R –128
16. x2 + 2x + 5
17. x2 – x – 6
18. –2x2 + 9x – 19, R 40
19. x + 1, R 4
20. 3x2 + 8x – 3
21. x2 – 3x + 9
22. 6x – 2, R –4
23. y = (x + 1)(x + 3)(x – 2)
6-3

36. 34. P(a) = 0; x – a is a factor of P(x).
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
24. y = (x + 3)(x – 4)(x – 3)
= x + 3 and h = x
26. 18
27. 0
28. 0
29. 12
31. 10
32. 51
33. 0
34. P(a) = 0; x – a is a factor of P(x).
35. x – 1 is not a factor of x3 – x2 – 2x
because it does not divide into
x3 – x2 – 2x evenly.
36. Answers may vary. Sample:
(x2 + x – 4) ÷ (x – 2)
37. x2 + 4x + 5
38. x3 – 3x2 + 12x – 35, R 109
39. x4 – x3 + x2 – x + 1
40. x + 4
41. x3 – x2 + 1
42. no
43. yes
44. yes
6-3

37. d. (x + 1)(x8 – x7 + x6 – x5 + x4 – x3 + x2 – x + 1)
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
45. no
46. no
47. yes
48. yes
49. yes
50. no
51. no
52. x3 – x2 + 1
53. x3 – 2x2 – 2x + 4, R –35
54. x3 – 2x2 – x + 6
55. x3 – 4x2 + x
56. a. x + 1
b. x2 + x + 1
c. x3 + x2 + x + 1
d. (x – 1)(x4 + x3 + x2 + x + 1)
57. a. x2 – x + 1
b. x4 – x3 + x2 – x + 1
c. x6 – x5 + x4 – x3 + x2 – x + 1
d. (x + 1)(x8 – x7 + x6 – x5 + x4 – x3 + x2 – x + 1)
58. By dividing it by a polynomial of degree 1,
you are reducing the degree-n polynomial
by one, to n – 1. The remainder will be constant
because it is not divisible by the variable.
59. x + 2i
6-3

38. 60. Yes; the graph could rise to the right
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
60. Yes; the graph could rise to the right
and fall to the left or it could fall to
the right and rise to the left.
61. D
62. I
63. B
64. [2] x2 – 6x – 7
x – 1 x3 – 7x2 – x + 7
x3 – x2
–6x2 – x
–6x2 + 6x
– 7x + 7
x2 – 6x – 7 = 0
(x + 1)(x – 7) = 0
x = 1, –1, 7
[1] incorrect illustration of division
or incorrect list of zeros
65. y = x2 + 2x – 15
66. y = x3 –9x2 + 8x
67. y = x3 – 6x2 + 3x + 10
68. y = x4 – 4x3 + 6x2 – 4x + 1
69. 24
70. 5
– 11i
6-3

39. Dividing Polynomials 72. –0.5 0 –1.5 0.5 1 1.5 0.5 0 0.5
ALGEBRA 2 LESSON 6-3
72. –0.5 0 –1.5
73. none exists
1 –4 –2
– –1.5
6-3

40. 1a. Divide using long division. (9x3 – 48x2 + 13x + 3) ÷ (x – 5)
Dividing Polynomials
ALGEBRA 2 LESSON 6-3
1a. Divide using long division.
(9x3 – 48x2 + 13x + 3) ÷ (x – 5)
1b. Is x – 5 a factor of 9x3 – 48x2 + 13x + 3?
2. Divide using synthetic division.
(6x3 – 4x2 + 14x – 8) ÷ (x + 2)
3. Use synthetic division and the given factor x – 4 to completely
factor x3 – 37x + 84.
4. Use synthetic division and the Remainder Theorem to find
P(–2) when P(x) = x4 – 2x3 + 4x2 + x + 1.
9x2 – 3x – 2, R –7 no
6x2 – 16x + 46, R –100
(x + 7)(x – 3)(x – 4) 47
6-3

41. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
(For help, go to Lessons 6-2 and 6-3.)
Graph each system. Find any points of intersection.
Factor each expression.
4. x2 – 2x – x2 – 9x x2 + 6x + 5
y = 3x + 1
y = –2x + 6
–2x + 3y = 0
x + 3y = 3
2y = – x + 8
x + 2y = –6
6-4

42. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
Solutions
1.   
intersection: (1, 4)
3.     Rewrite equations:
Rewrite equations:
intersection: (1, )
no points of intersection
4. Factors of –15 with a sum of –2: –5 and 3
x2 – 2x – 15 = (x – 5)(x + 3)
5. Factors of 14 with a sum of –9: –7 and –2
x2 – 9x + 14 = (x – 7)(x – 2)
6. Factors of 5 with a sum of 6: 5 and 1
x2 + 6x + 5 = (x + 5)(x + 1)
y = 3x + 1
y = –2x + 6
–2x + 3y = 0
x + 3y = 3
2y = – x + 8
x + 2y = –6
y = x
y = – x + 1 2 3 1 1 2
y = – x + 4
y = – x – 3 2 3 1 2
6-4

43. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
Graph and solve x3 – 19x = –2x
Step 1:  Graph y1 = x3 – 19x and y2 = –2x2 + 20
on a graphing calculator.
Step 2:  Use the Intersect feature to find the
x values at the points of intersection.
The solutions are –5, –1, and 4.
6-4

44. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
(continued)
Check: Show that each solution makes the original equation a true statement.
x3 – 19x = –2x x3 – 19x = –2x2 + 20
(–5)3 – 19(–5) –2(–5)2 + 20  (–1)3 – 19(–1) –2(–1)2 + 20
– – – –2 + 20
–30 = – = 18
x3 – 19x = –2x2 + 20
(4)3 – 19(4) –2(4)2 + 20
64 – 76 –
–12 = –12
6-4

45. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
The dimensions in inches of the cubicle area inside a doghouse can be expressed as width x, length x + 4, and height x – 3. The volume is 15.9 ft3. Find the dimensions of the doghouse.
15.9 ft3 • = in Convert the volume to cubic inches.
123 in.3
ft3
V = l • w • h Write the formula for volume.
= (x + 4)x(x – 3) Substitute.
Graph y1 = and y2 = (x + 4)x(x – 3).   
Use the Intersect option of the calculator.
When y = , x So x –
and x
The dimensions of the doghouse are about 30 in. by 27 in. by 34 in.
6-4

46. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
Factor x3 – 125.
x3 – 125 = (x)3 – (5)3 Rewrite the expression
as the difference of cubes.
= (x – 5)(x2 + 5x + (5)2) Factor.
= (x – 5)(x2 + 5x + 25) Simplify.
6-4

47. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
Solve 8x = 0. Find all complex roots.
8x = (2x)3 + (5)3 Rewrite the expression
as the difference of cubes.
= (2x + 5)((2x)2 – 10x + (5)2) Factor.
= (2x + 5)(4x2 – 10x + 25) Simplify.
Since 2x + 5 is a factor, x = – is a root. 5 2
The quadratic expression 4x2 – 10x + 25 cannot be factored, so use the
Quadratic Formula to solve the related quadratic equation 4x2 – 10x + 25 = 0.
6-4

48. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
(continued)
x = Quadratic Formula
=    Substitute 4 for a, –10 for b, and 25 for c.
–b ± b2 – 4ac 2a
–(–10) ± (–10)2 – 4(4)(25)
2(4)
= Use the Order of Operations.
– (–10) ± –300 8
= –1 = 1
10 ± 10i 3 8
= Simplify.
5 ± 5i 3 4
The solutions are – and 5 2
5 ± 5i 3 4
6-4

49. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
Factor x4 – 6x2 – 27.
Step 1:  Since x4 – 6x2 – 27 has the form of a quadratic expression, you can
factor it like one. Make a temporary substitution of variables.
x4 – 6x2 – 27 = (x2)2 – 6(x2) – 27 Rewrite in the form of a
quadratic expression.
= a2 – 6a – 27 Substitute a for x2.
Step 2:  Factor a2 – 6a – 27.
a2 – 6a – 27 = (a + 3)(a – 9)
Step 3:  Substitute back to the original variables.
(a + 3)(a – 9) = (x2 + 3)(x2 – 9) Substitute x2 for a.
= (x2 + 3)(x + 3)(x – 3) Factor completely.
The factored form of x4 – 6x2 – 27 is (x2 + 3)(x + 3)(x – 3).
6-4

50. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
Solve x4 – 4x2 – 45 = 0.
x4 – 4x2 – 45 = 0
(x2)2 – 4(x2) – 45 = 0    Write in the form of a
quadratic expression.
Think of the expression as
a2 – 4a – 45, which factors
as (a – 9)(a + 5).
(x2 – 9)(x2 + 5) = 0
(x – 3)(x + 3)(x2 + 5) = 0
x = 3 or x = –3 or x2 = –5 Use the factor theorem.
x = ± 3 or x = ± i Solve for x, and simplify.
6-4

51. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
pages 324–326  Exercises
1. –2, 1, 5
2. –1, 0, 3
3. 0, 1
4. 0, 8
5. 0, –1, –2
6. 0, –3.5, 1
7. 0, –0.5, 1.5
8. –0.5, 0, 3
9. 1, 7 %
11. about 5.78 ft  6.78 ft  1.78 ft
12. (x + 4)(x2 – 4x + 16)
13. (x – 10)(x2 + 10x + 100)
14. (5x – 3)(25x2 + 15x + 9)
15. 3,
16. –4, 2 ± 2i 3
17. 5,
18. –1, ,
20. – ,
21. (x2 – 7)(x – 1)(x + 1)
22. (x2 + 10)(x2 – 2)
–3 ± 3i 3 2
–5 ± 5i 3 2
1 ± i 3 2 1 2
–1 ± i 3 4 1 2
1 ± i 3 4
6-4

52. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
23. (x2 – 3)(x – 2)(x + 2)
24. (x – 2)(x + 2)(x – 1)(x + 1)
25. (x – 1)(x + 1)(x2 + 1)
26. 2(2x2 – 1)(x + 1)(x – 1)
27. ±3, ±1
28. ±2
29. ±4, ±2i
30. ±3i, ± 2
31. ± 2, ±i 6
32. ±i 5, ±i 3
33. –1, 3.24, –1.24
34. –9, 0
35. –2, –3, 1, 2
, 0.83
37. 0, 1.54, 8.46
38. 0, 1.27, 4.73
39. –1.04, 0, 6.04
40. (n – 1)(n)(n + 1) = 210; 5, 6, 7
41. about 3.58 cm, about 2.83 cm
42. – , ,
44. ±2 2, ±2i 2
45. ±5, ±i 2 6 5
3 ± 3i 3 5 4 3
–2 ± 2i 3 3
6-4

53. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
46. ±3i, ±i 3
47. 0, ±2, ±1
48. ± 10, ±i 10
49. 0, ±
50. 4, –2 ± 2i 3
51. 0, 3 ± 3
52. – , 0, 4
53. –1, 1, ±i 5
54. –3, –2, 2
55. –1, 3, 3
56. 0, 1, 3
57. 0, 0, 1, 6
58. ± , ±i
59. ± 2, ±i
60. Check students’ work.
61. V = x2(4x – 2), 4 in. by 4 in. by 16 in.
62. x = length, V = x(x – 1)(x – 2), 5 meters
63. – , 1; y = (2x + 5)(x – 1)
64. ±3, ±1; y = (x – 1)(x + 1)(x – 3)(x + 3)
65. –1, 2, 2; y = (x + 1)(x – 2)2
66. –2, 1, 3; y = (x + 2)(x – 1)(x – 3)
67. –4, –1, 3; y = (x + 4)(x + 1)(x – 3)
68. A cubic can only have 3 zeros. 3 2 1 2
265 10 5 2 3 2
6-4

54. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
69. a. Answers may vary. Sample:
x4 – 9 = 0, ± 3, ±i 3
b. No; two of the roots are imaginary.
70. Answers may vary. Sample:
The pink block has volume a2(a – 3),
the orange block has volume 9(a – 3),
the blue block has volume 3a(a - 3),
and the purple block has volume 27.
Thus a3 – 27 = a2(a – 3) + 3a(a – 3) +
9(a – 3) = (a2 + 3a + 9)(a – 3).
71. a. 10
b. 8 and 12
72. A
73. H
74. [2] =
[1] attempts to write each of the terms as a cube
75. [4] 8x3 – 27 = (2x)3 – (3)3 =
(2x – 3)(4x2 + 6x + 9).
If 2x – 3 = 0, then x = .
If 4x2 + 6x + 9 = 0,
then x = = =
= .
[3] minor computational errors
[2] (2x – 3) = 0 and 4x2 + 6x + 9 = 0
written and solved for x in the
first equation but not
in the second
[1] answer correct, without
work shown 3 2
–6 ± – 144 8
–6 ± 8
–6 ± 6i 3 8
–3 ± 3i 3 4 a3 b6 1 8 a b2 3 1 2 3
6-4

55. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
76. x2 – 3x – 10
77. 2x2 + x – 3, R 2
78. –2, 6
79. ±6
80. – , 3
81. 0 –5
82. no unique solution 1 2
6-4

56. Solving Polynomial Equations
ALGEBRA 2 LESSON 6-4
1. Solve x3 – 2x2 – 3 = x – 4 by graphing. Where necessary,
round to the nearest hundredth.
Factor each expression.
2. 216x3 – 1
3. 8x
4. x4 – 5x2 + 4
Solve each equation.
5. x = x4 + 3x2 – 28 = 0
–0.80, 0.55, 2.25
(6x – 1)(36x2 + 6x + 1)
(2x + 5)(4x2 – 10x + 25)
(x + 1)(x – 1)(x + 2)(x – 2)
5 ± 5i 3 4
–5,
±2, ±i 7
6-4

57. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
(For help, go to Lessons 1-1, 5-1, and 5-6.)
List all the factors of each number.
Multiply.
5. (x – 5)(x2 + 7) 6. (x + 2)(x )(x – 3)
Define each set of numbers.
7. rational 8. irrational 9. imaginary
6-5

58. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
Solutions
1. Factors of 12: ±1, ±2, ±3, ±4, ±6, ±12
2. Factors of 24: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24
3. Factors of 36: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36
4. Factors of 48: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48
5. (x – 5)(x2 + 7) = x3 + 7x – 5x2 – 35
= x3 – 5x2 + 7x – 35
6. (x + 2)(x )(x – 3) = (x + 2)(x – ( 3)2) = (x + 2)(x2 – 3) =
x3 – 3x + 2x2 – 6 = x3 + 2x2 – 3x – 6
7. Rational numbers are all the numbers that can be written as the quotient of
two integers: , where b = 0.
8. Irrational numbers are numbers that cannot be written as a quotient of integers.
9. An imaginary number is any number of the form a + bi, where b = 0 and i is
defined as the number whose square is –1. a b /
6-5

59. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
Find the rational roots of 3x3 – x2 – 15x + 5.
Step 1: List the possible rational roots.
The leading coefficient is 3. The constant term is 5. By the Rational
Root Theorem, the only possible rational roots of the equation have
the form
factors of 5
factors of 3
The factors of 5 are ±5 and ±1 and ±5. The factors of 3 are
±3 and ±1. The only possible rational roots are ±5, ± , ±1, ± . 5 3 1
6-5

60. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
(continued)
Step 2: Test each possible rational root.
5: 3(5)3 – (5)2 – 15(5) + 5 = 280 = 0
–5: 3(–5)3 – (–5)2 – 15(–5) + 5 = –320 = 0 /
: – – = –8.8 = 0
: – – = –13.3 = 0 / 5 3
( ) –
( )
1: 3(1)3 – (1)2 – 15(1) + 5 = –8 = 0
–1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 = 0 / 1 3
( ) –
: – – = 0 So is a root.
: – – = 9.7 = 0 /
( )
The only rational root of 3x3 – x2 – 15x + 5 = 0 is . 1 3
6-5

61. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
Find the roots of 5x3 – 24x2 + 41x – 20 = 0.
Step 1: List the possible rational roots.
The leading coefficient is 5. The constant term is 20. By the
Rational Root Theorem, the only possible roots of the equation
have the form
factors of – 20
factors of 5
The factors of –20 are ±1 and ±20, ±2 and ±10, and ±4 and ±5.
The only factors of 5 are ±1 and ±5. The only possible rational roots
are ± , ± , ± , ±1, ±2, ±4, ±5, ±10, and ±20. 1 5 2 4
6-5

62. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
(continued)
Step 2: Test each possible rational root until you find a root.
Test : – ± – 20 = –12.72 = 0
Test – : – ± – 2 = –29.2 = 0
Test : – ± – 20 = –7.12 = 0
Test – : – ± – 20 = –40.56 = 0
Test : – ± – 20 = 0 So is a root.
( ) 1 5 2 4
(– ) /
5 –24 41 –20
4 –
5 –
5x2 – 20x   Remainder 4 5
Step 3: Use synthetic division with the
root you found in Step 2
to find the quotient.
6-5

63. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
(continued)
Step 4: Find the roots of 5x2 – 20x + 25 = 0.
5x2 – 20x = 0
5(x2 – 4x + 5) = 0 Factor out the GCF, 5.
x2 – 4x + 5 = 0
x = Quadratic Formula
= Substitute 1 for a, –4 for b,
and 5 for c.
–b ± b2 – 4ac 2a
–(–4) ± (–4)2 – 4(1)(5)
2(1)
= Use order of operations.
=    –1 = i.
= 2 ± i Simplify.
4 ± –4 2
4 ± 2i
The roots of 5x3 – 24x2 + 41x – 20 = 0 are , 2 + i, and 2 – i. 4 5
6-5

64. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
A polynomial equation with rational coefficients has the roots 2 – 5 and Find two additional roots.
By the Irrational Root Theorem, if 2 – 5 is a root, then its conjugate
is also a root.
If is a root, then its conjugate – also is a root.
6-5

65. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
A polynomial equation and real coefficients has the roots
2 + 9i with 7i. Find two additional roots.
By the Imaginary Root Theorem, if 2 + 9i is a root, then its complex
conjugate 2 – 9i also is a root.
If 7i is a root, then its complex conjugate –7i also is a root.
6-5

66. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
Find a third degree polynomial with rational coefficients that has roots –2, and 2 – i.
Step 1: Find the other root using the Imaginary Root Theorem.
Since 2 – i is a root, then its complex conjugate 2 + i is a root.
Step 2: Write the factored form of the polynomial using the Factor Theorem.
(x + 2)(x – (2 – i))(x – (2 + i))
6-5

67. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
(continued)
Step 3: Multiply the factors.
(x + 2)[x2 – x(2 – i) – x(2 + i) + (2 – i)(2 + i)] Multiply
(x – (2 – i))
(x – (2 + i)).
(x + 2)(x2 – 2x + ix – 2x – ix + 4 – i 2) Simplify.
(x + 2)(x2 – 2x – 2x )
(x + 2)(x2 – 4x + 5) Multiply.
x3 – 2x2 – 3x + 10
A third-degree polynomial equation with rational coefficients and
roots –2 and 2 – i is x3 – 2x2 – 3x + 10 = 0.
6-5

68. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5 1 2
1 ± 3
1 ± 7 3 7
12. 1, –2,
13. – 5,
, – 3
, 2 – 2
16. 1 – i, 5i
17. 2 – 3i, –6i
i, 3 – 7i
19. x3 – x2 + 9x – 9 = 0
20. x3 + 3x2 – 8x + 10 = 0
21. x3 – 2x2 + 16x – 32 = 0
22. x3 – 3x2 – 8x + 30 = 0
23. x3 – 6x2 + 4x – 24 = 0
pages 333–334  Exercises
1. ±1, ±2; 1
2. ±1, ±2, ±3, ±6; 1, –2, –3
3. ±1, ±2, ±4; –1
4. ± , ±1, ±2, ±4, ±8; no rational roots
5. ±1, ±2, ±4, ±8, ±16; –2
6. ±1, ±3, ±5, ±15; no rational roots
7. 2, ±i 5
8. 5, ±i 7
9. –3, 1,
10. –5,
11. ± , ±3
6-5

69. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
24. x3 – x2 + 2 = 0
25. ± , ± , ± , ± , ± , ± , ± ,
±1, ± , ±2, ±3, ±6; , ,
26. ± , ± , ± , ± , ± , ±1, ±2, ± ,
±4, ±5, ±10, ±20; 2, ,
27. ± , ± , ± , ± , ± , ±1, ± , ±3,
± , ±7, ± , ±21; , – , 1, 3
28. ± , ± , ± , ± , ± , ± , ± , ± ,
± , ±1, ± , ± , ± , ±15,
±3, ±5; – , , 1 12 6 4 2
29. x4 – 6x3 + 14x2 – 24x + 40 = 0
30. x4 – 2x3 – x2 + 6x – 6 = 0
31. x4 – 6x3 + 2x2 + 30x – 35 = 0
32. Never true; 5 is not a factor of 8,
so by the Rational Root Theorem,
5 is not a root of the equation.
33. Sometimes true; since –2 is a factor
of 8, –2 is a possible root of
the equation.
34. Always true; use the Rational Root
Theorem with p = a and q = 1.
35. Sometimes true; since and – 5
are conjugates, they can be roots of
a polynomial equation with integer
coefficients. 3 10 5 7 21 8 15
6-5

70. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
36. Never true; since 2 + i and –2 – i
are not conjugates, they cannot be
the only imaginary roots of a
polynomial equation with integer
roots. If their conjugates were also
roots, there would be four roots and
the equation would have to be of
fourth degree.
37. If 2i is a root, then so is –2i.
38. Answers may vary. Sample:
x4 – x2 – 2 = 0; roots are ± 2 and ±i.
39. If b of a b were rational,
then the sum would also be rational.
Thus the roots would be rational,
which is not always the case.
40. a. 2 real, 2 imaginary;
4 imaginary; 4 real
b. 5 real; 3 real, 2 imaginary;
4 imaginary, 1 real
c. Answers may vary. Sample:
It has an odd number of real
solutions, but it must have at
least one real solution.
41. Answers may vary. Sample:
You cannot use the Irrational Root
Theorem unless the equation has
rational coefficients.
42. x2 + (–2 + i )x + 12 – 8i = 0
43. a–c. Answers may vary. Sample:
a. x – 1 – = 0
b. x2 – 2( )x + ( )2 = 0
c. –1
6-5

71. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
44. D
45. H
46. [2] p is a factor of 3, so p = ±1 or ±3
q is a factor of 2, so q = ±1 or ±2
[1] relates p and q to the coefficients
or the constant of the polynomial
47. [4] The three factors are (x + 4), (x + 4i ), (x – 4i ).
(x + 4)(x + 4i )(x – 4i ) = (x + 4)(x2 + 16) =
x3 + 4x2 + 16x The leading coefficient is ,
so the equation is x3 + 6x2 + 24x + 96 = 0.
[3] leaves the answer as x3 + 4x2 + 16x + 64 = 0,
OR makes minor errors
[2] identifies (x + 4), (x + 4i ), and (x – 4i ) as factors
[1] identifies the third root as 4i
48. – ,
49. ± 5, ±2i
50. ± 5, ±i 5
51. –1, 7
52. –2 ± 2i
± 5
54. (1, –4)
55. (–6, 2)
56. (5, 0) 3 2
3 ± 3i 3 4
6-5

72. Theorems About Roots of Polynomial Equations
ALGEBRA 2 LESSON 6-5
1. Use the Rational Root Theorem to list all possible rational roots of
3x3 + x2 – 15x – 5 = 0. Then find any actual rational roots.
2. Find the roots of 10x4 + x3 + 7x2 + x – 3 = 0.
3. A polynomial equation with rational coefficients has the roots and –5i.
Find two additional roots.
4. Find a third-degree polynomial equation with integer coefficients that has
the roots 8 and 3i.
±1, ±5, ± , ± ; – 1 3 5
– , , ±i 3 5 1 2
– 7, 5i
x3 – 8x2 + 9x – 72 = 0
6-5

73. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
(For help, go to Lessons 5-8 and 6-1.)
Find the degree of each polynomial.
1. 3x2 – x –x + 3 – x3 3. –4x5 + 1
Solve each equation using the quadratic formula.
4. x = 0 5. x2 – 2x + 3 = x2 + 5x + 4 = 0
6-6

74. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
Solutions
1. 3x2 – x + 5; degree 2
2. –x + 3 – x3; degree 3
3. –4x5 + 1; degree 5
4. x = 0
x = with a = 1, b = 0, and c = 16
x = = = = ± 4i
–b ± b2 – 4ac 2a
– 0 ± – 4(1)(16)
2(1)
0 ± – 64 2
± 8i 2
6-6

75. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
Solutions (continued)
5. x2 – 2x + 3 = 0
x = with a = 1, b = –2, and c = 3
x = = = =
= = 1 ± i 2
6. 2x2 + 5x + 4 = 0
x = with a = 2, b = 5, and c = 4
x = = = =
–b ± b2 – 4ac 2a
– (–2) ± (–2)2 – 4(1)(3)
2(1)
2 ± – 12 2
2 ± – 8 2
2 ± (–1) • 4 • 2 2
2 ± 2i 2 2
–b ± b2 – 4ac 2a
– 5 ± – 4(2)(4)
2(2)
–5 ± – 32 4
–5 ± – 7 4
–5 ± i – 7 4
6-6

76. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
For the equation x4 – 3x3 + 4x + 1 = 0, find the number of complex roots, the possible number of real roots, and the possible rational roots.
By the corollary to the Fundamental Theorem of Algebra,
x4 – 3x3 + 4x + 1 = 0 has four complex roots.
By the Imaginary Root Theorem, the equation has either no imaginary roots,
two imaginary roots (one conjugate pair), or four imaginary roots
(two conjugate pairs). So the equation has either zero real roots,
two real roots, or four real roots.
By the Rational Root Theorem, the possible rational roots of the
equation are ±1.
6-6

77. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
Find the number of complex zeros of
ƒ(x) = x5 + 3x4 – x – 3. Find all the zeros.
By the corollary to the Fundamental Theorem of Algebra, there are five
complex zeros. You can use synthetic division to find a rational zero.
Step 1: Find a rational root from the possible roots of ±1 and ±3.
Use synthetic division to test each possible until you get
a remainder of zero.
– –1 –3 –
–1 0
x4 –1
So –3 is one of the roots.
6-6

78. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
(continued)
Step 2: Factor the expression x4 – 1.
x4 – 1 = (x2 – 1)(x2 + 1)    Factor x2 – 1.
= (x – 1)(x + 1)(x2 + 1)
So 1 and –1 are also roots.
Step 3: Solve x = 0.
x2 + 1 = 0
x2 = –1
x = ± i
So i and –i are also roots.
The polynomial function ƒ(x) = x5 + 3x4 – x – 3 has three real zeros of
x = –3, x = 1, and x = –1, and two complex zeros of x = i, and x = –i.
6-6

79. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
pages 337–338  Exercises
1. 3 complex roots;
number of real roots: 1 or 3
possible rational roots: ±1
2. 2 complex roots;
number of real roots: 0 or 2
possible rational roots:
± , ± , ±1, ±7
3. 4 complex roots;
number of real roots: 0, 2, or 4
possible rational roots: 0
4. 5 complex roots;
number of real roots: 1, 3, or 5
± , ±1, ± , ±5
5. 7 complex roots;
number of real roots: 1, 3, 5, or 7
possible rational roots: ±1, ±3
6. 1 complex root;
number of real roots: 1
± , ± , ±1, ±2, ±4, ±8
7. 6 complex roots;
number of real roots: 0, 2, 4, or 6
± , ±1, ± , ±7
8. 10 complex roots;
number of real roots: 0, 2, 4, 6, 8, or 10 1 3 7 2 5 4
6-6

80. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
9. –1,
10. 3, ±i
11. 4,
12. 2, ± 3
13. ±2, ± 2
14. ±2, ±i
15. 0,
16. –6, ±i
17. 4 complex roots;
number of real roots: 0, 2, or 4
possible rational roots:
± , ±1, ±2, ± , ±13, ±26
1 ± i 7 4
1 ± i 3 2
3 ± 1 13
18. 5 complex roots;
number of real roots: 1, 3, or 5
±1, ±2, ±3, ±6, ±9, ±18
19. 3 complex roots;
number of real roots: 1 or 3
± , ± , ±1, ± , ±2, ±3, ±4, ±6, ±12
20. 6 complex roots;
number of real roots: 0, 2, 4, or 6
± , ± , ± , ±1, ± , ±2, ±3, ±4, ±6,
±8, ±12, ±24
21. 4, ±3i
22. –2, ± 5 3
6-6

81. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
23. –6,
24. – , –1 ± 2i
, ± 2i 5 ,
27. Answers may vary. Sample:
y = x4 + 3x2 + 2
28. ±0.75 1 4 2 5
–1 ± i 6
29. –3.24, 1.24
30. If you have no constant, then all terms
have an x that can be factored out.
The resulting expression will have a
constant that can be used in the
Rational Root Theorem.
31. Yes; for example, 2x2 – 11x + 5
has roots 0.5 and 5.
32. C
33. C
–1 ± i
6-6

82. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
34. A
35. C
36. x4 + 6x3 + 14x2 + 24x + 40 = 0
37. 3 (mult. 2)
38.
39.
–5 ± i 4
3 ± i
6-6

83. The Fundamental Theorem of Algebra
ALGEBRA 2 LESSON 6-6
For each equation, state the number of complex roots, the possible number
of real roots, and the possible rational roots.
1. x4 – 13x3 + x2 – 5 = 0         2. x7 – x5 + x2 – 3x + 3 = 0
Find all the zeros of each function.
3. x3 – 3x2 – 8x – 10 = x3 + 3x2 – 2x – 6 = 0
5. x4 – 29x = 0
4; 0, 2, or 4; ±1, ±5
7; 1, 3, 5, or 7; ±1, ±3
5, –1±i
–3, ± 2
±2, ±5
6-6

84. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
(For help, go to Lessons 6-2 and 6-5.)
Simplify each expression.
1. 10 • 9 • 8 • 7 •           3. 
Let a b = 2a(a + b). Evaluate each expression.
4. 3 • • • • 10
4 • 3 • 2
6 • 5
7 • 6 • 5 • 4 • 3 • 2 • 1
4 • 3 • 2 • 1 *
6-7

85. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
Solutions
1. 10 • 9 • 8 • 7 • 6 = 30,240
2. =
3. = 7 • 6 • 5 = 210
4. 3 • 4 with a • b = 2a(a + b):
2(3)(3 + 4) = 2(3)(7) = 6(7) = 42
5. 2 • 7 with a • b = 2a(a + b):
2(2)(2 + 7) = 2(2)(9) = 4(9) = 36
6. 5 • 1 with a • b = 2a(a + b):
2(5)(5 + 1) = 2(5)(6) = 10(6) = 60
7. 6 • 10 with a • b = 2a(a + b):
2(6)(6 + 10) = 2(6)(16) = 12(16) = 192
4 • 3 • 2
6 • 5
7 • 6 • 5 • 4 • 3 • 2 • 1
4 • 3 • 2 • 1
/ / / 4 5
6-7

86. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
In how many ways can 6 people line up from left to right for a group photo?
Since everybody will be in the picture, you are using all the items from the
original set. You can use the Multiplication Counting Principle or
factorial notation.
There are six ways to select the first person in line, five ways to select the
next person, and so on.
The total number of permutations is 6 • 5 • 4 • 3 • 2 • 1 = 6!.
6! = 720
The 6 people can line up in 720 different orders.
6-7

87. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
How many 4-letter codes can be made if no letter can be used twice?
Method 1:  Use the Multiplication Counting Principle.
26 • 25 • 24 • 23 = 358,800
Method 2:  Use the permutation formula. Since there are 26 letters
arranged 4 at a time, n = 26 and r = 4.
26P4 = = = 358,800
26!
(26 – 4)!
22!
There are 358,800 possible arrangements of 4-letter codes with no duplicates.
6-7

88. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
Evaluate 10C4.
10C4 =
10!
4!(10 – 4)! =
10!
4! • 6!
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
4 • 3 • 2 • 1 • 6 • 5 • 4 • 3 • 2 • 1 =
/ / / / / /
10 • 9 • 8 • 7
4 • 3 • 2 • 1 =
= 210
6-7

89. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
A disk jockey wants to select 5 songs from a new CD that contains 12 songs. How many 5-song selections are possible?
Relate: 12 songs chosen 5 songs at a time
Define: Let n = total number of songs.
Let r = number of songs chosen at a time.
Write: nCr = 12C5
Use the nCr feature of your calculator.
You can choose five songs in 792 ways.
6-7

90. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
A pizza menu allows you to select 4 toppings at no extra charge from a list of 9 possible toppings. In how many ways can you select 4 or fewer toppings?
You may choose 
4 toppings, 3 toppings, 2 toppings, 1 toppings, or none.
9C4 9C3 9C2 9C1 9C0
The total number of ways to pick the toppings is
= 256.
There are 256 ways to order your pizza.
6-7

91. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
pages 342–345  Exercises
1. 120
2. 3,628,800
3. 6,227,020,800
4. 720
5. 665,280
6. 120
7. 120
9. a. 24
b. 120
10. 8
11. 56
14. 6
16. 60,480
17. 60
18. 10,897,286,400
19. 4,151,347,200
20. 12
21. 15
22. 56
23. 1
24. 4
25. 35
26. 15
27. 35
28.
30. 21
33. true because of the
Comm. Prop. of Add. 5 18
6-7

92. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
34. true because of the Assoc. Prop. of Mult.
35. False; answers may vary. Sample:
(3 + 2)! = 120 and 3! + 2! = 8
36. False; answers may vary. Sample:
(3  2)! = 720 and 3!  2! = 12
37. False; answers may vary. Sample:
(3!)! = 720 and (3!)2 = 36
38. False; answers may vary. Sample:
(3!)2 = 36 and 3(2!) = 9
40. 60
41. 2 ways, because order matters
43. 84
44.
45. 5
46. permutation
47. permutation
48. combination
49. combination
,791,175
52. 12,650
53. 5 4
19,393 12 49
6-7

93. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
55. a. 56
b. 56
c. Answers may vary. Sample:
Each time you choose 3 of the
8 points to use as vertices of a ,
the 5 remaining points could be
used to form a pentagon.
59. 24
63. 0
64–67. Check students’ work.
68. a. 2048
b. Answers may vary. Sample:
No, because there are too
many possible solutions.
69. a. The graph for y = xCx–2 is
identical to the graph for
y = xC2 because 2C2–2 = 2C2,
3C3–2 = 3C2, 4C4–2 = 4C2,
5C5–2 = 5C2, etc.
The function is defined only at
discrete whole-number values of x,
and not over a smooth range of
points as in a continuous function.
6-7

94. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
70. a. 35
b. 6
c. 7C3 = , so 7C3 • 3! = ,
which is the permutation
formula for 7P3.
71. a. All the terms contain the factors
2 and 5. Since multiplication is
commutative, 2  5 = 10 and 10
times any integer ends in zero.
b. 24 zeros
72. Answers may vary. Sample: 99
73. a. 18 people
b. 8568
c. 658,008
d or 1.3% 7!
3!4! 4!
75. 1/336
76. 21
79. 47/60
81. 3 complex roots;
number of real roots: 1 or 3
possible rational roots:
± , ± , ± , ± , ± , ± , ±1, ±2
82. 2, ±i 6
83. –x3 – 3x2 + 6; cubic trinomial 1 12 6 4 3 2
6-7

95. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
84. 2x2 – 4x + 8; quadratic trinomial
85. 5t 2 – 3t; quadratic binomial
86. x4 – 100; quartic binomial
87. x3 – 4x; cubic binomial
88. t 4 – 2t 3 + t 2; quartic trinomial
89. 4(x – 1)2
90. –(x + 3)2
91. 3(x – 5)(x + 5)
92. minimum; –12
93. maximum; 4.125
94. minimum; –7
6-7

96. Permutations and Combinations
ALGEBRA 2 LESSON 6-7
Evaluate each expression.
1. 8! C3 4. 9P5
5. There are 12 students waiting to enter a van to go on a trip to the
science museum. In how many orders can they enter the van?
6. You have 8 shopping trips to make this week. You have time to make
any 4 of them today. In how many ways can you select the four that
you make today?
7. You want to hang 6 pictures in a row on a wall. You have 11 pictures
from which to choose. How many picture arrangements are possible?
40,320
14!
10!4!
1001 35
15,120
479,001,600 orders
70 ways
332,640 arrangements
6-7

97. The Binomial Theorem Multiply.
ALGEBRA 2 LESSON 6-8
(For help, go to Lessons 5-1 and 6-7.)
Multiply.
1. (x + 2)2 2. (2x + 3)2 3. (x – 3)3 4. (a + b)4
Evaluate.
5. 5C0 6. 5C1 7. 5C2 8. 5C3 9. 5C4
6-8

98. 2. (2x + 3)2 = (2x)2 + 2(3)(2x) + 32 = 4x2 + 12x + 9
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
Solutions
1. (x + 2)2 = x2 + 2(2)x + 22 = x2 + 4x + 4
2. (2x + 3)2 = (2x)2 + 2(3)(2x) + 32 = 4x2 + 12x + 9
3. (x – 3)3 = (x – 3)(x – 3)2 = (x – 3)(x2 – 2(3)x + 32) =
(x – 3)(x2 – 6x + 9) = x(x2 – 6x + 9) – 3(x2 – 6x + 9) =
x3 – 6x2 + 9x – 3x2 + 18x – 27 = x3 – 9x2 + 27x – 27
4. (a + b)4 = (a + b)2(a + b)2 = (a2 + 2ab + b2)(a2 + 2ab + b2) =
a2(a2 + 2ab + b2) + 2ab(a2 + 2ab + b2) + b2(a2 + 2ab + b2) =
a4 + 2a3b + a2b2 + 2a3b + 4a2b2 + 2ab3 + a2b2 + 2ab3 + b4 =
a4 + 4a3b + 6a2b2 + 4ab3 + b4
6-8

99. Solutions (continued)
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
Solutions (continued)
5. 5C0 = = = = 1
6. 5C1 = = = = = 5
7. 5C2 = = = = = 10
8. 5C3 = = = = = 10
9. 5C4 = = = = = 5
5 • 4 • 3 • 2 • 1
2 • 1 • 3 • 2 • 1 5!
0! (5 – 0)!
2! (5 – 2)!
1! (5 – 1)!
3! (5 – 3)!
4! (5 – 4)!
1 • 5!
1! • 4!
2! • 3!
3! • 2!
4! • 1! 1
1 • 4 • 3 • 2 • 1
/ / / /
/ / / 5 20 2
3 • 2 • 1 • 2 • 1
4 • 3 • 2 • 1 • 1 /
6-8

100. Use Pascal’s Triangle to expand (a + b)5.
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
Use Pascal’s Triangle to expand (a + b)5.
Use the row that has 5 as its second number.
The exponents for a begin with 5 and decrease.
     1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5
The exponents for b begin with 0 and increase.
In its simplest form, the expansion is
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.
6-8

101. Use Pascal’s Triangle to expand (x – 3)4.
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
Use Pascal’s Triangle to expand (x – 3)4.
First write the pattern for raising a binomial to the fourth power.
   Coefficients from
Pascal’s Triangle.
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Since (x – 3)4 = (x + (–3))4, substitute x for a and –3 for b.
(x + (–3))4 = x4 + 4x3(–3) + 6x2(–3)2 + 4x(–3)3 + (–3)4
= x4 – 12x3 + 54x2 – 108x + 81
The expansion of (x – 3)4 is x4 – 12x3 + 54x2 – 108x + 81.
6-8

102. Use the Binomial Theorem to expand (x – y)9.
ALGEBRA 2 LESSON 6-8
Use the Binomial Theorem to expand (x – y)9.
Write the pattern for raising a binomial to the ninth power.
(a + b)9 = 9C0a9 + 9C1a8b + 9C2a7b2 + 9C3a6b3
+ 9C4a5b4 + 9C5a4b5 + 9C6a3b6 + 9C7a2b7
+ 9C8ab8 + 9C9b9
Substitute x for a and –y for b. Evaluate each combination.
(x – y)9 = 9C0x9 + 9C1x8(–y) + 9C2x7(–y)2 + 9C3x6(–y)3
+ 9C4x5(–y)4 + 9C5x4(–y)5 + 9C6x3(–y)6
+ 9C7x2(–y)7 + 9C8x(–y)8 + 9C9(–y)9
= x9 – 9x8y + 36x7y2 – 84x6y x5y4
– 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9
The expansion of (x – y)9 is x9 – 9x8y + 36x7y2 – 84x6y x5y4
– 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9.
6-8

103. Since you want 7 successes (and 5 failures), use the term p7q5.
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
Dawn Staley makes about 90% of the free throws she attempts. Find the probability that Dawn makes exactly 7 out of 12 consecutive free throws.
Since you want 7 successes (and 5 failures), use the term p7q5.
This term has the coefficient 12C5.
Probability (7 out of 10) = 12C5 p7q5
= • (0.9)7(0.1)5 The probability p of success = 90%, or 0.9.
12!
5! •7!
= Simplify.
Dawn Staley has about a 0.4% chance of making exactly 7 out
of 12 consecutive free throws.
6-8

104. The Binomial Theorem pages 349–351 Exercises 1. a3 + 3a2b + 3ab2 + b3
ALGEBRA 2 LESSON 6-8
pages 349–351  Exercises
1. a3 + 3a2b + 3ab2 + b3
2. x2 – 2xy + y2
3. a4 + 4a3b + 6a2b2 + 4ab3 + b4
4. x5 – 5x4y + 10x3y2 – 10x2y3 +
5xy4 – y5
5. a6 – 6a5b + 15a4b2 – 20a3b3 +
15a2b4 – 6ab5 + b6
6. x7 – 7x6y + 21x5y2 – 35x4y3 +
35x3y4 – 21x2y5 + 7xy6 – y7
7. x8 + 8x7y + 28x6y2 + 56x5y3 +
70x4y4 + 56x3y5 + 28x2y6 + 8xy7 + y8
8. d9 + 9d8e + 36d7e2 + 84d6e d5e4 +
126d4e5 + 84d3e6 + 36d2e7 + 9de8 + e9
9. x3 – 9x2 + 27x – 27
10. a4 + 12a3b + 54a2b2 +
108ab3 + 81b4
11. x6 – 12x5 + 60x4 – 160x3 +
240x2 – 192x + 64
12. x8 – 32x x6 – 3584x5 +
17,920x4 – 57,344x3 +
114,688x2 – 131,072x + 65,536
13. x4 + 4x3y + 6x2y2 + 4xy3 + y4
14. w5 + 5w4 + 10w3 +
10w2 + 5w + 1
15. s2 – 2st + t 2
16. x6 – 6x5 + 15x4 – 20x3 +
15x2 – 6x + 1
6-8

105. The Binomial Theorem 17. x4 – 4x3y + 6x2y2 – 4xy3 + y4
ALGEBRA 2 LESSON 6-8
17. x4 – 4x3y + 6x2y2 – 4xy3 + y4
18. p7 + 7p6q + 21p5q2 + 35p4q3 +
35p3q4 + 21p2q5 + 7pq6 + q7
19. x5 – 15x4 + 90x3 – 270x x – 243
– 48x + 12x2 – x3
21. a. about 25%
b. about 21%
c. about 12%
22. about 66%
23. x7 + 7x6y + 21x5y2 + 35x4y3 +
35x3y4 + 21x2y5 + 7xy6 + y7
24. x8 – 40x7y + 700x6y2 – 7000x5y3 +
43,750x4y4 – 175,000x3y5 +
437,500x2y6 – 625,000xy ,625y8
25. 81x4 – 108x3y + 54x2y2 –
12xy3 + y4
26. x5 – 20x4y + 160x3y2 – 640x2y3 +
1280xy4 – 1024y5
,649 – 201,684x + 144,060x2 –
54,880x3 + 11,760x4 –
1344x5 + 64x6
28. 8x3 + 36x2y + 54xy2 + 27y3
29. x4 + 2x2y2 + y4
30. x6 – 6x4y + 12x2y2 – 8y3
31. x6 + 6x5 + 15x4 + 20x3 +
15x2 + 6x + 1
32. x6 – 6x5 + 15x4 – 20x3 +
15x2 – 6x + 1
6-8

106. The Binomial Theorem 33. x5 + 10x4 + 40x3 + 80x2 + 80x + 32
ALGEBRA 2 LESSON 6-8
33. x5 + 10x4 + 40x3 + 80x2 + 80x + 32
34. x5 – 10x4 + 40x3 – 80x2 + 80x – 32
35. 16x4 + 96x3y + 216x2y2 +
216xy3 + 81y4
36. 27x x2y + 225xy y3
37. 64x x5y + 960x4y2 +
1280x3y x2y4 +
384xy5 + 64y6
38. 81x x3y + 216x2y2 + 96xy3 + 16y4
39. 32x5 + 80x4y + 80x3y2 + 40x2y3 +
10xy4 + y5
x x6y x5y2 +
2835x4y x3y4 +
189x2y5 + 21xy6 + y7
41. x6 + 18x5y + 135x4y x3y3 +
1215x2y xy y6
42. x3 + 15x2y + 75xy y3
43. a. about 31%
b. about 16%
c. about 16%
44. a. 6
b. 84
45. 8C4x4y4
46. 9
47. 7, 7r 6s
x10
49. 80x2
6-8

107. 58. Answers may vary. Sample: Since one of the terms is negative
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
50. 27x8
x10
52. x11
53. 64y6
54. –823,680x8y7
,928x7
56. 29,568x10y6
x12y14
58. Answers may vary. Sample:
Since one of the terms is negative
and it is alternately raised to odd
and even powers, the term is
negative when raised to an odd
power and positive when raised
to an even power.
59. a. (s + 0.5)3
b. s s s
60. The exponent of q should be 5 because
the exponent of q should be the degree
(7) minus the exponent of p.
61. 13, d12, 12d11e
62. 16, x15, –15x14y
63. 6, 32a5, 80a4b
64. 8, x7, –21x6y
65. a. –4
b. (1 – i )4 = 1 – 4i – 6 + 4i + 1 = –4 
66. (– • i )3 =
–1 + 3i – 3i = 8  
6-8

108. 68. a. (k + 1)! = (k + 1) • (k) • (k - 1) • . . . • 1 =
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
67. Answers may vary. Sample: A coin is tossed five times with the probability
of heads on each toss 0.5. Write an expression for the probability of exactly
2 heads being tossed.
68. a. (k + 1)! = (k + 1) • (k) • (k - 1) • • 1 =
(k + 1)[(k) • (k - 1) • • 1] = (k + 1) • k!
b. The derivation below finds a common denominator for the fractions that
represent nCk and nCk+1, and then uses algebra to show that nCk + nCk+1 =
n+1Ck+1. In addition, the identity from part (a) is used three times.
nCk + nCk+1 = = =
= = = = n+1Ck+1
c. If you consider the row of Pascal’s Triangle containing just 1 to be row zero,
4C2 is 6, the third entry in the fourth row. 4C3 is 4, the fourth entry in the
fourth row. 5C3 is 10, the fourth entry in the fifth row.
4C2 + 4C3 = = 10 = 5C3 n!
k!(n – k)!
(k + 1)!(n – k – 1)!
(k + 1)n!
(k + 1)!(n – k)!
(n – k)n!
(k n – k)n!
(n + 1)n!
(n + 1)!
(k + 1)!((n + 1) – (k +1))!
6-8

109. [1] provides some expression in x
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
69. D
70. I
71. D
72. G
73. [2] 7C1x6y1 or 7x6y
[1] provides some expression in x
and y without the correct coefficients
74. [4] The sixth term is 7C5(2x)2(–3y)5 =
21(4x2)(–243y5) = –20,412x2y5
[3] incorrect coefficient but correct
exponents for x and y
[2] 7C6(2x)2(–3y)5 correct,
but not the rest of the answer
[1] presents some expression in x and y
75. 20
76. 24
77. 35
78. 39,916,800
79. 35
80. 20
; –12.60; –3, 1, 4
; –32.49; –5, –1, 5
83. y = (x – 3)2 – 7
84. y = (x + 3.5)2 – 13.25
85. y = –4(x – 0)2 + 9
6-8

110. Use Pascal’s Triangle to expand each binomial. 1. (x – y5)3
The Binomial Theorem
ALGEBRA 2 LESSON 6-8
Use Pascal’s Triangle to expand each binomial.
1. (x – y5)3
2. (2a + b)5
3. Use the Binomial Theorem to expand (3x – 2y)4.
4. What is the coefficient of a4b6 in the expansion of (a + b)10?
x3 – 3x2y5 + 3xy10 – y15
32a5 + 80a4b + 80a3b2 + 40a2b3 + 10ab4 + b5
81x4 – 216x3y + 216x2y2 – 96xy3 + 16y4
210
6-8

111. Polynomials and Polynomial Functions
ALGEBRA 2 CHAPTER 6
1. –8x4 + 3x2 + 9; quartic trinomial
2. 8x x; quadratic binomial
3. 2x3 – 2x2 – 12x; cubic trinomial
4. t 3 – 3t – 2; cubic trinomial
5. –1.62, 0, 0.62
6. –5, –2, –1, 1
, –3.04
8. –1.26, 1
9. –0.73, 1, 2.73 3 8
6-A

112. Polynomials and Polynomial Functions
ALGEBRA 2 CHAPTER 6
10. y = x3 – x x –
11. y = x4 + 2x3 – 3x2 – 6x
12. y = x3 + 12x2 + 48x + 64
13. y = x3 – x2 – x + 1
14. y = x4 – x2 – 2
15. y = x4 – 8x3 + 18x2 + 4x – 40
16. Answers may vary. Sample:
Using ±i and ± 2, y = x4 – x2 – 2.
17. –2, – , ,
18. ± 3, –4, 1
19. – ,
20. 0, 1 18 5 19 6 2 3 7 8
–5 ±
, –4
22. x + 4
23. x2 + 5x – 15, R 24
24. 3x – 6, R 10
25. x2 – x + 3, R –20
26. 0
27. 0
28. 0
29. 84
30. 3
32. 15
1 ± 5 2
6-A

113. Polynomials and Polynomial Functions
ALGEBRA 2 CHAPTER 6
33. 35
34. 20
35. 19,958,400
36. 9
37. 7
38. combination, 126
39. combination, 455
40. permutation, 120
41. x5 + 5x4z + 10x3z2 + 10x2z3 + 5xz4 + z5
42. 1 – 4t + 4t 2 %
44. Combinations can be used by
starting with nC0 for the first term.
Then the second term would be nC1,
the third nC2, and so forth until you
reach nCn–1, and finally nCn.
6-A