Download - Polynomial Functions
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
(For help, go to Lesson 1-2.)
Polynomial FunctionsPolynomial Functions
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 – 912
6-1
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
12
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
Solutions
6-1
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
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 polynomial is a quartic trinomial.
The term with the largest degree is x3,so the polynomial is degree 3.
It has two terms.The polynomial is a cubic binomial.
The term with the largest degree is –3x4, so the polynomial is degree 4. It has three terms.
6-1
x y0 2.82 54 66 5.58 4
Using a graphing calculator, determine whether a linear,
quadratic, or cubic model best fits the values in the table.
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
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.
The quadratic model appears to best fit the given values.
Linear model Quadratic model Cubic model
6-1
To estimate the number of employees for 1988, you can use the Table function option of a graphing calculator to find that ƒ(13) 62.72. According to the model, there were about 62 employees in 1988.
The table shows data on the number of employees that a small
company had from 1975 to 2000. Find a cubic function to model the data.
Use it to estimate the number of employees in 1998. Let 0 represent 1975.
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
To find a cubic model, use the CubicReg option of a graphing calculator.
The function ƒ(x) = 0.0096x3 – 0.375x2 + 3.541x + 58.96 is an approximate model for the cubic function.
1975 60
1980 65
1985 70
1990 60
1995 55
2000 64
Number ofEmployeesYear
Enter the data.Graph the model.
6-1
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
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 – 10x2 + 100
17. a. males: y = –0.002571x2 + 0.2829x + 67.21females: y = –0.002286x2 + 0.2514x + 74.82
b. males: y = 0.00008333x3 – 0.007571x2 + 0.3545x + 67.11females: y = 0.00008333x3 – 0.007286x2 + 0.3231x + 74.72
c. The cubic model is a better fit.
6-1
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
18. y = x3 – 2x2; 4335
19. y = x3 – 10x2; 2023
20. y = –0.5x3 + 10x2; 433.5
21. y = –0.03948x3 + 2.069x2 – 17.93x + 106.9; 206.07
22. y = –0.007990x3 + 0.4297x2 – 6.009x + 43.57; 26.34
23. y = 0.01002x3 – 0.3841x2 + 5.002x + 2.132; 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
30. x5 + x; quintic binomial
31. a. V = 10 r 2
b. V = r 3
c. V = r 3 + 10 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
23
12
2323
6-1
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
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; polynomial of 4 terms
43. 80x3 – 109x2 + 7x – 75; polynomial of 4 terms
44. 2x3 – 2x2 + 8x – 27; polynomial of 4 terms
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; polynomial of 4 terms
51. x4 + 2x2 + 1; trinomial
52. 8x3 + 60x2 + 150x + 126; polynomial of 4 terms
53. a3 – a2b – b2a + b3; polynomial of 4 terms
6-1
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
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 = 0.7166x + 47.61y = 0.0009365x3 – 0.0744x2 +
2.2929x + 41.4129y = –0.00004789x4 + 0.004666x3 –
0.1647x2 + 2.9797x + 40.7831
b.
The quartic model fits best.
c. 72.2 1015
6-1
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
60. 2.5 108 cm3
61. a. up 4 unitsb. 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)
70. 1 –3 1 5–8 –3 0 –3
71. 2 –3 –65 –2 0
72. –2 –1 –2 –1–3 –3 –4 –4
6-1
Write each polynomial in standard form. Then classify it by degree and by number of terms.
1. –x2 + 2x + x2 2. 7x2 + 10 + 4x3 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.
ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1
Polynomial FunctionsPolynomial Functions
x 1 2 3 5 8
y 3 2.2 3 7 5
2x; degree 1, linear monomial
2x4 + 12x2; degree 4, quartic binomial
4x3 + 7x2 + 10; degree 3, cubic trinomial
the cubic model
–49.3
y = –0.14401x3 + 1.7805x2 – 5.29x + 6.62,y = –0.1319x2 + 1.682x + 0.37
6-1
Factor each quadratic expression.
1. x2 + 7x + 12 2. x2 + 8x – 20 3. x2 – 14x + 24
Find each product.
4. x(x + 4) 5. (x + 1)2 6. (x – 3)2(x + 2)
(For help, go to Lessons 5-1 and 5-4.)
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
6-2
Solutions
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
1. Factors of 12 with a sum of 7: 4 and 3x2 + 7x + 12 = (x + 4)(x + 3)
2. Factors of –20 with a sum of 8: 10 and –2x2 + 8x – 20 = (x + 10)(x – 2)
3. Factors of 24 with a sum of –14: –12 and –2x2 – 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
Write (x – 1)(x + 3)(x + 4) as a polynomial in standard form.
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
(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
Write 3x3 – 18x2 + 24x in factored form.
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
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
Another airline has different carry-on luggage regulations. The sum of the length, width, and depth may not exceed 50 in.
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
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 + 10 = 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
(continued)
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
b. Describe a realistic domain for V(x).
c. What is the maximum possible volume of the box? What are the corresponding dimensions of the box?
Since the volume must be positive, x < 20. A realistic domain is 0 < x < 20.
Then the length is about 22.2 in. and the width is about 15.6 in.
The function has values over the set of all real numbers x.Since x represents the depth of the luggage, x > 0.
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.
6-2
Find the zeros of y = (x + 1)(x – 1)(x + 3). Then graph the
function using a graphing calculator.
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
Using the Zero Product Property, find a zero for each linear factor.
x + 1 = 0 or x – 1 = 0 or x + 3 = 0x = –1 x = 1 x = –3
The zeros of the function are –1, 1, –3.
Now sketch and label the function.
6-2
Write a polynomial in standard form with zeros at 2, –3, and 0.
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
= (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.
2 –3 0 Zeros
ƒ(x) = (x – 2)(x + 3)(x) Write a linear factor for each zero.
6-2
Find any multiple zeros of ƒ(x) = x5 – 6x4 + 9x3 and state the
multiplicity.
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
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
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
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)
13. 24.2, –1.4, 0, –5, 1
14. 5.0, –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
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
20. –1, 1, 2
21. y = x3 – 18x2 + 107x – 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)
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
12
32
6-2
72
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
40. a. h = x + 3; w = xb.
0 < –3 < 2; where the volume is zeroc. 0 < x < 2d. 4.06 ft3
41. y = –2x3 + 9x2 – x – 12
42. y = 5x4 – 23x3 – 250x2 + 1164x + 504
43. y = 3x(x – 8)(x – 1)
44. y = –2x(x + 5)(x – 4)
45. y = x2(x + 4)(x – 1)
46. y = x x – x +
47. 2.5, 5.1; , 4, 6
48. 0.9, –6.9, –1.4; 0, –3, –1, 1
49. 2.98, –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 – 21x2 + 147x – 343
53. y = x4 – 4x3 – 7x2 + 22x + 24
54. –4, 5 (mult. 3)
55. 0 (mult. 2), –1 (mult. 2)
12
12
12
6-2
32
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
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 + 4xb. 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
6-2
78
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
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
ALGEBRA 2 LESSON 6-2ALGEBRA 2 LESSON 6-2
Polynomials and Linear FactorsPolynomials and Linear Factors
1. Find the zeros of y = (x + 7)(x – 3)(x – 2). Then write the polynomialin 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
Answers may vary. Sample: ƒ(x) = x3 + 6x2 – 7x – 60
The number 0 is a zero with multiplicity 2.
6-2
x(x + 3)(x – 5); , –3640027
(For help, go to Lessons 5-1 and 6-1.)
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
Simplify each expression.
1. (x + 3)(x – 4) + 2 2. (2x + 1)(x – 3)
3. (x + 2)(x + 1) – 11 4. –3(2 – x)(x + 5)
Write each polynomial in standard form. Then list the coefficients.
5. 5x – 2x2 + 9 + 4x3 6. 10 + 5x3 – 9x2
7. 3x + x2 – x + 7 – 2x2 8. –4x4 – 7x2 + x3 + x4
6-3
Solutions
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
1. (x + 3)(x – 4) + 2 = x2 – 4x + 3x – 12 + 2= 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 – 2x2 + 9 + 4x3 = 4x3 – 2x2 + 5x + 9 coefficients: 4, –2, 5, 9
6. 10 + 5x3 – 9x2 = 5x3 – 9x2 + 10 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
Divide x2 + 2x – 30 by x – 5.
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
– 30 Subtract: (x2 + 2x) – (x2 – 5x) = 7x. Bring down –30.
x Divide = x.x – 5 x2 + 2x – 30
x2
x
x2 – 5x Multiply: x(x – 5) = x2 – 5x 7x
Repeat the process of dividing, multiplying, and subtracting.
5 Subtract: (7x – 30) – (7x – 35) = 5.
The quotient is x + 7 with a remainder of 5, or simply x + 7, R 5.
7x – 35 Multiply: 7(x – 5) = 7x – 35.
x + 7 Divide = 7.x – 5 x2 + 2x – 30
x2 – 5x7x – 30
7xx
6-3
Determine whether x + 2 is a factor of each polynomial.
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
a. x2 + 10x + 16 b. x3 + 7x2 – 5x – 6
Since the remainder is zero,x + 2 is a factor of x2 + 10x + 16.
x + 8x + 2 x2 + 10x + 16
x2 + 2x8x + 16 8x + 16
0
x2 + 5x – 15x + 2 x3 + 7x2 – 5x – 6
x3 + 2x2
5x2 – 5x5x2 + 10x
–15x – 6–15x – 30
24
Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6.
/
6-3
Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
Step 1: Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form.
Step 2: Bring down the coefficient.
Write x – 3 5x3 – 6x2 + 4x – 1
as 3 5 –6 4 –1
Bring down the 5.3 5 –6 4 –1 This begins the quotient.
5
6-3
(continued)
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
Multiply 3 by 5. Write3 5 –6 4 –1 the result under –6.x 15
5 9 Add –6 and 15.
Step 3: Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add.
Step 4: Repeat the steps of multiplying and adding until the remainder is found.
The quotient is 5x2 + 9x + 31, R 92.
3 5 –6 4 –115 27 93
5 9 31 92
5x2 + 9x + 31 Remainder
6-3
The volume in cubic feet of a shipping carton is V(x) = x3 – 6x2 + 3x + 10. The height is x – 5 feet.
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
a. Find linear expressions for the other dimensions. Assume that the length is greater than the width.
b. If the width of the carton is 4 feet, what are the other two dimensions?
x2 – x – 2 = (x – 2)(x + 1) Factor the quotient.The length and the width are x + 1 and x – 2, respectively.
x – 2 = 4 Substitute 4 into the expression for width. Find x.x = 6Since the height equals x – 5 and the length equals x + 1, the height is 1 ft. and the length is 7 ft.
5 1 –6 3 10 Divide. 5 –5 –10
1 –1 –2 0
x2 – x – 2 Remainder
6-3
Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3.
The remainder is 21, so P(3) = 21.
3 1 –2 0 1 –9 3 3 9 30
1 1 3 10 21
6-3
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
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
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
24. y = (x + 3)(x – 4)(x – 3)
25. = x + 3 and h = x
26. 18
27. 0
28. 0
29. 12
30. 168
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
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
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
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
60. Yes; the graph could rise to the right and fall to the left or it could fall tothe right and rise to the left.
61. D
62. I
63. B
64. [2] x2 – 6x – 7x – 1 x3 – 7x2 – x + 7 x3 – x2
–6x2 – x –6x2 + 6x
– 7x + 7 – 7x + 7
0
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
71. 23 – 11i
6-3
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
72. –0.5 0 –1.50.5 1 1.50.5 0 0.5
73. none exists
74. 0 2 11 –4 –20.5 –3.5 –1.5
6-3
ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3
Dividing PolynomialsDividing Polynomials
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 findP(–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
(For help, go to Lessons 6-2 and 6-3.)
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
Graph each system. Find any points of intersection.
1. 2. 3.
Factor each expression.
4. x2 – 2x – 15 5. x2 – 9x + 14 6. x2 + 6x + 5
y = 3x + 1y = –2x + 6
–2x + 3y = 0 x + 3y = 3
2y = – x + 8x + 2y = –6
6-4
Solutions
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
1. 2.
intersection: (1, 4) 3. Rewrite equations:
Rewrite equations:
intersection: (1, ) no points of intersection4. 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 + 1y = –2x + 6
–2x + 3y = 0 x + 3y = 3
2y = – x + 8x + 2y = –6 y = x
y = – x + 1
23
13
y = – x + 4
y = – x – 3
1212
6-4
23
Graph and solve x3 – 19x = –2x2 + 20.
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
Step 2: Use the Intersect feature to find thex values at the points of intersection.
Step 1: Graph y1 = x3 – 19x and y2 = –2x2 + 20 on a graphing calculator.
The solutions are –5, –1, and 4.
6-4
(continued)
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
Check: Show that each solution makes the original equation a true statement.
x3 – 19x = –2x2 + 20 x3 – 19x = –2x2 + 20 (–5)3 – 19(–5) –2(–5)2 + 20 (–1)3 – 19(–1) –2(–1)2 + 20
–125 + 95 –50 + 20 –1 + 19 –2 + 20–30 = –30 18 = 18
x3 – 19x = –2x2 + 20(4)3 – 19(4) –2(4)2 + 20
64 – 76 –32 + 20–12 = –12
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.
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
V = l • w • h Write the formula for volume.
27475.2 = (x + 4)x(x – 3) Substitute.
15.9 ft3 • = 27475.2 in.3 Convert the volume to cubic inches.123 in.3
ft3
The dimensions of the doghouse are about 30 in. by 27 in. by 34 in.
6-4
Graph y1 = 27475.2 and y2 = (x + 4)x(x – 3). Use the Intersect option of the calculator.When y = 27475.2, x 30. So x – 3 27and x + 4 34.
Factor x3 – 125.
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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
Solve 8x3 + 125 = 0. Find all complex roots.
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
8x3 + 125 = (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.
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
Since 2x + 5 is a factor, x = – is a root.52
(continued)
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
x = Quadratic Formula
= Substitute 4 for a, –10 for b, and 25 for c.
–b ± b2 – 4ac2a
–(–10) ± (–10)2 – 4(4)(25)2(4)
= Use the Order of Operations. – (–10) ± –300
8
6-4
= –1 = 110 ± 10i 3
8
= Simplify.5 ± 5i 3
4
The solutions are – and .52
5 ± 5i 34
Factor x4 – 6x2 – 27.
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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
Solve x4 – 4x2 – 45 = 0.
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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 5 Solve for x, and simplify.
6-4
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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
10. 4.8%
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,
19. ,
20. – ,
21. (x2 – 7)(x – 1)(x + 1)
22. (x2 + 10)(x2 – 2)
–3 ± 3i 32
–5 ± 5i 32
1 ± i 32
1 ± i 34
12
12
–1 ± i 34
6-4
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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
36. 1.71, 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. – ,
43. ,
44. ±2 2, ±2i 2
45. ±5, ±i 2
3 ± 3i 35
–2 ± 2i 33
65
43
6-4
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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
32
12
26510
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.
32
52
6-4
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
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. 10b. 8 and 12
72. A
73. H74. [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
–6 ± 6i 38
32
–3 ± 3i 34
–6 ± 36 – 1448
–6 ± 1088
18
a3
b6
a b2
3 312
6-4
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
76. x2 – 3x – 10
77. 2x2 + x – 3, R 2
78. –2, 6
79. ±6
80. – , 3
81. 0
–5
82. no unique solution
12
6-4
ALGEBRA 2 LESSON 6-4ALGEBRA 2 LESSON 6-4
Solving Polynomial EquationsSolving Polynomial Equations
1. Solve x3 – 2x2 – 3 = x – 4 by graphing. Where necessary, round to the nearest hundredth.
Factor each expression.
2. 216x3 – 1
3. 8x3 + 125
4. x4 – 5x2 + 4
Solve each equation.
5. x3 + 125 = 0 6. 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)
±2, ±i 7
6-4
5 ± 5i 34
–5,
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
(For help, go to Lessons 1-1, 5-1, and 5-6.)
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
List all the factors of each number.
1. 12 2. 24 3. 36 4. 48
Multiply.
5. (x – 5)(x2 + 7) 6. (x + 2)(x + 3)(x – 3)
Define each set of numbers.
7. rational 8. irrational 9. imaginary
6-5
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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 + 3)(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.
ab /
/
6-5
Find the rational roots of 3x3 – x2 – 15x + 5.
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
Step 1: List the possible rational roots.
6-5
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 5factors 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, ± .53
13
(continued)
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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
//
1: 3(1)3 – (1)2 – 15(1) + 5 = –8 = 0–1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 = 0
//
6-5
: 3 3 – 2 – 15 + 5 = –8.8 = 0
: 3 3 – 2 – 15 + 5 = –13.3 = 0
/
/
5353
( )53
( )53
( )53
– ( )53
– ( )53
– ( )53
–
13 ( )1
3– ( )1
3– ( )1
3–
: 3 3 – 2 – 15 + 5 = 0 So is a root.
: 3 3 – 2 – 15 + 5 = 9.7 = 0/
13
( )13
( )13
( )13
13
–
The only rational root of 3x3 – x2 – 15x + 5 = 0 is .13
Find the roots of 5x3 – 24x2 + 41x – 20 = 0.
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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 – 20factors 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 ±20.15
25
45
6-5
(continued)
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
Step 2: Test each possible rational root until you find a root.
Test : 5 3 – 24 2 ± 41 – 20 = –12.72 = 0
Test – : 5 3 – 24 2 ± 41 – 2 = –29.2 = 0
Test : 5 3 – 24 2 ± 41 – 20 = –7.12 = 0
Test – : 5 3 – 24 2 ± 41 – 20 = –40.56 = 0
Test : 5 3 – 24 2 ± 41 – 20 = 0 So is a root.
( )15
15
15
25
25
45
( )25
( )45
( )15
( )25
( )45
( )45
( )25
( )15
(– )15
(– )15
(– )15
(– )25
(– )25
(– )25
45
/
/
/
/
Step 3: Use synthetic division with the root you found in Step 2to find the quotient.
5 –24 41 –20 4 –16 205 –20 25 0
5x2 – 20x + 25 Remainder
45
6-5
(continued)
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
Step 4: Find the roots of 5x2 – 20x + 25 = 0.
5x2 – 20x + 25 = 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 – 4ac2a
–(–4) ± (–4)2 – 4(1)(5)2(1)
= Use order of operations.
= –1 = i.
= 2 ± i Simplify.
4 ± –42
4 ± 2i 2
6-5
The roots of 5x3 – 24x2 + 41x – 20 = 0 are , 2 + i, and 2 – i.45
A polynomial equation with rational coefficients has the roots 2 – 5 and 7 . Find two additional roots.
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
By the Irrational Root Theorem, if 2 – 5 is a root, then its conjugate 2 + 5 is also a root.
If 7 is a root, then its conjugate – 7 also is a root.
6-5
A polynomial equation and real coefficients has the roots 2 + 9i with 7i. Find two additional roots.
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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
Find a third degree polynomial with rational coefficients that has roots –2, and 2 – i.
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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
(continued)
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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 + 4 + 1) (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
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
6-5
12
12
1 ± 32
1 ± 73
72
12. 1, –2,
13. – 5, 13
14. 4 + 6, – 3
15. 1 + 10, 2 – 2
16. 1 – i, 5i
17. 2 – 3i, –6i
18. 4 + 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
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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
16
14
12
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 5 and – 5 are conjugates, they can be roots of a polynomial equation with integer coefficients.
13
23
34
32
12
32
23
1 10
15
25
12
45
52
25
52
73
16
12
13
76
32
72
212
13
72
14
54
12
34
18
58
38
154
52
32
158
152
12
32
52
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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 – 2 = 0b. x2 – 2(1 + 2 )x + (1 + 2 )2 = 0c. –1
6-5
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
6-5
44. D
45. H
46. [2] p is a factor of 3, so p = ±1 or ±3q 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 + 64. 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
53. ± 5
54. (1, –4)
55. (–6, 2)
56. (5, 0)
32
32
3 ± 3i 34
32
32
ALGEBRA 2 LESSON 6-5ALGEBRA 2 LESSON 6-5
Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations
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 7 and –5i. Find two additional roots.
4. Find a third-degree polynomial equation with integer coefficients that has the roots 8 and 3i.
– 7, 5i
x3 – 8x2 + 9x – 72 = 0
±1, ±5, ± , ± ; –13
53
13
– , , ±i35
12
6-5
(For help, go to Lessons 5-8 and 6-1.)
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
Find the degree of each polynomial.
1. 3x2 – x + 5 2. –x + 3 – x3 3. –4x5 + 1
Solve each equation using the quadratic formula.
4. x2 + 16 = 0 5. x2 – 2x + 3 = 0 6. 2x2 + 5x + 4 = 0
6-6
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
Solutions
1. 3x2 – x + 5; degree 2
2. –x + 3 – x3; degree 3
3. –4x5 + 1; degree 5
4. x2 + 16 = 0
x = with a = 1, b = 0, and c = 16
x = = = = ± 4i
–b ± b2 – 4ac2a
– 0 ± 02 – 4(1)(16)2(1)
0 ± 0 – 642
± 8i 2
6-6
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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 – 4ac2a
– (–2) ± (–2)2 – 4(1)(3)2(1)
2 ± 4 – 122
–b ± b2 – 4ac2a
2 ± – 82
2 ± (–1) • 4 • 22
2 ± 2i 22
– 5 ± 52 – 4(2)(4)2(2)
–5 ± 25 – 324
–5 ± – 74
–5 ± i – 74
Solutions (continued)
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.
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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
Find the number of complex zeros of ƒ(x) = x5 + 3x4 – x – 3. Find all the zeros.
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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.
–3 1 3 0 0 –1 –3–3 0 0 0 3
1 0 0 0 –1 0
x4 –1
So –3 is one of the roots.
6-6
(continued)
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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 x2 + 1 = 0. x2 + 1 = 0
x2 = –1 x = ± iSo 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
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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 2possible rational roots: ± , ± , ±1, ±7
3. 4 complex roots; number of real roots: 0, 2, or 4possible rational roots: 0
4. 5 complex roots; number of real roots: 1, 3, or 5possible rational roots: ± , ±1, ± , ±5
5. 7 complex roots; number of real roots: 1, 3, 5, or 7possible rational roots: ±1, ±3
6. 1 complex root; number of real roots: 1possible rational roots: ± , ± , ±1, ±2, ±4, ±8
7. 6 complex roots; number of real roots: 0, 2, 4, or 6possible rational roots: ± , ±1, ± , ±7
8. 10 complex roots; number of real roots: 0, 2, 4, 6, 8, or 10possible rational roots: ±1
13
73
12
52
14
12
12
72
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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 4possible rational roots: ± , ±1, ±2, ± , ±13, ±26
1 ± i 74
1 ± i 32
3 ± 3 52
12
132
18. 5 complex roots; number of real roots: 1, 3, or 5possible rational roots: ±1, ±2, ±3, ±6, ±9, ±18
19. 3 complex roots; number of real roots: 1 or 3possible rational roots: ± , ± , ±1, ± , ±2, ±3, ±4, ±6, ±12
20. 6 complex roots; number of real roots: 0, 2, 4, or 6possible rational roots: ± , ± , ± , ±1, ± , ±2, ±3, ±4, ±6,
±8, ±12, ±24
21. 4, ±3i
22. –2, ± 5
13
23
43
14
12
34
32
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
6-6
23. –6,
24. – , –1 ± 2i
25. , ± 2i 5
26. ,
27. Answers may vary. Sample: y = x4 + 3x2 + 2
28. ±0.75
14
1225
–1 ± i 116
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 2
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
6-6
34. A
35. C
36. x4 + 6x3 + 14x2 + 24x + 40 = 0
37. 3 (mult. 2)
38.
39.
–5 ± i 474
3 ± i 234
ALGEBRA 2 LESSON 6-6ALGEBRA 2 LESSON 6-6
The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra
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 = 0 4. x3 + 3x2 – 2x – 6 = 0
5. x4 – 29x2 + 100 = 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
(For help, go to Lessons 6-2 and 6-5.)
ALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
Permutations and CombinationsPermutations and Combinations
6-7
Simplify each expression.
1. 10 • 9 • 8 • 7 • 6 2. 3.
Let a b = 2a(a + b). Evaluate each expression.
4. 3 • 4 5. 2 • 7 6. 5 • 1 7. 6 • 10
4 • 3 • 26 • 5
7 • 6 • 5 • 4 • 3 • 2 • 14 • 3 • 2 • 1
*
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
Solutions
6-7
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 • 26 • 5
7 • 6 • 5 • 4 • 3 • 2 • 14 • 3 • 2 • 1
/ // /
/ //
45
/ // /
In how many ways can 6 people line up from left to right for a
group photo?
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
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
How many 4-letter codes can be made if no letter can be used twice?
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
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.
There are 358,800 possible arrangements of 4-letter codes with no duplicates.
26P4 = = = 358,800 26! (26 – 4)!
26!22!
6-7
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
Evaluate 10C4.
10C4 = 10! 4!(10 – 4)!
= 10! 4! • 6!
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 14 • 3 • 2 • 1 • 6 • 5 • 4 • 3 • 2 • 1= / / / / / /
/ / / / / /
10 • 9 • 8 • 74 • 3 • 2 • 1=
= 210
6-7
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 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 timeDefine: Let n = total number of songs.
Let r = number of songs chosen at a time.
Write: nCr = 12C5
You can choose five songs in 792 ways.
Use the nCr feature of your calculator.
6-7
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 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?
The total number of ways to pick the toppings is 126 + 84 + 36 + 9 + 1 = 256.
There are 256 ways to order your pizza.
You may choose 4 toppings, 3 toppings, 2 toppings, 1 toppings, or none.
9C4 9C3 9C2 9C1 9C0
6-7
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
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
8. 3003
9. a. 24b. 120
10. 8
11. 56
12. 336
13. 1680
14. 6
15. 120
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.
29. 4368
30. 21
31. 2600
32. 126
33. true because of the Comm. Prop. of Add.
5 18
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
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
39. 3125
40. 60
41. 2 ways, because order matters
42. 0.000206
43. 84
44.
45. 5
46. permutation
47. permutation
48. combination
49. combination
50. 330,791,175
51. 210
52. 12,650
53. 5
4 19,393
1249
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
54. 9504
55. a. 56b. 56c. 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.
56. 120
57. 3024
58. 360
59. 24
60. 1680
61. 840
62. 5040
63. 0
64–67. Check students’ work.
68. a. 2048b. 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.
b. Answers may vary. Sample: 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
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
6-7
70. a. 35b. 6c. 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 peopleb. 8568c. 658,008d. 0.013 or 1.3%
7! 3!4!
7!4!
74. 5040
75. 1/336
76. 21
77. 118
78. 351
79. 47/60
80. 600
81. 3 complex roots; number of real roots: 1 or 3possible rational roots: ± , ± , ± , ± , ± , ± , ±1, ±2
82. 2, ±i 6
83. –x3 – 3x2 + 6; cubic trinomial
1 12
16
14
13
12
23
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 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
Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7
Evaluate each expression.
1. 8! 2. 3. 7C3 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?
14!10!4!
40,320 1001 35 15,120
479,001,600 orders
70 ways
332,640 arrangements
6-7
(For help, go to Lessons 5-1 and 6-7.)
ALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
The Binomial TheoremThe Binomial Theorem
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
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 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
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
6-8
Solutions (continued)
5. 5C0 = = = = 1
6. 5C1 = = = = = 5
7. 5C2 = = = = = 10
8. 5C3 = = = = = 10
9. 5C4 = = = = = 5
5 • 4 • 3 • 2 • 12 • 1 • 3 • 2 • 1
5! 0! (5 – 0)!
5! 2! (5 – 2)!
5! 1! (5 – 1)!
5! 3! (5 – 3)!
5! 4! (5 – 4)!
5! 1 • 5!
5! 1! • 4!
5! 2! • 3!
5! 3! • 2!
5! 4! • 1!
11
5 • 4 • 3 • 2 • 11 • 4 • 3 • 2 • 1
/ / / /
/ / /
51
/ / /202
202
51
5 • 4 • 3 • 2 • 13 • 2 • 1 • 2 • 1
5 • 4 • 3 • 2 • 14 • 3 • 2 • 1 • 1
/ / // / /
/ / / /
/ / / // / / /
//
Use Pascal’s Triangle to expand (a + b)5.
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
Use the row that has 5 as its second number.
In its simplest form, the expansion is a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.
The exponents for a begin with 5 and decrease.
1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5
The exponents for b begin with 0 and increase.
6-8
Use Pascal’s Triangle to expand (x – 3)4.
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
First write the pattern for raising a binomial to the fourth power.
1 4 6 4 1 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
The Binomial TheoremThe Binomial Theorem
Use the Binomial Theorem to expand (x – y)9.
ALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
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 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9
The expansion of (x – y)9 is x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4
– 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9.
6-8
The Binomial TheoremThe Binomial Theorem
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.
ALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
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
= 0.0037881114 Simplify.
Dawn Staley has about a 0.4% chance of making exactly 7 out of 12 consecutive free throws.
= • (0.9)7(0.1)5 The probability p of success = 90%, or 0.9. 12! 5! •7!
6-8
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 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 + 84d6e3 + 126d5e4 + 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 – 32x7 + 448x6 – 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
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
17. x4 – 4x3y + 6x2y2 – 4xy3 + y4
18. p7 + 7p6q + 21p5q2 + 35p4q3 + 35p3q4 + 21p2q5 + 7pq6 + q7
19. x5 – 15x4 + 90x3 – 270x2 + 405x – 243
20. 64 – 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 – 625,000xy7 + 390,625y8
25. 81x4 – 108x3y + 54x2y2 – 12xy3 + y4
26. x5 – 20x4y + 160x3y2 – 640x2y3 + 1280xy4 – 1024y5
27. 117,649 – 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
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 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. 27x3 + 135x2y + 225xy2 + 125y3
37. 64x6 + 384x5y + 960x4y2 + 1280x3y3 + 960x2y4 + 384xy5 + 64y6
38. 81x4 + 216x3y + 216x2y2 + 96xy3 + 16y4
39. 32x5 + 80x4y + 80x3y2 + 40x2y3 + 10xy4 + y5
40. 2187x7 + 5103x6y + 5103x5y2 + 2835x4y3 + 945x3y4 + 189x2y5 + 21xy6 + y7
41. x6 + 18x5y + 135x4y2 + 540x3y3 + 1215x2y4 + 1458xy5 + 729y6
42. x3 + 15x2y + 75xy2 + 125y3
43. a. about 31%b. about 16%c. about 16%
44. a. 6b. 84
45. 8C4x4y4
46. 9
47. 7, 7r 6s
48. 594x10
49. 80x2
6-8
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
50. 27x8
51. 264x10
52. x11
53. 64y6
54. –823,680x8y7
55. 314,928x7
56. 29,568x10y6
57. 1716x12y14
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. s3 + 1.5s2 + 0.75s + 0.125
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. –4b. (1 – i )4 = 1 – 4i – 6 + 4i + 1 = –4
66. (–1 + 3 • i )3 = –1 + 3i 3 + 9 – 3i 3 = 8
6-8
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8
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 = 6 + 4 = 10 = 5C3
n! k!(n – k)!
n! (k + 1)!(n – k – 1)!
(k + 1)n! (k + 1)!(n – k)!
(n – k)n! (k + 1)!(n – k)!
(k + 1 + n – k)n!(k + 1)!(n – k)!
(n + 1)n! (k + 1)!(n – k)!
(n + 1)! (k + 1)!(n – k)!
(n + 1)! (k + 1)!((n + 1) – (k +1))!
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 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
81. 20.75; –12.60; –3, 1, 4
82. 65.67; –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
The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 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
ALGEBRA 2 CHAPTER 6ALGEBRA 2 CHAPTER 6
Polynomials and Polynomial FunctionsPolynomials and Polynomial Functions
1. –8x4 + 3x2 + 9; quartic trinomial
2. 8x2 + 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
7. 0.65, –3.04
8. –1.26, 1
9. –0.73, 1, 2.73
6-A
38
ALGEBRA 2 CHAPTER 6ALGEBRA 2 CHAPTER 6
Polynomials and Polynomial FunctionsPolynomials and Polynomial Functions
21. , –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
31. 720
32. 15
1 ± 52
6-A
10. y = x3 – x2 + 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
185
195
65
23
78
32
23
–5 ± 212
ALGEBRA 2 CHAPTER 6ALGEBRA 2 CHAPTER 6
Polynomials and Polynomial FunctionsPolynomials and Polynomial Functions
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
43. 1.7%
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.
45. 2 + 2
6-A