5-5 Solving Polynomial Equations
Factor completely. If the polynomial is not factorable, write prime .
2. 2k x + 4mx – 2nx – 3k y – 6my + 3ny
SOLUTION: 3
4. 16g + 2h
3
SOLUTION: 2
2
2
2
2
6. 3w + 5x – 6y + 2z + 7a – 9b
2
SOLUTION: There is no GCF and the polynomial cannot be factored by quadratic or cubic methods so it is prime
3
3
3
5
4
8. x y2 – 8x y + 16x + y – 8y + 16y
3
SOLUTION: 10. 6bx + 12cx + 18dx – by – 2cy – 3dy
SOLUTION: Solve each equation.
3
12. x – 64 = 0
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SOLUTION: Page 1
5-5 Solving Polynomial Equations
Solve each equation.
3
12. x – 64 = 0
SOLUTION: Use the quadratic formula to factor
. This gives a solutions of
Therefore, the solution are x = 4,
4
.
.
2
14. x – 33x + 200 = 0
SOLUTION: 2
Let y = x .
Therefore, the solutions are 5, –5,
.
Write each expression in quadratic form, if possible.
6
3
16. 4x – 2x + 8
SOLUTION: 3
Let y = (2x )
Solve each equation.
4
2
18. x – 6x + 8 = 0
SOLUTION: Let y = x
2
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Page 2
3
Let y = (2x )
5-5 Solving Polynomial Equations
Solve each equation.
4
2
18. x – 6x + 8 = 0
SOLUTION: Let y = x
2
Therefore, the solutions are
.
Factor completely. If the polynomial is not factorable, write prime .
3
20. 8c – 27d
3
SOLUTION: 8
2 6
22. a – a b
SOLUTION: 6
24. 18x + 5y
6
SOLUTION: Page 3
The polynomial cannot be factored using the sum of two cubes pattern or quadratic methods. Therefore it is prime.
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2
2
2
2
2
2
5-5 Solving Polynomial Equations
6
24. 18x + 5y
6
SOLUTION: The polynomial cannot be factored using the sum of two cubes pattern or quadratic methods. Therefore it is prime.
2
2
2
2
2
2
26. gx – 3hx – 6fy – gy + 6fx + 3hy
SOLUTION: 3 2
3
3
3 2
3
28. a x – 16a x + 64a – b x + 16b x – 64b
3
SOLUTION: Solve each equation.
4
2
30. x + x – 90 = 0
SOLUTION: 2
Let y = x .
Therefore,
solutions
are
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by Cognero
4
2
32. x – 7x – 44 = 0
.
Page 4
5-5 Solving Polynomial Equations
Solve each equation.
4
2
30. x + x – 90 = 0
SOLUTION: 2
Let y = x .
Therefore, the solutions are
4
.
2
32. x – 7x – 44 = 0
SOLUTION: 2
Let y = x .
Therefore, the solutions are
.
3
34. x + 216 = 0
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Page 5
5-5 Solving Polynomial Equations
Therefore, the solutions are
.
3
34. x + 216 = 0
SOLUTION: Therefore, the solutions are
.
Write each expression in quadratic form, if possible.
4
2
36. x + 12x – 8
SOLUTION: 2
Let y = x .
6
3
38. 8x + 6x + 7
SOLUTION: 3
Let y = 2x .
8
4
40. 9x – 21x + 12
SOLUTION: 4
Let y = 3x .
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Solve each equation.
4
2
42. x + 6x + 5 = 0
Page 6
Let y = 2x .
5-5 Solving Polynomial Equations
8
4
40. 9x – 21x + 12
SOLUTION: 4
Let y = 3x .
Solve each equation.
4
2
42. x + 6x + 5 = 0
SOLUTION: 2
Let y = x .
The solutions are
4
.
2
44. 4x – 14x + 12 = 0
SOLUTION: 2
Let y = 2x .
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The solutions are
Page 7
.
5-5 Solving
Polynomial
Equations
The solutions
are
.
4
2
44. 4x – 14x + 12 = 0
SOLUTION: 2
Let y = 2x .
The solutions are
4
.
2
46. 4x – 5x – 6 = 0
SOLUTION: 2
Let y = x .
The solutions are
.
48. ZOOLOGY A species of animal is introduced to a small island. Suppose the population of the species is represented
4
2
by P(t) = –t + 9t + 400, where t is the time in years. Determine when the population becomes zero.
SOLUTION: eSolutions
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Substitute 0 for P(t) and solve for t.
Page 8
5-5 Solving
Polynomial
Equations
The solutions
are
.
48. ZOOLOGY A species of animal is introduced to a small island. Suppose the population of the species is represented
4
2
by P(t) = –t + 9t + 400, where t is the time in years. Determine when the population becomes zero.
SOLUTION: Substitute 0 for P(t) and solve for t.
2
Let x = t .
Therefore, x = 25 or –16.
2
–16 is irrelevant because t cannot be negative.
2
So, t = 25
–5 is irrelevant because t cannot be negative.
Therefore, the population becomes zero in 5 years.
Factor completely. If the polynomial is not factorable, write prime.
6
50. x – 64
SOLUTION: 5 2
2 5
52. 8x y – 27x y
SOLUTION: 2 2
2 2
2 2
2 3
2 3
2 3
2 2
2 2
2 2
54. 6a x – 24b x + 18c x – 5a y + 20b y – 15c y + 2a z – 8b z + 6c z
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SOLUTION: Page 9
5-5 Solving Polynomial Equations
2 2
2 2
2 2
2 3
4
3
2 3
2 3
2 2
2 2
2 2
54. 6a x – 24b x + 18c x – 5a y + 20b y – 15c y + 2a z – 8b z + 6c z
SOLUTION: 6
5
4
56. 20x – 7x – 6x – 500x + 175x + 150x
2
SOLUTION: Solve each equation.
4
2
58. 8x + 10x – 3 = 0
SOLUTION: 2
Let u = 2x .
The solutions are
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4
2
60. 20x – 53x + 18 = 0
.
Page 10
5-5 Solving Polynomial Equations
Solve each equation.
4
2
58. 8x + 10x – 3 = 0
SOLUTION: 2
Let u = 2x .
The solutions are
4
.
2
60. 20x – 53x + 18 = 0
SOLUTION: 2
Let u = x .
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Page 11
The solutions
are
.
5-5 Solving
Polynomial
Equations
4
2
60. 20x – 53x + 18 = 0
SOLUTION: 2
Let u = x .
The solutions are
4
.
2
62. 8x – 18x + 4 = 0
SOLUTION: 2
Let u = x .
The solutions are
6
.
3
64. x + 7x – 8 = 0
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SOLUTION: Page 12
5-5 Solving
Polynomial
Equations
The solutions
are
.
6
3
64. x + 7x – 8 = 0
SOLUTION: 3
Let u = x .
Solve each equation for x.
The solutions are 1, –2,
6
, and
3
66. 8x + 999x = 125
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3
Let u = x .
Page 13
5-5 Solving
Polynomial
Equations , and
The solutions
are 1, –2,
6
3
66. 8x + 999x = 125
SOLUTION: 3
Let u = x .
Solve each equation for x.
The solutions are –5,
6
4
,
, and
.
2
68. x – 9x – x + 9 = 0
SOLUTION: Manual - Powered by Cognero
eSolutions
2
Let u = x .
Page 14
5-5 Solving
Polynomial
The solutions
are –5, Equations
,
6
4
, and
.
2
68. x – 9x – x + 9 = 0
SOLUTION: 2
Let u = x .
By Zero Product Property:
The solutions are
.
70. CCSS SENSE-MAKING A rectangular prism with dimensions x – 2, x – 4, and x – 6 has a volume equal to 40x
cubic units.
a. Write out a polynomial equation using the formula for volume.
b. Use factoring to solve for x.
c. Are any values for x unreasonable? Explain.
d. What are the dimensions of the prism?
SOLUTION: a. The volume of the prism is
.
Therefore,
b. Solve for x.
By Zero Product Property:
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solutions
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and 12.
c. Sample answer: ±2i because they are imaginary numbers.
Page 15
5-5 Solving
Polynomial Equations
The solutions are
.
70. CCSS SENSE-MAKING A rectangular prism with dimensions x – 2, x – 4, and x – 6 has a volume equal to 40x
cubic units.
a. Write out a polynomial equation using the formula for volume.
b. Use factoring to solve for x.
c. Are any values for x unreasonable? Explain.
d. What are the dimensions of the prism?
SOLUTION: a. The volume of the prism is
.
Therefore,
b. Solve for x.
By Zero Product Property:
The solutions are
and 12.
c. Sample answer: ±2i because they are imaginary numbers.
d. x – 2 =12 – 2 = 10
x – 4 = 12 – 4 = 8
x – 6 = 12 – 6 = 6
The dimensions are 6, 8 and 10 units.
3
2
72. BIOLOGY During an experiment, the number of cells of a virus can be modeled by P(t) = –0.012t – 0.24t + 6.3t
+ 8000, where t is the time in hours and P is the number of cells. Jack wants to determine the times at which there
are 8000 cells.
a. Solve for t by factoring.
b. What method did you use to factor?
c. Which values for t are reasonable and which are unreasonable? Explain.
d. Graph the function for
using your calculator.
SOLUTION: a. Substitute 8000 for P(t) and solve for x.
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d. x – 2 =12 – 2 = 10
x – 4 = 12 – 4 = 8
x – 6 = 12 – 6 = 6
5-5 Solving
Polynomial Equations
The dimensions are 6, 8 and 10 units.
3
2
72. BIOLOGY During an experiment, the number of cells of a virus can be modeled by P(t) = –0.012t – 0.24t + 6.3t
+ 8000, where t is the time in hours and P is the number of cells. Jack wants to determine the times at which there
are 8000 cells.
a. Solve for t by factoring.
b. What method did you use to factor?
c. Which values for t are reasonable and which are unreasonable? Explain.
d. Graph the function for
using your calculator.
SOLUTION: a. Substitute 8000 for P(t) and solve for x.
Therefore, t = 0, 15 and –35.
b. Sample answer: Subtract 8000 from both sides. Then convert the decimals to integers and factor out 120t, then
factor the remaining trinomial.
c. 15 and 0 are reasonable, and –35 is unreasonable because time cannot be negative.
d.
3
2
74. BIOLOGY A population of parasites in an experiment can be modeled by f (t) = t + 5t – 4t – 20, where t is the
time in days.
a. Use factoring by grouping to determine the values of t for which f (t) = 0.
b. At what times does the population reach zero?
c. Are any of the values of t unreasonable? Explain.
SOLUTION: a. Substitute 0 for x and solve for x.
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Therefore, the values of t are 2, –2 and –5.
b. The population reaches zero in 2, –2 and –5 days.
Page 17
5-5 Solving Polynomial Equations
3
2
74. BIOLOGY A population of parasites in an experiment can be modeled by f (t) = t + 5t – 4t – 20, where t is the
time in days.
a. Use factoring by grouping to determine the values of t for which f (t) = 0.
b. At what times does the population reach zero?
c. Are any of the values of t unreasonable? Explain.
SOLUTION: a. Substitute 0 for x and solve for x.
Therefore, the values of t are 2, –2 and –5.
b. The population reaches zero in 2, –2 and –5 days.
c. –2 and –5 are unreasonable because time cannot be negative.
Factor completely. If the polynomial is not factorable, write prime.
9
6
6
3
3
76. y – y – 2y + 2y + y – 1
SOLUTION: 3
Let x = y .
78. CCSS SENSE-MAKING Fredo’s corral, an enclosure for livestock, is currently 32 feet by 40 feet. He wants to
enlarge the area to 4.5 times its current area by increasing the length and width by the same amount.
a. Draw a diagram to represent the situation.
b. Write a polynomial equation for the area of the new corral. Then solve the equation by factoring.
c. Graph the function.
d. Which solution is irrelevant? Explain.
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a.
Page 18
enlarge the area to 4.5 times its current area by increasing the length and width by the same amount.
a. Draw a diagram to represent the situation.
b. Write a polynomial equation for the area of the new corral. Then solve the equation by factoring.
5-5 Solving
c. GraphPolynomial
the function.Equations
d. Which solution is irrelevant? Explain.
SOLUTION: a.
2
b. The area of the corral is 32 × 40 = 1280 ft .
The area of the new corral in terms of x is
.
2
The area of the new corral is 1280 ×4.5 = 5760 ft .
Solve for x.
Therefore, x = –56 or 20.
c.
d. –56 is irrelevant because length cannot be negative.
80. CHALLENGE Solve
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Let
.
Page 19
5-5 Solving
Polynomial Equations
d. –56 is irrelevant because length cannot be negative.
80. CHALLENGE Solve
SOLUTION: Let
.
By Zero Product Property:
The solutions are
.
3
2
82. OPEN ENDED The cubic form of an equation is ax + bx + cx + d = 0. Write an equation with degree 6 that can
be written in cubic form
SOLUTION: 2 3
Sample answer: A polynomial with degree 6 that can be written in cubic form will have a first term in the form (x ) .
2 2
Then the second term will be in the form of (x ) . 6
4
2
2 3
2 2
2
12x + 6x + 8x + 4 = 12(x ) + 6(x ) + 8(x ) + 4
84. SHORT RESPONSE Tiles numbered from 1 to 6 are placed in a bag and are drawn to determine which of six
tasks will be assigned to six people. What is the probability that the tiles numbered 5 and 6 are the last two drawn?
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Page 20
2 3
Sample answer: A polynomial with degree 6 that can be written in cubic form will have a first term in the form (x ) .
2 2
Then the second term will be in the form of (x ) . 5-5 Solving
Polynomial Equations
6
4
2
2 3
2 2
2
12x + 6x + 8x + 4 = 12(x ) + 6(x ) + 8(x ) + 4
84. SHORT RESPONSE Tiles numbered from 1 to 6 are placed in a bag and are drawn to determine which of six
tasks will be assigned to six people. What is the probability that the tiles numbered 5 and 6 are the last two drawn?
SOLUTION: 2
86. Which of the following most accurately describes the translation of the graph y = (x + 4) – 3 to the graph of y = (x
2
– 1) + 3?
F down 1 and to the right 3
G down 6 and to the left 5
H up 1 and to the left 3
J up 6 and to the right 5
SOLUTION: 2
2
The graph y = (x + 4) – 3 is translated 6 units up and 5 units to the right and positioned at y = (x – 1) + 3.
Therefore, option J is the correct answer.
Graph each polynomial function. Estimate the x-coordinates at which the relative maxima and relative
minima occur.
3
2
88. f (x) = 2x – 4x + x + 8
SOLUTION: 3
2
Graph the polynomial f (x) = 2x – 4x + x + 8.
The x-coordinate at the relative maximum at
The x-coordinate at the relative minimum at
3
.
2
90. f (x) = –x + 3x + 4x – 6
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3
2
Graph the polynomial f (x) = –x + 3x + 4x – 6.
Page 21
The x-coordinate at the relative maximum at
5-5 Solving
Polynomial
Equations
The x-coordinate
at the
relative minimum at
3
.
2
90. f (x) = –x + 3x + 4x – 6
SOLUTION: 3
2
Graph the polynomial f (x) = –x + 3x + 4x – 6.
The x-coordinate at the relative maximum at
The x-coordinate at the relative minimum at
.
State the degree and leading coefficient of each polynomial in one variable. If it is not a polynomial in one
variable, explain why.
5
4
2
92. f (x) = –2x + 5x + 3x + 9
SOLUTION: The degree of the polynomial is 5.
The leading coefficient of the polynomial is –2.
94. ELECTRICITY The impedance in one part of a series circuit is 3 + 4j ohms, and the impedance in another part of
the circuit is 2 – 6j . Add these complex numbers to find the total impedance of the circuit.
SOLUTION: 96. GEOMETRY The sides of an angle are parts of two lines whose equations are 2y + 3x = –7 and 3y – 2x = 9. The
angle’s vertex is the point where the two sides meet. Find the coordinates of the vertex of the angle.
SOLUTION: Solve the system of equations 2y + 3x = –7 and 3y – 2x = 9.
The solution is (–3, 1).
Therefore, the coordinates of the vertex of the angle is (–3, 1).
Divide.
2
98. (2x + 8x – 10) ÷ (2x + 1)
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Page 22
SOLUTION: Solve the system of equations 2y + 3x = –7 and 3y – 2x = 9.
5-5 Solving
Polynomial
The solution
is (–3, 1).Equations
Therefore, the coordinates of the vertex of the angle is (–3, 1).
Divide.
2
98. (2x + 8x – 10) ÷ (2x + 1)
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Page 23