8-4
Solving Polynomial Equations
TEKS FOCUS
VOCABULARY
TEKS (7)(E) Determine linear and quadratic factors of a polynomial
expression of degree three and of degree four, including factoring the
sum and difference of two cubes and factoring by grouping.
ĚDifference of cubes – A difference of
TEKS (1)(C) Select tools, including real objects, manipulatives, paper and
pencil, and technology as appropriate, and techniques, including mental
math, estimation, and number sense as appropriate, to solve problems.
ĚSum of cubes – A sum of cubes is an
cubes is an expression of the form
a3 - b3 .
expression of the form a3 + b3 .
ĚNumber sense – the understanding of what
Additional TEKS (1)(D), (7)(D)
numbers mean and how they are related
ESSENTIAL UNDERSTANDING
If (x - a) is a factor of a polynomial, then the polynomial has value 0 when x = a. If a
is a real number, then the graph of the polynomial has (a, 0) as an x-intercept.
Concept Summary
Polynomial Factoring Techniques
Techniques
Examples
Factoring Out the GCF
Factor out the greatest common
factor of all the terms.
15x4 - 20x3 + 35x2
= 5x2(3x2 - 4x + 7)
Quadratic Trinomials
For ax2 + bx + c find factors with
product ac and sum b.
6x2 + 11x - 10
= (3x - 2)(2x + 5)
Perfect Square Trinomials
a2 + 2ab + b2 = (a + b)2
a2 - 2ab + b2 = (a - b)2
x2 + 10x + 25 = (x + 5)2
x2 - 10x + 25 = (x - 5)2
Difference of Squares
a2 - b2 = (a + b)(a - b)
4x2 - 15 = (2x + 115 )(2x - 115 )
Factoring by Grouping
ax + ay + bx + by
= a(x + y) + b(x + y)
= (a + b)(x + y)
x3 + 2x2 - 3x - 6
= x2(x + 2) + ( -3)(x + 2)
= (x2 - 3)(x + 2)
Sum or Difference of Cubes
a3 + b3 = (a + b)(a2 - ab + b2)
a3 - b3 = (a - b)(a2 + ab + b2)
8x3 + 1 = (2x + 1)(4x2 - 2x + 1)
8x3 - 1 = (2x - 1)(4x2 + 2x + 1)
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Problem 1
P
Solving Polynomial Equations Using Factors
What are the real or imaginary solutions of each polynomial equation?
What does it mean if
x is a common factor
of every term in P(x)?
You can write P(x) as
xQ(x), so 0 will be a
solution of P(x) = 0.
A 2x3 − 5x2 = 3x
2x3 - 5x2 - 3x = 0
Rewrite in the form P(x) = 0.
x(2x2
- 5x - 3) = 0
Factor out the GCF, x.
x(2x + 1)(x - 3) = 0
Factor 2x 2 - 5x - 3.
x=0
2x + 1 = 0
or
or
x = - 12
x=0
The solutions are 0,
- 12,
x-3=0
x=3
Zero Product Property
Solve each equation for x.
and 3.
B 3x 4 + 12x 2 = 6x 3
3x4 - 6x3 + 12x2 = 0
How will the solution
be similar to the
solution of the
equation in part (A)?
Both equations have 0 as
a solution, but here it will
have a multiplicity of 2.
x4
-
2x3
+
4x2
Rewrite in the form P(x) = 0.
Multiply by 13 to simplify.
=0
x2(x2 - 2x + 4) = 0
x2 = 0
or
x2 - 2x + 4 = 0
x=0
or
x=
Factor out the GCF, x2.
Zero Product Property
- ( - 2) { 2( - 2)2 - 4(1)(4)
= 1 { i23
2(1)
Use the Quadratic Formula
to solve x 2 - 2x + 4 = 0.
The solutions are 0, 1 + i 13, and 1 - i 13.
Problem
bl
2
TEKS Process Standard (1)(D)
Solving Polynomial Equations by Factoring
How can you write
the polynomial in
quadratic form?
Write in terms of x 2 :
(x 2)2 - 3(x 2) - 4 = 0,
which shows the
factorable quadratic form
a2 - 3a - 4 = 0.
What are the real or imaginary solutions of x4 − 3x2 = 4?
W
x4 - 3x2 - 4 = 0
a2 - 3a - 4 = 0
(a - 4)(a + 1) = 0
(x2 - 4)(x2 + 1) = 0
(x + 2)(x - 2)(x2 + 1) = 0
Rewrite in the form P(x) = 0.
Let a = x 2 .
Factor.
Replace a with x 2 .
Factor x 2 - 4 as a difference of squares.
It follows from the Zero Product Property that x = 2, x = -2, or x2 = -1.
Solving x2 = -1 yields two imaginary roots: x = i or x = -i.
Check
Graph the related function y = x4 - 3x2 - 4.
The graph shows zeros at x = 2 and x = -2. It also
shows three turning points. This means that there are
imaginary roots, which do not appear on the graph.
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Lesson 8-4
Solving Polynomial Equations
Problem 3
TEKS Process Standard (1)(C)
Factoring the Sum of Two Cubes
A Which tool would you use to find the factors of x3 + 27: paper and pencil or
technology? Why?
What technique can
be used to factor a
sum of cubes?
The sum of cubes formula
is the most efficient
technique since a
quadratic factor may not
have real zeros.
You could use a graphing calculator to find the linear factors since they correspond
to zeros of the function. But when a quadratic factor exists, the graph does not give
an easy way to find it. Paper and pencil is the best tool in this case.
B How can you use mental math, estimation, or number sense to factor x3 + 27?
You use number sense to recognize that each term of the expression is a perfect
cube and to select the appropriate formula. To find the factors of the expression,
you use mental math to identify the value of each variable in the formula.
C What are the linear and quadratic factors for the polynomial expression x3 + 27?
Select the formula: a3 + b3 = (a + b)(a2 - ab + b2).
Identify the value of each variable in the formula:
a3 = x3 , so a = x
b3 = 27, so b = 3
Substitute the values into the sum of two cubes formula and simplify:
x3 + 27 = (x + 3)(x2 - 3x + 32) = (x + 3)(x2 - 3x + 9)
The linear factor of the expression x3 + 27 is x + 3.
The quadratic factor of the expression x3 + 27 is x2 - 3x + 9.
Problem
bl
4
Sums and Differences of Cubes
Is there a common
factor in every term?
Yes; the greatest common
factor is 3x.
A Factor 3x4 − 3x. What are the linear and quadratic factors of the expression?
3x4 - 3x
3x(x3 - 1)
3x(x -
1)(x2
Factor out the GCF.
+ x + 1)
Factor x 3 - 1 as a difference of cubes.
The linear factors of 3x4 - 3x are 3x and x - 1.
The quadratic factor of 3x4 - 3x is x2 + x + 1.
B Factor 8x4 + 27x. What are the linear and quadratic factors of the expression?
8x4 + 27x
x(8x3 + 27)
x(2x +
3)(4x2
Factor out the GCF.
- 6x + 9)
Factor 8x 3 + 27 as a sum of cubes.
The linear factors of 8x4 + 27x are x and 2x + 3.
The quadratic factor of 8x4 + 27x is 4x2 - 6x + 9.
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Problem 5
Factoring Polynomials by Grouping
What are the linear and quadratic factors of each polynomial?
A x3 + 3x2 − 16x − 48
What is a common
factor?
A common factor is a
factor that two numbers
or expressions share.
A common factor may
be a number, a variable,
or even a polynomial
expression.
x3 + 3x2 - 16x - 48 = (x3 + 3x2) + ( -16x - 48)
Group the first two and
last two terms.
= x2(x + 3) + ( -16)(x + 3)
Factor out the common
factor of each group.
= (x2 - 16)(x + 3)
Factor the GCF, x + 3.
= (x + 4)(x - 4)(x + 3)
Factor the difference of
squares.
B x4 − 2x3 + 8x − 16
x4 - 2x3 + 8x - 16 = (x4 - 2x3) + (8x - 16)
Group the first two and
last two terms.
= x3(x - 2) + 8(x - 2)
Factor out the GCF of
each group.
= (x3 + 8)(x - 2)
Factor out the common
factor x - 2.
= (x + 2)(x2 - 2x + 4)(x - 2)
Factor x 3 + 8 as a sum
of cubes.
Problem
P
bl
6
What factoring
technique can you
use to find all the
solutions?
The difference of cubes
can be used to find the
linear and quadratic
factors. Then the Zero
Product Property can be
used with each factor
to find the real and
imaginary solutions.
Solving Polynomial Equations Using a Difference of Two Cubes
S
What are the real or imaginary solutions of 27x3 = 343?
W
27x3 = 343
27x3 - 343 = 0
(3x - 7)(9x2 + 21x + 49) = 0
Rewrite in the form P(x) = 0.
Factor the difference of cubes.
IIt follows from the Zero Product Property that x = 73 or 9x2 + 21x + 49 = 0.
U the Quadratic Formula to solve 9x2 + 21x + 49 = 0.
Use
x=
- (21) { 2(21)2 - 4(9)(49) - 21 { 1 - 1323 - 7 { 7i 13
=
=
18
6
2(9)
7 13
7 13
The three solutions of 27x3 = 343 are 73 , - 76 + i 6 , and - 76 - i 6 .
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Lesson 8-4
Solving Polynomial Equations
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PRACTICE and APPLICATION EXERCISES
Scan page for a Virtual Nerd™ tutorial video.
Factor the polynomial. Identify the linear and quadratic factors of each expression.
For additional support when
completing your homework,
go to PearsonTEXAS.com.
1. x3 + 64
2. 343x3 + 8y 3
3. 2 - 16m3
4. 7x4 - 7x
5. 27x3y + 216y 4
6. 250x4 + 128x
Find the real or imaginary solutions of each equation by factoring.
7. 64x3 - 1 = 0
8. x3 + 2x2 + 5x + 10 = 0
9. 0 = x3 - 27
10. 0 = x3 - 64
11. 8x3 = 1
12. x4 - 10x2 = -9
13. x4 - 8x2 = -16
14. x4 - 12x2 = 64
15. x4 + 4x2 = 12
Factor the polynomial by grouping. Identify the linear and quadratic factors of
each expression.
16. x3 + 3x2 + 4x + 12
17. 6x2y + 8x2 + 3xy + 4x
18. x4 + 4x3 - 9x2 - 36x
19. Apply Mathematics (1)(A) The width of a
plastic storage box is 1 ft longer than the height.
The length is 4 ft longer than the height. The
volume is 36 ft3. What are the dimensions of
the box?
20. Explain Mathematical Ideas (1)(G) A student
claims that 1, 2, 3, and 4 are the zeros of a cubic
polynomial function. Explain why the student
is mistaken.
21. The width of a box is 2 m less than the length.
The height is 1 m less than the length. The
volume is 60 m3. What is the length of the box?
Solve each equation.
22. x3 + 13x = 10x2
23. x3 - 6x2 + 6x = 0
24. 12x3 = 60x2 + 75x
25. 125x3 + 216 = 0
26. x4 - 64 = 0
27. -2x2 - 100 = 0
28. 27 = -x4 - 12x2
29. 5x3 = 5x2 + 12x
30. x3 + 1 = x2 + x
Graph each function to find the zeros. Rewrite the function with the polynomial in
factored form.
31. y = 2x2 + 3x - 5
32. y = x4 - 10x2 + 9
33. y = x3 - 3x2 + 4
34. Use Multiple Representations to Communicate Mathematical Ideas (1)(D)
To solve a polynomial equation, you can use any combination of graphing,
factoring, and the Quadratic Formula. Write and solve an equation to illustrate
each method.
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35. Select Techniques to Solve Problems (1)(C) How can you use mental math,
estimation, or number sense to help you factor x3 + 106 ? Explain.
36. Connect Mathematical Ideas (1)(F) Your teacher asks the class to factor x4 - x
completely, and to apply the difference of cubes formula. Your friend says this
can’t be done, because the expression is not a difference of cubes. Is your friend
correct? Explain.
37. Select Tools to Solve Problems (1)(C) Which tool would you use to factor
x4 + 4x3 + 9x2 + 36x completely: paper and pencil or technology? Why?
38. The geometric figure below has volume a3 + b3. You can split it into three
rectangular blocks (including the long one with side a + b). Explain how
to use this figure to prove the factoring formula for the sum of cubes,
a3 + b3 = (a + b)(a2 - ab + b2).
a
a
a–b
b
a+b
b
39. Connect Mathematical Ideas (1)(F) Find equations for two different
polynomial functions whose zeros include -12, 0, 14, and 16.
40. What are the complex solutions of x5 + x3 + 2x = 2x4 + x 2 + 1?
TEXAS Test Practice
T
41. Which value is NOT a solution to the equation x4 - 3x2 - 54 = 0?
A. -3
B. 3
C. -3i
D. -i16
42. Ava drove 3 hours at 45 miles per hour. How many miles did she drive?
F. 45 miles
G. 48 miles
H. 90 miles
J. 135 miles
43. Which polynomial has the complex roots 1 + i12 and 1 - i12?
A. x2 + 2x + 3
B. x2 - 2x + 3
C. x2 + 2x - 3
D. x2 - 2x - 3
44. Sam has only quarters and dimes in his pocket. He has a total of 12 coins,
totaling $1.95. How many of each coin does Sam have?
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Lesson 8-4
Solving Polynomial Equations