Lesson 6–6
Example 1
GCF
Factor –18ax3 – 12bx2 + 6x2.
–18ax3 – 12bx2 + 6x2
= (–1 • 2 • 3 • 3 • a • x • x • x) + (–1 • 2 • 2 • 3 • b • x • x) + (2 • 3 • x • x)
2
The GCF is 6x .
= (6x2 • –3ax) + (6x2 • –2b) + (6x2 • 1)
2
The remaining polynomial cannot
= 6x (–3ax – 2b + 1)
be factored.
Example 2
Grouping
Factor x + 5x – 7x – 35.
3
2
x3 + 5x2 – 7x – 35 = (x3 + 5x2) + (– 7x – 35)
= x2(x + 5) + (–7)(x + 5)
= (x + 5)(x2 – 7)
Group to find a GCF.
Factor the GCF of each binomial.
Distributive Property
Example 3
Two or Three Terms
Factor each polynomial.
a. 6a2 - 13a - 5
To find the coefficients of the a-terms, you must find two numbers whose product is 6(-5) or -30, and
whose sum is -13. The coefficients must be -15 and 2 since -15(2) = -30 and -15 + 2 = -13. Rewrite the
expression using -15a and 2a and factor by grouping.
Substitute -15a + 2a for -13a.
6a2 - 13a - 5 = 6a2 - 15a + 2a - 5
2
Associative Property
= (6a - 15a) + (2a - 5)
Factor out the GCF of each group.
= 3a(2a - 5) + 1(2a - 5)
Distributive Property
= (3a + 1)(2a - 5)
b. z3 – 8x3
z3 = (z)3 and (2x)3 = 8x3. Thus, this is the difference of two cubes.
Difference of two cubes formula with a = z and b = 2x
z3 – 8x3 = (z – 2x)[z2 + z(2x) + (2x)2]
2
2
Simplify.
= (z – 2x)(z + 2zx + 4x )
Example 4
6
Write Expressions in Quadratic Form
3
a. x + 2x + 5
x 6 + 2x 3 + 5 = (x 3 ) 2 + 2(x 3 ) + 5


(x 3 ) 2 = x 6
b. 3x 9 - 4x 3 + 1

This cannot be written in quadratic form since x 9  (x 3 ) 2 .
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Example 5
Solve Polynomial Equations
Solve each equation.
4
a. x - 26x 2 + 25 = 0
4
Original equation
x - 26x 2 + 25 = 0
2 2
2
Write the expression on the left in quadratic form.
(x ) - 26(x ) + 25 = 0
2
2
Factor the trinomial.
(x - 1)(x - 25) = 0

 (x – 1)(x + 1)(x – 5)(x + 5) = 0 Factor each difference of squares.

 the Zero Product Property.
Use
 (x – 
1) = 0
or
(x + 1) = 0
or
(x – 5) = 0
or
 x =1
x = -1
x=5
The solutions are -5, -1, 1, and 5.
Check
(x + 5) = 0
x = -5
4
The graph of f(x) = x - 26x 2 + 25 shows that the graph intersects the x-axis at -5, -1, 1, and 5.
[put in art ee_alg2_0606.1]
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
b. x 3 - 27 = 0
Original equation
x 3 - 27 = 0
3
3
This is a difference of two cubes.
(x) - 3 = 0
2
2
Difference of two cubes with a = x and b = 3
(x - 3)(x + x(3) + 3 ) = 0

2
Simplify.
(x - 3)(x + 3x + 9) = 0

2
Zero Product Property
(x – 3) = 0 or x + 3x + 9 = 0
 
The solution of the first equation is 3. The second equation can be solved by using the Quadratic Formula.






b  b 2  4ac
 x 
2a

2
3  3  4(1)(9)

2(1)
3  27

2
3  i 27
3  3i 3
or

2
2
Quadratic Formula
Replace a with 1, b with 3, and c with 9.
Simplify.
27 
1 = 3i 3
Thus the solutions of the original equation are 3,
3  3i 3
3  3i 3
, and
.
2
2

Check
The graph of f(x) = x 3 - 27 confirms the solution.
[put in art ee_alg2.0606.2]
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