M098
Carson Elementary and Intermediate Algebra 3e
Section 6.4
Objectives
1.
2.
3.
Factor perfect square trinomials.
Factor a difference of squares.
Factor a difference and sum of cubes.
Vocabulary
conjugates
Binomials that differ only in the sign separating the terms.
Prior Knowledge
2
2
2
(a + b) = (a + b)(a + b) = a + 2ab + b
2
2
(a + b)(a – b) = a – b
Multiplying polynomials
The first 20 perfect squares and 5 perfect cubes.
New Concepts
This section shows several special cases for factoring. This is really a memorizing activity. You must
recognize the special case and then know that it will factor a certain way.
These are the special cases:
2
Perfect Square Trinomial:
2
2
x + 2xy + y = (x + y)
2
2
2
x – 2xy + y = (x – y)
When you see that the first and last terms are perfect squares, there’s a good chance the trinomial is a
perfect square trinomial that will factor by this rule. Always foil back to be sure it works.
2
Example 1:
y + 6y + 9
Example 2:
4a – 20ab + 25b
Example 3:
9m – 24m + 16
Example 4:
y + 6y + 36
Example 5:
4y – 12by + 9b
Example 6:
54q – 72q + 24
(y + 3)
2
2
(2a – 5b)
2
(3m – 4)
2
2
2
2
2
Not a perfect square
2
(2y – 3b)
2
2
6(3q – 2)
2
2
2
a – b = (a – b)(a + b)
Difference of two squares:
What you need to recognize in this special case is that two perfect squares are subtracted. There is no
middle term. The only way the middle term of the trinomial will disappear is if, when you foil, the inner
and outer terms are the same but with different signs. We call these conjugates.
2
2
The sum of two squares is prime and cannot be factored: a + b is prime.
2
Example 7:
a – 121
(a + 11)(a – 11)
Example 8:
25x – 16
2
(5x – 4)(5x + 4)
Example 9:
49 – a
Example 10:
64m – 36y
V. Zabrocki 2011
2
2
(7 – a)(7 + a)
2
4(4m – 3y)(4m + 3y)
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M098
Carson Elementary and Intermediate Algebra 3e
2
2
Example 11:
16x – 36y
Example 12:
x – 16
Section 6.4
4(2x – 3y)(2x + 3y)
4
2
(x + 4)(x – 2)(x + 2)
3
3
2
2
x + y = (x + y)(x – xy + y )
3
3
2
2
x – y = (x – y)(x + xy + y )
Sum and Difference of Two Cubes:
Because these are cubes, the last factor will be a trinomial (3 terms).
To find the first factor (the binomial), take the cube root of the first term, keep the sign from the original
problem, and take the cube root of the second term.
The first sign of the trinomial factor will always be the opposite of the sign in the binomial and the last
sign in the trinomial factor will always be +. Once the signs for the trinomial are determined, you can
find the terms of the trinomial by squaring the first term of the binomial, multiplying the two terms of the
binomial together and squaring the last term of the binomial.
3
Example 13:
3
8a + 27b
This is the sum of two cubes so there will be a binomial factor and a trinomial factor.
Find the binomial factor:
3
the cube root of 8a = 2a
the sign in the original problem is +
3
the cube root of 27b = 3b
So the first factor (the binomial) is (2a + 3b)
Determine the signs for the trinomial.
Since the sign in the binomial is +, the first sign in the trinomial must be – .
The last sign is always +.
So the signs for the trinomial are ( –
+ ).
To find the terms of the trinomial factor, look at the binomial factor (2a + 3b):
2
2
Square the first term: (2a) = 4a
Multiply the terms together: (2a)(3b) = 6ab
2
2
Square the last term: (3b) = 9b
2
2
So the trinomial is (4a – 6ab + 9b )
3
3
2
2
8a + 27b = (2a + 3b) (4a – 6ab + 9b )
3
2
Example 14:
y – 64
Example 15:
1000a – 27b
Example 16:
8p + q z
(y – 4)(y + 4y + 16)
3
3
3
3 3
2
2
(10a – 3b)(100a + 30ab + 9b )
2
2 2
(2p + qz)(4p – 2pqz + q z
Always look for a GCF and remove it first. Even if the problem looks complicated, the patterns in the special
cases will always work.
Example 17:
V. Zabrocki 2011
25y 2 
1
36
1 
1

 5y   5y  
6 
6

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M098
Carson Elementary and Intermediate Algebra 3e
8y 3 
Example 19:
25a – 16(b + c)
V. Zabrocki 2011
2
1  2 2
1 

 2y   4y  y 

5
5
25



1
125
Example 18:
Section 6.4
2
(5a – 4b – 4c)(5a + 4b + 4c)
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