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Chapter 6 Review Part III Page 1
Chapter 6 Review - Part III
I.
Factoring Difference of Squares
A.
Remember that the product of two conjugates is:
2
2
(F + L)(F – L) = F – L
B.
We are going to use this pattern to factor the difference of two squares.
C.
Remember that a perfect square is any number that is the product of a number
times itself.
D. Remember also that for a variable term to be a perfect square, the exponent on the
variable must be an even number.
E. Procedure for factoring Difference of Squares
1.
Factor out the GCF
2.
Set up two parentheses, where
a.
The First numbers are the square root of the First term
b.
The Last numbers are the square root of the Last term
c.
The signs in between the first and the last terms are “+” in one
parenthesis and “-“ in the other.
3.
Check by multiplying.
F.
Examples – Factor the following completely.
3
3
1.
2xy – 8x y
First, notice that the GCF is 2xy, so we will factor that out to get:
2
2
2xy(y – 4x )
2
2
We notice that “y ” is a perfect square, as is “4x ”, and this is a difference. So
we set up parentheses to get:
Answer: 2xy(y + 2x)(y – 2x)
4
2.
81 – m
There is no GCF, so we can immediately set up parentheses to get:
2
2
(9 – m )(9 + m )
Notice that the first parenthesis is still a difference of squares, so we need to
continue factoring.
2
Answer: (3 – m)(3 + m)(9 + m )
Note that a sum of squares is not factorable over the rationals.
2
3. x – 121
Again, no GCF, so we jump right into setting up parentheses:
Answer: (x – 11)(x + 11)
4.
Now you try one: 25 – x
Answer: (5 – x)(5 + x)
2
III. Sum and Difference of Cubes
A.
This is a formula that you just have to learn, it is not at all intuitive.
B.
Procedure
1.
Factor out the GCF.
2.
In the first set of parentheses, the First number will be the cube root of the
First term
a.
Take the cube root of the coefficient, but divide the exponent on the
variable by 3 since perfect cubes will always have exponents that are
multiples of 3.
b.
Take the sign in the middle
1.
“-“ if you have a difference of cubes
2.
+” if you have a sum of cubes
c.
The Last number will be the cube root of the Last term.
© Copyright 2005 by John Fetcho. All rights reserved
Chapter 6 Review Part III Page 2
C.
3.
The second set of parentheses depends entirely on what is in the first
parentheses, we don't care about the original problem anymore.
a.
The First number will be the square of the First number in the first set of
parentheses.
b.
The middle sign will be the opposite of the middle sign in the first set of
parentheses.
c.
Multiply the First and Last terms from the first set of parentheses (don't
worry about the sign).
d.
Add the square of the Last term from the first set of parentheses.
4.
For those of you that like to see formulas, here they are:
3
3
2
2
a.
F – L = (F – L)(F + FL + L )
3
3
2
2
b.
F + L = (F + L)(F - FL + L )
Examples – Factor each of the following completely.
3
1.
8x + 125
First, we notice that there is no GCF.
3
Next, the cube root of 8x is 2x; the sign in the middle is "+", the cube root of
125 is 5. So the first parenthesis will be:
(2x + 5)
2
To get the terms of the second parenthesis, we square 2x to get 4x .
The middle sign will be “-“.
The middle term will be 2x times 5, or 10x.
To get the last term, we square 5 to get 25.
Putting this all together, we get:
2
Answer: (2x + 5)(4x – 10x + 25)
2.
3
3
125x – 27y
Again, no GCF.
3
The cube root of 125x is 5x.
The sign will be “-“.
3
The cube root of 27y is 3y.
So the first parenthesis will be:
(5x – 3y)
2
Now we square 5x to get 25x .
The middle sign will be “+”.
The middle term will be 5x times 3y, which is 15xy.
2
The last term will be the square of 3y, which is 9y .
Putting this all together, we get:
2
2
Answer: (5x – 3y)(25x + 15xy + 9y )
3
3.
Now you try one: x + 64
2
Answer: (x + 4)(x – 4x + 16)
III.
Perfect Square Trinomials
A.
This is really just an extension of the Trial and Error method.
B.
Remember that when a factor is multiplied by itself, we can say that it is squared.
C.
Examples - Factor completely.
2
2
1.
9t + 24tr + 16r
2
There is no GCF, so let’s remember that the square root of 9t is 3t; the last term is
positive, so the signs have to be the same as the middle term, in this case + and +;
2
the square root of 16r is 4r. So putting this all together, we get:
(3t + 4r)(3t + 4r)
Checking:
2
2
2
2
9t + 12tr + 12tr + 16r = 9t + 24tr + 16r
Which is what we want.
© Copyright 2005 by John Fetcho. All rights reserved
Chapter 6 Review Part III Page 3
Answer:
(3t + 4r)
2
2
2
2.
Now you try one: -50h + 40hy - 8y
2
Answer:
-2(5h - 2y)
© Copyright 2005 by John Fetcho. All rights reserved