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Algebra 2 Dien
Name:
Period:
Special Case Factoring
Before we go deeper into factoring let’s go over some more vocabulary.
Polynomial: is a group of monomial terms separated by + or – signs.
Trinomial: is a group of 3 terms.
Binomial: is a group of 2 terms.
Now most of the factoring that we are doing will result in a pair of binomials or a binomial and
trinomial, both being multiplied.
1st case Perfect Square of a Binomial
Usually of the form: a 2  2ab  b 2   a  b 
2
or a 2  2ab  b 2   a  b 
2
If you have a polynomial and the first term ( x 2 term) and the last term (the constant term) are
both perfect squares figure out what a and b are and see if they produce the 2ab term.
Note about variables that are perfect squares: if you have even powers then you can break
them down into perfect squares as long as the coefficients are perfect squares.
Example 1:
a) Factor 25x² − 20x + 4
Notice that 25x 2 and 4 are perfect squares. If a 2  25 x 2 then square rooting both sides
you get a  5x . As for the 4 it too is a perfect square although we need to use the
negative version which gives us (-2), since bx is a negative and
a 2  2ab  b 2   a  b  . So the factored form is  5 x  2 
2
2
b) Factor 25x² + 30x + 9
Before you do any work see if you can tell what the answer might be, (a  b)2 or
(a  b)2 . How can you tell?
What would “a” be?
What would “b” be?
What would the factor form be?
c) Factor x 2  3 x  9
2nd case of special factoring: “Difference of Two Squares”
We start with the form a2  b2   a  b  a  b  so if you see the first term ( x 2 term) and the
last term (the constant term) are both perfect squares. Notice that there is no middle term and that
the a 2 and b 2 are subtracting. Usually for us to use the special case factoring a 2 and b 2 are
perfect squares. Figure out what a and b are and put them in the factored form and you are done.
Example 2:
a) Factor 25m 2  9n 2
See how 25 and 9 are a perfect square and the powers of the variables are even? So
a 2  25m2 and b 2  9n 2 *Notice I don’t do anything with the minus sign, it is taken care of in
the formula. So square rooting we get a  5m and b  3n . Plug that in and you get
5m  3n5m  3n as your answer.
b) Factor 1 − 4z² and state what a and b are.
c) Factor 169 x 2  144 y 2 and state what a and b are.
d) Factor x2 y 4  z 6 and state what a and b are.
e) Factor 4 x8  64 y 50 z 34 and state what a and b are.
f) Factor 49 x 4 y 5  81z10 and state what a and b are.
Sometimes a Difference of Two Squares or a Perfect Square is hidden and you have to
first pull out common factors before you can see it. Or you might just be able to factor
more even after you factor the first time.
a) Factor xy 2  xz 2 Now we can’t factor this until we pull out the common factor x.
Then we get y 2  z 2 now we can factor it.
b) Factor 8 x 2  72
c) Factor x 4  81
Homework
Name:
Period:
Factor as a perfect square trinomial, if possible
1) x² − 4x + 4 =
12) 16m² − 40mn+ 25n² =
2) x² + 6x + 9 =
13) x 4  2 x 2 y 2  y 4 =
3) x² − 18x + 36
14) 4 x6  10 x3 y 4  25 y8 =
4) x² − 12x + 36 =
5) x² − 3x + 9 =
6) x² + 10x + 25 =
7) 25x² − 10x + 4 =
8) 25x² + 30x + 9 =
9) 4x² − 28x + 49 =
10) 25x² − 20x + 4 =
11) 1 − 16y + 64y² =
15) x² − 100 =
16) y² − 1 =
17) 1 − 4z² =
18) 25m² − 9n² =)
19) x6 − 36 =
20) y4 − 144 =
21) x8 − y10 =
22) x2n − 1 =
Part 2: Pull out a common factor before using one of the special cases.
1) 64z − z3 =
2) rs3 − r3s =
3) 32m²n − 50n3 =
4) 5x4y5 − 5y5 =