 # Math 95 Quiz 7 Solutions

1. Multiply the following polynomials to derive the perfect square and difference of squares formulae.
You will want to use these formulas for the rest of the questions!

(a) (A − B)(A + B) = A2 − B2

(b) (A + B)2 = A2 + 2AB + B2

(c) (A − B)2 = A2 − 2AB

2. Instructions: Completely factor all of the polynomials below . Show all of your work, including
noting any formulas you use . You may use any of the formulas above and the following:

A3 − B3 = (A − B)(A2 + AB + B2)

A3 + B3 = (A + B)(A2 − AB + B2)

(a) 81x3 − 3
First we factor out the GCF = 3 to get 3(27x3 −1). Now, since 27x3 −1 is a binomial with
a minus sign it is either a difference of squares or a difference of cubes, in this case it is a
difference of cubes with A = 3x and B = 1. Just apply the formula:

27x3 − 1
= (3x − 1)((3x)2 + 3x + 1)
= (3x − 1)(9x2 + 3x + 1)

Hence

81x3 − 3 = 3(3x − 2)(9x2 + 3x + 1)

(b) 4x9 − 400x
The GCF is 4x so 4x9 − 400 = 4x(x8 − 100). Again, x8 − 100 is a binomial and so either
the difference of squares of the difference of cubes. In this case it is a difference of squares
where A = x4 and B = 10 so

x8 − 100 = (x4 − 10)(x4 + 10)

Since 10 isn’t a perfect square each of the above factors are prime. So finally,

4x9 − 400x = 4x(x4 + 10)(x4 − 1)

(c) 27x2 + 36xy + 12y2

First take out the GCF, 3 to get 3(9x2 + 12xy + 4y2). Before trying any of our longer
strategies for factoring this notice that the first and last terms are perfect squares, i.e.
(3x)2 and (2y)2. A quick check show that the middle term, 12xy, is the same as 2(3x)(2y)
and therefore 9x2 + 12xy + 4y2 is of the form A2 + 2AB + B2 where A = 3x and B = 2y.
Applying the formula we get

27x2 + 36xy + 12y2
= 3(9x2 + 12xy + 4y2)
= 3(3x + 2y)2

(d) 3y2 + 22y − 16
This is a little harder as the GCF is 1 and 3y2 isn’t a perfect square so we need to use one
of our strategies. Let’s use the ac method .

ac = −48

Now we need to find two numbers that multiply to −48 and add to 22. The two numbers
are 24 and −2.

Now we rewrite our middle term , 22y as 22y = 24y − 2y and then factor by grouping :
3y2 + 22y − 16
= 3y2 + 24y − 2y − 16
= (3y2 + 24y) − (2y + 16)
= 3y(y + 8) − 2(y + 8)
= (y + 8)(3y − 2)

 Prev Next