# 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) = A^{2} − B^{2}

(b) (A + B)^{2} = A^{2} + 2AB + B^{2}

(c) (A − B)^{2} = A^{2} − 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:

A^{3} − B^{3} = (A − B)(A^{2} + AB + B^{2})

A^{3} + B^{3} = (A + B)(A^{2} − AB + B^{2})

(a) 81x^{3} − 3

First we factor out the GCF = 3 to get 3(27x^{3} −1). Now, since 27x^{3} −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:

27x^{3} − 1

= (3x − 1)((3x)^{2} + 3x + 1)

= (3x − 1)(9x^{2} + 3x + 1)

Hence

81x^{3} − 3 = 3(3x − 2)(9x^{2} + 3x + 1)

(b) 4x^{9} − 400x

The GCF is 4x so 4x^{9} − 400 = 4x(x^{8} − 100). Again, x^{8} − 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 = x^{4} and B = 10 so

x^{8} − 100 = (x^{4} − 10)(x^{4} + 10)

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

4x^{9} − 400x = 4x(x^{4} + 10)(x^{4} − 1)

(c) 27x^{2} + 36xy + 12y^{2}

First take out the GCF, 3 to get 3(9x^{2} + 12xy + 4y^{2}). 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 9x^{2} + 12xy + 4y^{2} is of the form A^{2} + 2AB + B^{2} where A = 3x and
B = 2y.

Applying the formula we get

27x^{2} + 36xy + 12y^{2}

= 3(9x^{2} + 12xy + 4y^{2})

= 3(3x + 2y)^{2}

(d) 3y^{2} + 22y − 16

This is a little harder as the GCF is 1 and 3y^{2} 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 :

3y^{2} + 22y − 16

= 3y^{2} + 24y − 2y − 16

= (3y^{2} + 24y) − (2y + 16)

= 3y(y + 8) − 2(y + 8)

= (y + 8)(3y − 2)

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