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)