Factoring Differences of Perfect Squares
Notice that
(u + v)(u – v) = (u + v)(u) + (u + v)(v)
= (u)(u + v) + (v)(u + v)
= (u)(u) + (u)(v) + (v)(u) + (v)(v)
= u^{ 2} + uv – uv – v^{ 2}
= u^{ 2} – v^{ 2 }
So, an expression of the form u^{ 2} – v^{ 2}
(called a difference of two perfect squares ) can always be
factored into the form
u^{ 2}
– v^{ 2} = (u +
v)(u – v)
The formula in the box is a pattern. The symbols u and v may
represent numbers, other algebraic symbols, or even algebraic
expressions. It is the precise pattern which must be satisfied in
each case.
Example 1:
Factor 4x^{ 2} – y^{ 2} .
solution:
First, we check, and find that the two terms contain no
monomial factors in common. Then, we note that this is a twoterm
expression, so the methods for factoring trinomials that were
demonstrated in detail in the previous document in the series are
not likely to apply.
Whenever the expression to be factored contains just two
terms, it is worth checking to see if the difference of squares
pattern above can be used. You need to identify that the
expression has two important features:
(i) it is a difference of two terms – the second term must
be subtracted from the first term. In this example, we do have a
difference of two terms.
(ii) it must be possible to write each of the two terms
as a square of some expression. In this example, we see
that
4x^{ 2} = (2x)^{ 2} and y^{ 2} = (y)^{
2} ,
so this condition is met.
Matching the parts of this expression with the parts of the
pattern in the box above, we get
4x^{ 2} – y^{ 2} = (2x)^{ 2}
– (y)^{ 2} u^{ 2} – v^{ 2}
Thus,
u^{ 2} = (2x)^{ 2} giving u = 2x
and
v^{ 2} = (y)^{ 2} giving v = y
Then
4x^{ 2} – y^{ 2} = u^{ 2} –
v^{ 2} = (u + v)(u – v) = (2x + y)(2x – y)
or, directly
4x^{ 2} – y^{ 2} = (2x + y)(2x – y)
as the required factorization.
It is each to check this result directly.
(2x + y)(2x – y) = (2x + y)(2x) + (2x + y)(y)
= (2x)(2x + y) + (y)(2x + y)
= (2x)(2x) + (2x)(y) + (y)(2x) + (y)(y)
= 4x^{ 2} + 2xy – 2xy – y^{ 2}
= 4x^{ 2} – y^{ 2}
Thus our factorization checks.
Example 2:
Factor 9a^{ 4} – 36b^{ 2} .
solution:
Again, we check that there are no common monomial factors in
the two terms (there are none). Then, recognizing the possibility
that this is a difference of two perfect squares (since this
expression obviously is the difference of precisely two terms),
we note that:
9a^{ 4} = (3a^{ 2} )^{ 2} and 36b^{
2} = (6b)^{ 2}
So, each of the two terms are perfect squares. So, we can
apply the pattern formula given in the box above:
u^{ 2} = (3a 2 )^{ 2} or u = 3a^{ 2}
and
v^{ 2} = (6b)^{ 2} or v = 6b
to get
9a^{ 4} – 36b^{ 2} = (3a^{ 2} +
6b)(3a^{ 2} – 6b)
You can easily verify by multiplication that this
factorization is correct
Example 3:
Factor 9x^{ 2} + 4y^{ 2} .
solution:
This expression consists of two terms, both of which are
perfect squares:
9x^{ 2} + 4y^{ 2} = (3x)^{ 2} + (2y)^{
2}
However, it is a sum of two perfect squares rather than a
difference of two perfect squares. We might be tempted to try
9x^{ 2} + 4y^{ 2} ? (3x + 2y)(3x + 2y)
thinking that since a minus has become a plus on the lefthand
side, perhaps the thing to do is to change the minus to a plus on
the righthand side. However, we should not simply assume this is
the best we can do and move on, but rather, we must check to see
if the assumption made here is correct. That is easy to do, as
usual, by multiplication:
(3x + 2y)(3x + 2y) = (3x + 2y)(3x) + (3x + 2y)(2y)
= (3x)(3x + 2y) + (2y)(3x + 2y)
= (3x)(3x) + (3x)(2y) + (2y)(3x) + (2y)(2y)
= 9x^{ 2} + 6xy + 6xy + 4y^{ 2}
= 9x^{ 2} + 12xy + 4y^{ 2}
which is not the same as the original expression, 9x^{ 2}
+ 4y^{ 2} .
In fact, there is no way to factor the sum of
two perfect squares. So, in answer to this example, we need to
simply state that the given expression cannot be factored.
