Factoring Expressions
In order to understand the following lesson, you should
already know:
(i.) a general method for removing monomial factors from an
expression
(ii.) two methods for factoring expressions with very special
(but quite commonly occurring) forms:
 trinomials (ax^{ 2} + bx + c)
 difference of two perfect squares (u^{ 2} –
v^{ 2} )
All we want to point out here is that the overall strategy for
factoring any algebraic expression is to apply the methods listed
above in order. So,
(i) first remove all monomial factors from the expression
then
(ii) check whatever is left against any of the special
patterns that you know
By following this strategy, the more difficult, tedious, and
intuitive analysis of the second type in the list above will
always be done on the simplest possible expressions.
We’ve already alluded to this strategy in the previous
documents on factoring, and shown one or two brief examples. Here
is a few more short examples to help you understand the overall
strategy better.
Example 1:
Factor completely: 3x^{ 4} y + 6x^{ 3} y
– 45x^{ 2} y.
solution:
We see that 3, x^{ 2} , and y, are each monomial
factors of all three terms. So, step (i) in the general strategy
leads to
3x^{ 4} y + 6x^{ 3} y – 45x^{ 2}
y = 3x^{ 2} y(x^{ 2} + 2x – 15)
Now, we need to apply step (ii) of the general strategy to the
expression, x^{ 2} + 2x – 15, remaining after we
have accounted for the common monomial factors. This is a
trinomial in x, which it may be possible to rewrite in the form
x^{ 2} + 2x – 15 = (x + a)(x + b)
if we can find two whole numbers, a and b, such that
ab = 15 and a + b = 2.
You can easily verify that a = 5 and b = 3 works. Thus
x^{ 2} + 2x – 15 = (x + 5)(x  3)
Thus, for the original expression we now have
3x^{ 4}y + 6x^{ 3}y – 45x^{ 2}y =
3x^{ 2}y(x + 5)(x – 3)
which is factored as completely as is possible. We know that
no further factorization is possible, because the factors here
which are not monomial factors contain no products at all.
Notice that when it came time to check if any of our special
pattern formulas might apply, we needed to check them against the
quite simple expression, x^{ 2} + 2x – 15, rather
than the much more complicated original expression. To be sure,
(x + 5) and (x – 3) are factors of the original expression,
but its complexity (due to the presence of common monomial
factors in each term) would make it very difficult to recognize
these binomial factors in that expression.
