Monomial Factors
The first thing you should know before factoring an algebraic
expression is to identify any factors which are monomials.
Let's see what this is all about in the following example.
Example:
Remove all common monomial factors from 42b^{ 2}y
– 28by^{ 2}.
solution:
This example is very similar to the previous one, and so you
should try it as a practice problem on your own first, before
looking at our brief outline of a solution. First, write out the
factorization of each of the two terms explicitly:
42b^{ 2}y = 2 Â· 3 Â· 7 Â· b^{ 2} Â· y^{ 1}
and
28by^{ 2} = 2^{ 2} Â· 7 Â· b^{ 1} Â· y^{
2}
Notice that we’ve written exponents of symbols
explicitly, even if those exponents are 1 (just as a visual cue
when we check now for common factors between the two terms). It
is also helpful to sort the factors of the individual terms in a
common order. Here numerical prime factors are sorted from
smallest to largest (going left to right) and symbolic factors
are sorted alphabetically.
Comparison of these two factorizations indicates immediately
that the common factors are 2, 7, b, and y, all to the first
power. Thus
42b^{ 2}y – 28by^{ 2} = 2 Â· 7 Â· b Â· y
(3b – 2y)
= 14by (3b – 2y).
The terms in the expression in brackets on the right here are
obtained by taking what’s left of each of the original terms
when the common factors are removed. Thus, since
42b^{ 2}y = 2 Â· 3 Â· 7 Â· b^{ 2} Â· y^{ 1}
when we remove the factors 2, 7, b^{ 1} , and y^{ 1},
all that’s left is the 3 and one of the factors b, or 3b. A
similar inspection indicates that after removal of these four
common factors from 28by^{ 2}, all that is left is the
factors 2 and y, each to the first power.
Thus, the required factorization here is
42b^{ 2}y – 28by^{ 2} = 14by (3b –
2y).
We’ll leave it up to you to verify that this is correct
by multiplication.
