Monomial Factors
If we were asked to identify all of the common monomial
factors in
25x^{ 4}y^{ 3}zw^{ 2} + 150x^{ 3}y^{
3}zr^{ 4} + 30x^{ 2}y^{ 3}zw
we would reach the answer:
25x^{ 4}y^{ 3}zw^{ 2} + 150x^{ 3}y^{
3}zr^{ 4} + 30x^{ 2}y^{ 3}zw = (5x^{
2}y^{ 3}z)(5x^{ 2}w^{ 2} + 30xr^{
4} + 6w).
However, a couple more observations about this example may
help you solve other problems of this sort.
Notice that the second term in the example contained a power,
r^{ 4}, of r. Yet our detailed analysis of potential
monomial factors did not consider powers of r at all. We
didn’t consider powers of r, because r did not occur in the
term 25x^{ 4}y^{ 3}zw^{ 2} on which we
were basing our analysis. But since r did not occur in this term
at all, a power of r could not possibly be a common factor of all
three terms. Because any common monomial factors must be a factor
of the term on which we base our analysis, any symbols which do
not appear in that term need not be considered at all. Hence the
value in keying the analysis on a simplerlooking rather than
more complicatedlooking term in the original expression. Once
you understand the strategy of the method displayed in great
detail above, you may be able to accomplish the same end with
much less writing without sacrificing the systematic approach.
You still key the analysis on one of the terms in the expression,
but common powers of factors are identified and removed from the
original expression in a stepwise fashion.
25x^{ 4}y^{ 3}zw^{
2} + 150x^{ 3}y^{ 3}zr^{ 4}
+ 30x^{ 2}y 3zw^{ 5} 
is a factor of the first term, and each
of the remaining terms contain a factor of 5, so remove a
factor of 5 from all three terms. 
= 5(5x^{ 4}y^{ 3}zw^{
2} + 30x^{ 3}y^{ 3}zr^{ 4} +
6x^{ 2}y^{ 3}zw) 
The only remaining numerical factor in
the first term in the brackets is 5, but not all of the
remaining terms in brackets contain a factor of 5, so
there are no more numerical factors to remove. The first
symbolic factor is a power of x. All three terms have a
factor of at least x^{ 2}, so remove a factor of
x^{ 2} from the bracketed expression. 
=5x^{ 2}(5x 2y^{
3}zw^{ 2} + 30xy^{ 3}zr^{ 4}
+ 6y^{ 3}zw) 
The next symbolic factor in the first
term in the brackets is a power of y. All three terms
contain a factor of y^{ 3}, so remove it. 
=5x^{ 2}y^{ 3}(5x^{
2}zw^{ 2} + 30xzr^{ 4} + 6zw) 
The next symbolic factor in the first
term in the brackets is a z. Each of the three terms in
brackets contains a factor z, so remove it. 
=5x^{ 2}y^{ 3}z(5x^{
2}w^{ 2} + 30xr^{ 4} + 6w) 
The next symbolic factor in the first
term in brackets is a power of w. However not all of the
remaining terms in the brackets contain a factor which is
a power of w, so there is no power of w to be removed
from the bracketed expression. Since we have run out of
factors in the first term in brackets, we are done. The
resultant expression has all common monomial factors
identified. 
Notice that at each step, we needed to deal only with the
expression left in the brackets, which is also generally getting
simpler with each step (at least with those steps in which
we’ve been able to remove a factor). You can also check your
work at each step because if you remultiply by the most recently
removed factor, you must obtain the expression from the previous
step.
This has been a rather lengthy discussion of issues and
approaches to factoring one fairly complicated expression. To
conclude this document, we’ll briefly demonstrate the
identification of monomial factors for a couple of additional
expressions.
