Monomial Factors
Example:
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
solution:
Our goal is to rewrite this trinomial in the form:
( a monomial) ( whatever else)
Note that this pattern is a single term which is the product
of two parts. The “a monomial” part will be identified
here, in the process determining what the "whatever
else” part is.
We start by selecting any of the three terms to guide the
process – it doesn’t matter which term you choose,
though there may be a slight savings in work if a simplerlooking
term is used. So, we decide to base this analysis on the parts of
the first term, the 25x^{ 4}y^{ 3}zw^{ 2}.
Next, identify all of the simple factors of the selected term,
including prime factors in the numerical coefficient. The term we
have chosen to work from has simple factors 5, x, y, z and w.
(Note that if we had chosen to key our analysis to the third
term, the 30x^{ 2}y^{ 3}zw, our list of simple
factors would be 2, 3, 5, x, y, z, w – slightly longer. Don’t
spend a lot of time making your selection here.)
Now, go through the list of these simple factors in your key
term one at a time, and determine the highest power to which each
of them occurs in all (three, in this case) terms of the entire
expression. This highest power will then be a common monomial
factor of the entire expression. (Of course, if an item in this
list doesn’t occur at all in one of the terms, then this
highest power will be the zero power – that item is not a
common factor of all of the terms in the expression.)
So, for our example, we have five items in the list of
potential common monomial factors to check.
(i) 5:
25x^{ 4}y^{ 3}zw^{ 2} contains 5^{
2}
150x^{ 3}y^{ 3}zr^{ 4} contains 5^{
2}
30x^{ 2}y^{ 3}zw contains 5^{ 1}
The highest power of 5 common to all three terms (which is
actually the lowest power that occurs in this list) is 5^{ 1}
= 5. Thus, 5 is a common factor of all three terms.
(ii) x:
25x^{ 4}y^{ 3}zw^{ 2} contains x^{
4}
150x^{ 3}y^{ 3}zr^{ 4} contains x^{
3}
30x^{ 2}y^{ 3}zw contains x^{ 2}
Thus, the highest power of x common to all three terms is x^{
2}, and so x^{ 2} is a common monomial factor of the
entire expression.
(iii) y:
25x^{ 4}y^{ 3}zw^{ 2} contains y^{
3}
150x^{ 3}y^{ 3}zr^{ 4} contains y^{
3}
30x^{ 2}y^{ 3}zw containsy^{ 3}
Thus, the highest power of y common to all three terms is y^{
3}, and so y^{ 3} is a common monomial factor of the
entire expression.
(iv) z:
25x^{ 4}y^{ 3}zw^{ 2} contains z^{
1}
150x^{ 3}y^{ 3}zr^{ 4} contains z^{
1}
30x^{ 2}y^{ 3}zw contains z^{ 1}
Thus, the highest power of z common to all three terms is z^{
1} = z, and so z is a common monomial factor of the entire
expression.
(v) w:
25x^{ 4}y^{ 3}zw^{ 2} contains w^{
2}
150x^{ 3}y^{ 3}zr^{ 4} does not
contain w (or contains w^{ 0})
30x^{ 2}y^{ 3}zw contains w^{ 1}
Thus the highest power of w common to all three terms is w^{
0} = 1. This is the same thing as saying that there is no
power of w common to all three terms, and so there is no power of
w which is a common monomial factor of all three terms.
So, the common monomial factors of the entire threeterm
expression have been identified as 5, x^{ 2}, y^{ 3},
and z. Our strategy has guaranteed that the product of these, 5x^{
2}y^{ 3}z makes up the greatest monomial factor
common to all three terms. Thus, going back to the template of
our original goal, we can now write that
25x^{ 4}y^{ 3}zw^{ 2} + 150x^{ 3}y^{
3}zr^{ 4} + 30x^{ 2}y^{ 3}zw = (5x^{
2}y^{ 3}z)( whatever else)
All that needs determining yet is the form of the “
whatever else” part. This we do in the same way as was done
in the previous examples. Identify what is left of each term
after the common monomial factor is removed. For the first term,
25x^{ 4}y^{ 3}zw^{ 2} = 5 2x^{ 4}y
3zw^{ 2} = (5x^{ 2}y^{ 3}z)(5x^{ 2}w^{
2})
since to get 25x^{ 4}y^{ 3}zw^{ 2} = 5
2x^{ 4}y^{ 3}zw^{ 2} from 5x^{ 2}y^{
3}z, we need an additional factor of 5 (to make the 5^{ 2}),
an additional factor of x 2 (to make the x 4), and an additional
factor of w^{ 2} (to make the w^{ 2}).
Multiplying the two factors in brackets on the right above is
seen to regenerate the original term on the left.
Now, repeat this process with each of the remaining two terms:
150x^{ 3}y^{ 3}zr^{ 4} = 2 Â· 3 Â· 5^{
2} x^{ 3}y^{ 3}zr^{ 4} = (5x^{ 2}y^{
3}z)(2 Â· 3 Â· 5xr^{ 4}) = (5x^{ 2}y^{ 3}z)(30xr^{
4})
and
30x^{ 2}y^{ 3}zw = 2 Â· 3 Â· 5x^{ 2}y^{
3}zw = (5x^{ 2}y^{ 3}z)(2 Â· 3w) = (5x^{
2}y^{ 3}z)(6w)
So, 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)
This completes the operation of identifying the common
monomial factors in the original expression. You should verify
that multiplying to remove the brackets on the righthand side of
this result gives precisely the expression on the lefthand side,
confirming that the two forms are mathematically equivalent.
