Monomial Factors
Examples with solutions
Example 1:
Remove all common monomial factors from the expression
3r 4s 3 + 6r 3s 4
– 105r 2s 5
solution:
The first term in this expression has factors of 3, r, and s.
So, we need to check whether 3 is a common factor in all three
terms, and whether there are common powers of r and s in all
three terms.
| 3r 4s 3
+ 6r 3s 4 – 105r 2s
5 |
3 is a factor of the first term, and upon
inspection is found to be a factor of the remaining two
terms as well. So, remove a factor of 3 from all three
terms. |
| = 3(r 4s 3
+ 2r 3s 4 – 35r 2s
5) |
Each term contains a factor of r 2.
Remove it. |
| = 3r 2(r 2s
3 + 2rs 4 – 35s 5) |
Each term in the brackets now contains a
factor of s 3. Remove it. |
| = 3r 2s 3(r
2 + 2rs – 35s 2) |
Since the first term in brackets contains
only powers of r, but the third term contains no factor
of r, there cannot be any further common monomial factors
in the three terms of the expression in brackets. The
required factorization of monomial factors is thus
complete. |
Thus, with all fo the common monomial factors identified, we
can write the final result as:
3r 4s 3 + 6r 3s 4
– 105r 2s 5 = 3r 2s 3(r
2 + 2rs – 35s 2)
(Once you’ve read the next section in these notes,
you’ll be able to see that the expression in brackets in the
final answer of this last example can, in fact, be written as a
product, though not monomial factors. So, this is a final answer
only as far as the methods presented so far here concerned.)
Example 2:
Remove all common monomial factors from the expression
54x 3y 2z – 150xy 4z
solution:
Looking at the first term here, we see that in addition to a
possible common numerical factor, need to check for common powers
of x, y, and z in the two terms.
The hardest part of this example is in determining the common
numerical factor in the two terms. However, writing the two
numerical coefficients as products of prime factors (we described
how do this in the note called “Prime Factors” in the
section on numerical fractions), we get
54 = 2 × 3 × 3 × 3
and
150 = 2 × 3 × 5 × 5
Thus, both terms share a factor of 2 × 3 = 6. Proceeding as
in previous examples, we now get:
54x 3y 2z – 150xy 4z
| = 6(9x 3y 2z –
25xy 4z) |
removing the already identified common
numerical factor of 6. Now we note that both terms share
a factor of x 1 = x, so remove this factor. |
| = 6x(9x 2y 2z
– 25y 4z) |
Next, there is a common factor of y
2 shared by the two terms, so remove it. |
| = 6xy 2(9x 2z
– 25y 2z) |
Finally, both terms share a factor of z,
so remove it. |
| = 6xy 2z(9x 2
– 25y 2) |
|
Thus, the final answer here is
54x 3y 2z – 150xy 4z =
6xy 2z(9x 2 – 25y 2)
Shortly, we’ll demonstrate that the bracketed expression
in the answer given above in Example can be factored into the
product of two binomials, so that the most factored form of the
original expression is
54x 3y 2z – 150xy 4z =
6xy 2z(3x + 5y)(3x – 5y).
However, the identification of all common monomial factors is
the necessary first step before more specialized types of
factorization can be attempted.
