College Algebra & Trigonometry, 2018a

1.2. FACTORING 15
Example:
Factor 42x 2 y 6 +98xy 3 − 210x 3 y 2
This expression has three terms. It’s not immediately clear what the greatest
common factor of the coefficients is, but they’re all even numbers, so we could at
least divide them all by 2. The 98xy 3 term has an x 1 , which means that this is the
hightest power of x that we could factor out of all the terms. The 210x 3 y 2 has a
y 2 , which is the highest power of y that can be factored out of all the terms. So we
can at least proceed with these factors:
42x 2 y 6 +98xy 3 − 210x 3 y 2 =2xy 2 ∗ 21xy 4 +2xy 2 ∗ 49y − 2xy 2 ∗ 105x 2
=2xy 2 (21xy 4 +49y − 105x 2 )
Now, we didn’t try very hard to find the greatest common factor in the beginning
of this problem, so it’s important that we continue to question whether or not
there are any remaining common factors. The 21 and 49 clearly share a common
factor of 7, so it would make sense to see if 105 is divisible by 7 as well. If we
divide 105 by 7, we see that 105 = 7*15. So, we can also factor out a common
factor of 7 from the remaining terms in the parentheses.
2xy 2 (21xy 4 +49y − 105x 2 )=2xy 2 (7 ∗ 3xy 4 +7∗ 7y − 7 ∗ 15x 2 )
=7∗ 2xy 2 (3xy 4 +7y − 15x 2 )
=14xy 2 (3xy 4 +7y − 15x 2 )