SOLUTION: Factor a^2(b-c)-16b^2(b-c)

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Question 78520: Factor a^2(b-c)-16b^2(b-c)
Answer by mathdoc314(58) About Me  (Show Source):
You can put this solution on YOUR website!
This expression can be made longer or shorter
The %28b-c%29 is "like terms" so one thing to do with it is combine them:
%28a%5E2+-+16+b%5E2%29+%28b-c%29
Maybe it can be made even more compact if we multiply out the %28b-c%29 first
a%5E2+b+-+a%5E2+c+-+16+b%5E3+%2B+16+b%5E2+c
and then factor it in terms of b since b appears as a cube, a square, and a linear term.
16+b%5E3+%2B+16+c+%2A+b%5E2+%2B+a%5E2+%2A+b+%2B+a%5E2+c
I don't think that is going to factor any better than we had it to start with
Let's go back to the first idea. We got it to
%28a%5E2+-+16+b%5E2%29+%28b-c%29
I just noticed we can factor the first quantity there.
a%5E2+-+16+b%5E2 is the same as %28a%2B4b%29%28a-4b%29.
So I think the answer is: %28a%2B4b%29%28a-4b%29%28b-c%29.
Now it appears to me to be factored completely into nice linear terms.