How do you factor 3u^3w^4- 3u^3?

1 Answer
Mar 23, 2016

3u^3w^4-3u^3=3u^3(w^2+1)(w+1)(w-1)

Explanation:

Start with a partial factorization:

3u^3w^4-3u^3=3u^3(w^4-1)=3u^3[(w^2)^2-1]=

Now you can factor [(w^2)^2-1] with the remarkable identity:

a^2-b^2=(a+b)(a-b) where a=w^2 and b=1

=3u^3[(w^2+1)(w^2-1)]=

Now (w^2-1) can more be factorizide with the same identity

=3u^3[(w^2+1)(w+1)(w-1)]