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

1 Answer
Mar 23, 2016

Answer:

#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)]#