Question 788072
{{{6w^4-54w^2}}}
The coefficients are both multiples of {{{6}}}, because {{{-54=6*(-9)}}}, so we could take out {{{6}}} as a common factor.
Both terms have powers of {{{w}}}, and the smallest exponent is {{{2}}}, and we can write {{{w^4}}} as {{{w^2*w^2=w^4}}}, so we could take out {{{w^2}}} as a common factor.
We can take out both actors at the same time, as {{{6w^2}}}
{{{6w^4=6w^2*w^2}}} and {{{-54w^2=6W^2*(-9)}}}
{{{6w^4-54w^2=6w^2(w^2-9)}}}
Next, we realize that {{{w^2-9}}} is a difference of squares, a special product of the form
{{{a^2-b^2=(a+b)(a-b)}}}.
Specifically, {{{w^2-9=(w+3)(w-3)}}}.
So the complete factorization is
{{{6w^4-54w^2=6w^2(w^2-9)=highlight(6w^2(w+3)(w-3))}}}