jmc

Since $w^4$ and 16 are both perfect squares, we can use our difference of squares factorization: $$w^4-16=(w^2{)}^2-4^2=(w^2-4)(w^2+4)$$. We're not finished! The expression $w^2-4$ is also a difference of squares, which we can factor as $w^2-4=(w-2)(w+2)$. So, we have $$w^4-16=(w^2-4)(w^2+4)=\boxed{(w-2)(w+2)(w^2+4)}$$.