Solve 2w^4y^4-32w^4

$$2 w ^ { 4 } y ^ { 4 } - 32 w ^ { 4 }$$

Factor

$$2\left(y-2\right)\left(y+2\right)\left(y^{2}+4\right)w^{4}$$

Solution Steps

Factor out 2.

$$2\left(w^{4}y^{4}-16w^{4}\right)$$

Consider w^{4}y^{4}-16w^{4}. Factor out w^{4}.

$$w^{4}\left(y^{4}-16\right)$$

Consider y^{4}-16. Rewrite y^{4}-16 as \left(y^{2}\right)^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(y^{2}-4\right)\left(y^{2}+4\right)$$

Consider y^{2}-4. Rewrite y^{2}-4 as y^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(y-2\right)\left(y+2\right)$$

Rewrite the complete factored expression. Polynomial y^{2}+4 is not factored since it does not have any rational roots.

$$2w^{4}\left(y-2\right)\left(y+2\right)\left(y^{2}+4\right)$$

Evaluate

$$2w^{4}\left(y^{4}-16\right)$$