Solve 2x^4-32y^4

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

Factor

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

Solution Steps

Factor out 2.

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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