Solve 32y^4-2x^4y^4

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

Factor

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

Solution Steps

Factor out 2.

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

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

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

Consider 16-x^{4}. Rewrite 16-x^{4} as 4^{2}-\left(-x^{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).

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

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

Evaluate

$$32y^{4}-2\left(xy\right)^{4}$$