Solve x^6+2x^4-16x^2-32

$${ x }^{ 6 } +2 { x }^{ 4 } -16 { x }^{ 2 } -32$$

Factor

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

Solution Steps

Do the grouping x^{6}+2x^{4}-16x^{2}-32=\left(x^{6}+2x^{4}\right)+\left(-16x^{2}-32\right), and factor out x^{4} in the first and -16 in the second group.

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

Factor out common term x^{2}+2 by using distributive property.

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

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

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{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).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}+2,x^{2}+4.

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

Evaluate

$$x^{6}+2x^{4}-16x^{2}-32$$