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

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

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}-2\right)\left(x^{2}+2\right)$$

Solution Steps

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

$$x^{4}\left(x^{2}-1\right)-4\left(x^{2}-1\right)$$

Factor out common term x^{2}-1 by using distributive property.

$$\left(x^{2}-1\right)\left(x^{4}-4\right)$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)$$

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

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

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

$$\left(x^{2}-2\right)\left(x-1\right)\left(x+1\right)\left(x^{2}+2\right)$$

Evaluate

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