Solve x^8-4x^6-9x^4+36x^2

$${ x }^{ 8 } -4 { x }^{ 6 } -9 { x }^{ 4 } +36 { x }^{ 2 }$$

Factor

$$\left(x-2\right)\left(x+2\right)x^{2}\left(x^{2}-3\right)\left(x^{2}+3\right)$$

Solution Steps

Factor out x^{2}.

$$x^{2}\left(x^{6}-4x^{4}-9x^{2}+36\right)$$

Consider x^{6}-4x^{4}-9x^{2}+36. Do the grouping x^{6}-4x^{4}-9x^{2}+36=\left(x^{6}-4x^{4}\right)+\left(-9x^{2}+36\right), and factor out x^{4} in the first and -9 in the second group.

$$x^{4}\left(x^{2}-4\right)-9\left(x^{2}-4\right)$$

Factor out common term x^{2}-4 by using distributive property.

$$\left(x^{2}-4\right)\left(x^{4}-9\right)$$

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).

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

Consider x^{4}-9. Rewrite x^{4}-9 as \left(x^{2}\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x^{2}-3\right)\left(x^{2}+3\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-3,x^{2}+3.

$$x^{2}\left(x-2\right)\left(x+2\right)\left(x^{2}-3\right)\left(x^{2}+3\right)$$

Evaluate

$$x^{2}\left(x^{6}-4x^{4}-9x^{2}+36\right)$$