Solve x^12-8x^10+16x^8

$$x ^ { 12 } - 8 x ^ { 10 } + 16 x ^ { 8 }$$

Factor

$$\left(x^{2}-4\right)^{2}x^{8}$$

Solution Steps

Factor out x^{8}.

$$x^{8}\left(x^{4}-8x^{2}+16\right)$$

Consider x^{4}-8x^{2}+16. Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{4} and m divides the constant factor 16. One such factor is x^{2}-4. Factor the polynomial by dividing it by this factor.

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

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.

$$x^{8}\left(x-2\right)^{2}\left(x+2\right)^{2}$$

Evaluate

$$\left(x^{2}-4\right)^{2}x^{8}$$