Solve x^8-2x^4+1

$$x ^ { 8 } - 2 x ^ { 4 } + 1$$

Factor

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

Solution Steps

Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{8} and m divides the constant factor 1. One such factor is x^{4}-1. Factor the polynomial by dividing it by this factor.

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

Consider x^{4}-1. Rewrite x^{4}-1 as \left(x^{2}\right)^{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).

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

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}-1. Rewrite x^{4}-1 as \left(x^{2}\right)^{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).

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

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).

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

Rewrite the complete factored expression. Polynomial x^{2}+1 is not factored since it does not have any rational roots.

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

Evaluate

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