Solve 2x^8-x^4-1

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

Factor

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

Solution Steps

Find one factor of the form kx^{m}+n, where kx^{m} divides the monomial with the highest power 2x^{8} and n divides the constant factor -1. One such factor is 2x^{4}+1. Factor the polynomial by dividing it by this factor.

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

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).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}+1,2x^{4}+1.

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

Evaluate

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