Solve 8x^4-128

$$8 x ^ { 4 } - 128$$

Factor

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

Solution Steps

Factor out 8.

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

Consider x^{4}-16. Rewrite x^{4}-16 as \left(x^{2}\right)^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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. Polynomial x^{2}+4 is not factored since it does not have any rational roots.

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

Evaluate

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