Solve 16x^4-81=

$$16 x ^ { 4 } - 81 =$$

Factor

$$\left(2x-3\right)\left(2x+3\right)\left(4x^{2}+9\right)$$

Solution Steps

Rewrite 16x^{4}-81 as \left(4x^{2}\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider 4x^{2}-9. Rewrite 4x^{2}-9 as \left(2x\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).

$$\left(2x-3\right)\left(2x+3\right)$$

Rewrite the complete factored expression. Polynomial 4x^{2}+9 is not factored since it does not have any rational roots.

$$\left(2x-3\right)\left(2x+3\right)\left(4x^{2}+9\right)$$

Evaluate

$$16x^{4}-81$$