Solve 16a^4-81

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

Factor

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

Solution Steps

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

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

Consider 4a^{2}-9. Rewrite 4a^{2}-9 as \left(2a\right)^{2}-3^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

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

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

Evaluate

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