Solve 64a^4-4

$$64 a ^ { 4 } - 4$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(16a^{4}-1\right)$$

Consider 16a^{4}-1. Rewrite 16a^{4}-1 as \left(4a^{2}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Consider 4a^{2}-1. Rewrite 4a^{2}-1 as \left(2a\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2a-1\right)\left(2a+1\right)$$

Rewrite the complete factored expression. Polynomial 4a^{2}+1 is not factored since it does not have any rational roots.

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

Evaluate

$$64a^{4}-4$$