Solve 16p^4-1=

$$16 p ^ { 4 } - 1 =$$

Factor

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

Solution Steps

Rewrite 16p^{4}-1 as \left(4p^{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).

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

Consider 4p^{2}-1. Rewrite 4p^{2}-1 as \left(2p\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).

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

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

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

Evaluate

$$16p^{4}-1$$