Solve 1-16p^4

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

Factor

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

Solution Steps

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

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

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