Solve 1-16a^4

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

Factor

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

Solution Steps

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

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

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

Evaluate

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