## Precalculus (6th Edition) Blitzer

$\text{Prime}$
Recall that $a^2-b^2=(a+b)(a-b)$. Since $x^2+16$ is in this form, we would use this method to factor. However, as $\sqrt{-16}$ is imaginary, there are no real factors, and thus the expression is $\text{prime}$.