Answer
$(x^{2}+4)(x-2)(x+2)$
Work Step by Step
This is a difference of squares, $ x^{4}=(x^{2})^{2}$
$16=4^{2}$
We apply the formula $(A-B)(A+B)=A^{2}-B^{2}$
$=(x^{2}+4)(x^{2}-4)$
The first parentheses are a sum of squares, which is prime; the second parentheses hold another difference of squares $ x^{2}-2^{2}$,
= $(x^{2}+4)(x-2)(x+2)$
