## Precalculus (6th Edition) Blitzer

$(x^{2}+4)(x-2)(x+2)$
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 $x^{2}-2^{2}$, = $(x^{2}+4)(x-2)(x+2)$