## Precalculus (6th Edition)

$(b+4+a)(b+4-a)$
Recognize $b^{2}+8b+16=(b+4)^{2}$ $b^{2}+8b+16-a^{2}=(b+4)^{2}-a^{2}$ ... and this is a difference of squares, $A^{2}-B^{2}=(A-B)(A+B)$ $=[(b+4)+a][(b+4)-a]$ $=(b+4+a)(b+4-a)$