## Precalculus (6th Edition) Blitzer

$(2x+3)(2x+5)$
$4x^{2}+16x+15=$ two factors of $ac=+60$ whose sum is $b=+16$ are $+6$ and $+10.$ Rewrite the middle term and factor in pairs. $4x^{2}+16x+15=4x^{2}+6x+10x+15$ $=2x(2x+3)+5(2x+3)$ $=(2x+3)(2x+5)$