## Precalculus (6th Edition) Blitzer

$(x+y)(4a+5)(4a-5)$
$16a^{2}x-25y-25x+16a^{2}y$= ... Group the terms containing $b^{2}$ and factor in pairs $=(16a^{2}x+16a^{2}y)+(-25y-25x)$ $=16a^{2}(x+y)-25(x+y)$ $=(x+y)(16a^{2}-25)$ ...The second parentheses hold a difference of squares $(4a)^{2}-5^{5}$ = $(x+y)(4a+5)(4a-5)$