## Intermediate Algebra for College Students (7th Edition)

$(9y-5), (9y+5)$
Given: $81y^2-25$ Since, $a^2-b^2=(a-b)(a+b)$ This implies, $81y^2-25=(9y)^2-(5)^2$ we have $=(9y-5)(9y+5)$ Hence, factors are:$(9y-5), (9y+5)$