## Intermediate Algebra (6th Edition)

$3(2y-5)(2y-3)$
$12y^{2}-48y+45$ Factor out greatest common factor $3$ $=3(4y^{2}-16y+15)$ In this trinomial $(4y^{2}-16y+15)$, $a=4$, $b=-16$ and $c=15$. Splitting the middle term $b$ into two numbers whose product is $60$ $(a \times c)$ and whose sum is $(b)$, $-16$. The numbers are $-6$ and $-10$. $=3(4y^{2}-6y-10y+15)$ Factor out by grouping. $=3((4y^{2}-6y)+(-10y+15))$ $=3(2y(2y-3)-5(2y-3))$ $=3(2y-5)(2y-3)$