Answer
$3(2y-5)(2y-3)$
Work Step by Step
$12y^{2}-48y+45$
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.
$=3((4y^{2}-6y)+(-10y+15))$
$=3(2y(2y-3)-5(2y-3))$
$=3(2y-5)(2y-3)$
