Answer
$(2y-5)(2y-3)$
Work Step by Step
Rewrite $-16y$ as $-10y-6y$.
$4y^2-16y+15=4y^2-10y-6y+15$
Group the first two terms together and group the last two terms together.
$=(4y^2-10y)+(-6y+15)$
Factor out the GCF in each group.
$=2y(2y-5)+(-3)(2y-5)$
$=2y(2y-5)-3(2y-5)$
Factor out $(2y-5)$.
$=(2y-5)(2y-3)$
Hence, the completely factored form of the given expression is $(2y-5)(2y-3)$.
