Answer
$(3y-2)(3y-4)$
Work Step by Step
When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$
and, if we find them, we rewrite $bx$ and proceed to factor in groups.
Here, factors of $9\times 8=72$ that add to $-18$ are $-12$ and $-6.$
$9y^{2}-18y+8=9y^{2}-6y-12y+8$
$=\left(9y^{2}-6y\right)+(-12y+8)$
$=3y(3y-2)-4(3y-2)$
$=(3y-2)(3y-4)$
