Answer
$(y-7)(y-1)$
Work Step by Step
The 2 numbers whose product is $ac=
1(-8)=-8
$ and whose sum is $b=
-2
$ are $\left\{
-4,2
\right\}$. Using these numbers to decompose the middle term results to
\begin{array}{l}
(y-3)^2-4(y-3)+2(y-3)-8
\\\\=
[(y-3)^2-4(y-3)]+[2(y-3)-8]
\\\\=
[(y-3)(y-3-4)]+[2(y-3-4)]
\\\\=
(y-3-4)(y-3+2)
\\\\=
(y-7)(y-1)
.\end{array}
