#### Answer

$(y-5)(y-4)$

#### Work Step by Step

Let $z=(y-6)$. Then the given expression, $
(y-6)^2+3(y-6)+2
$, is equivalent to
\begin{array}{l}
z^2+3z+2
.\end{array}
The two numbers whose product is $ac=
1(2)=2
$ and whose sum is $b=
3
$ are $\{
1,2
\}$. Using these two numbers to decompose the middle term of the expression, $
z^2+3z+2
$, results to
\begin{array}{l}
z^2+1z+2z+2
\\\\=
z(z+1)+2(z+1)
\\\\=
(z+1)(z+2)
.\end{array}
Since $z=(y-6)$, then,
\begin{array}{l}
(z+1)(z+2)
\\\\=
(y-6+1)(y-6+2)
\\\\=
(y-5)(y-4)
.\end{array}