#### Answer

$(c-3)(c+1)$

#### Work Step by Step

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