#### Answer

$(3r+16)(r+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
3(r+5)^2-11(r+5)-4
,$ use substitution to simplify the expression. Then use factoring of trinomials by finding two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. The use back-substitution to have the factored form of the original expression.
$\bf{\text{Solution Details:}}$
Let $z=(r+5).$ Then the given expression is equivalent to
\begin{array}{l}\require{cancel}
3z^2-11z-4
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
3(-4)=-12
$ and the value of $b$ is $
-11
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-12\}, \{2,-6\}, \{3,-4\},
\\
\{-1,12\}, \{-2,6\}, \{-3,4\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,-12
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3z^2+z-12z-4
.\end{array}
Grouping the first and second terms and the third and fourth terms of the trinomial, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(3z^2+z)-(12z+4)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
z(3z+1)-4(3z+1)
.\end{array}
Factoring the $GCF=
(3z+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3z+1)(z-4)
.\end{array}
Since $z=r+5,$ then by back-substitution, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(3(r+5)+1)((r+5)-4)
\\\\=
(3r+15+1)(r+5-4)
\\\\=
(3r+16)(r+1)
.\end{array}