#### Answer

$(p+2)^2(p-2)(p+3)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
p^2(p+2)^2+p(p+2)^2-6(p+2)^2
,$ factor first the $GCF.$ Then factor the resulting trinomial 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.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
(p+2)^2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(p+2)^2(p^2+p-6)
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
1(-6)=-6
$ and the value of $b$ is $
1
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-6\}, \{2,-3\},
\\
\{-1,6\}, \{-2,3\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-2,3
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
(p+2)^2(p^2-2p+3p-6)
.\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}
(p+2)^2[(p^2-2p)+(3p-6)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
(p+2)^2[p(p-2)+3(p-2)]
.\end{array}
Factoring the $GCF=
(p+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(p+2)^2[(p-2)(p+3)]
\\\\=
(p+2)^2(p-2)(p+3)
.\end{array}