#### Answer

$\dfrac{6+2\sqrt{6p}}{3}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\dfrac{6p+\sqrt{24p^3}}{3p}
,$ simplify the radicand that contains a factor that is a perfect power of the index Then, find the $GCF$ of all the terms and express all terms as factors using the $GCF.$ Finally, cancel the $GCF$ in all the terms.
$\bf{\text{Solution Details:}}$
Writing the radicand as an expression containing a factor that is a perfect power of the index and extracting the root of that factor result to
\begin{array}{l}\require{cancel}
\dfrac{6p+\sqrt{4p^2\cdot6p}}{3p}
\\\\=
\dfrac{6p+\sqrt{(2p)^2\cdot6p}}{3p}
\\\\=
\dfrac{6p+2p\sqrt{6p}}{3p}
.\end{array}
The $GCF$ of the constants of the terms $\{
6,2,3
\}$ is $
1
$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
p,p,p
\}$ is $
p
.$ Hence, the entire expression has $GCF=
1p \text{ or } p
.$ Writing the given expression as factors using the $GCF$ results to
\begin{array}{l}\require{cancel}
\dfrac{p\cdot6+p\cdot2\sqrt{6p}}{p\cdot3}
.\end{array}
Cancelling the $GCF$ in every term results to
\begin{array}{l}\require{cancel}
\dfrac{\cancel{p}\cdot6+\cancel{p}\cdot2\sqrt{6p}}{\cancel{p}\cdot3}
\\\\=
\dfrac{6+2\sqrt{6p}}{3}
.\end{array}