Answer
$-r \left( r^{2}-3r-5 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
First, get the $GCF$ of the given expression, $
-r^3+3r^2+5r
.$ Then, divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. Finally, factor out the $-1$.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
-1,3,5
\}$ is $
1
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
r^3,r^2,r
\}$ is $
r
.$ Hence, the entire expression has $GCF=
r
.$
Factoring the $GCF=
r
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
r \left( \dfrac{-r^3}{r}+\dfrac{3r^2}{{r}}+\dfrac{5r}{{r}} \right)
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
r \left( -r^{3-1}+3r^{2-1}+5r^{1-1} \right)
\\\\=
r \left( -r^{2}+3r^{1}+5r^{0} \right)
\\\\=
r \left( -r^{2}+3r+5(1) \right)
\\\\=
r \left( -r^{2}+3r+5 \right)
.\end{array}
Factoring out $-1$ from the second factor results to
\begin{array}{l}\require{cancel}
-r \left( r^{2}-3r-5 \right)
.\end{array}
