Answer
$5(m+p)^2 \left( m+p-2-3m^2-6mp-3p^2 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Get the $GCF$ of the given expression, $
5(m+p)^3-10(m+p)^2-15(m+p)^4
.$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
5,-10,-15
\}$ is $
5
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
(m+p)^3,(m+p)^2,(m+p)^4
\}$ is $
(m+p)^2
.$ Hence, the entire expression has $GCF=
5(m+p)^2
.$
Factoring the $GCF=
5(m+p)^2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
5(m+p)^2 \left( \dfrac{5(m+p)^3}{5(m+p)^2}-\dfrac{10(m+p)^2}{5(m+p)^2}-\dfrac{15(m+p)^4}{5(m+p)^2} \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}
5(m+p)^2 \left( (m+p)^{3-2}-2(m+p)^{2-2}-3(m+p)^{4-2} \right)
\\\\=
5(m+p)^2 \left( (m+p)^{1}-2(m+p)^{0}-3(m+p)^{2} \right)
\\\\=
5(m+p)^2 \left( (m+p)-2(1)-3(m+p)^{2} \right)
.\end{array}
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the expression above is equivalent to
\begin{array}{l}\require{cancel}
5(m+p)^2 \left( (m+p)-2(1)-3(m^2+2mp+p^2) \right) .\end{array}
Using $a(b+c)=ab+ac$ or the Distributive Property and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
5(m+p)^2 \left( m+p-2-3m^2-6mp-3p^2 \right)
.\end{array}
