Answer
$(t-s) \left( 2+4t-4s-t^2+2st-s^2 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Get the $GCF$ of the given expression, $
2(t-s)+4(t-s)^2-(t-s)^3
.$ 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 $\{
2,4,-1
\}$ is $
1
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{
(t-s),(t-s)^2,(t-s)^3
\}$ is $
(t-s)
.$ Hence, the entire expression has $GCF=
(t-s)
.$
Factoring the $GCF=
(t-s)
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(t-s) \left( \dfrac{2(t-s)}{(t-s)}+\dfrac{4(t-s)^2}{(t-s)}-\dfrac{(t-s)^3}{(t-s)} \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}
(t-s) \left( 2(t-s)^{1-1}+4(t-s)^{2-1}-(t-s)^{3-1} \right)
\\\\=
(t-s) \left( 2(t-s)^{0}+4(t-s)^{1}-(t-s)^{2} \right)
\\\\=
(t-s) \left( 2(1)+4(t-s)-(t-s)^{2} \right)
\\\\=
(t-s) \left( 2+4(t-s)-(t-s)^{2} \right)
.\end{array}
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the expression above is
\begin{array}{l}\require{cancel}
(t-s) \left( 2+4(t-s)-(t^2-2st+s^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}
(t-s) \left( 2+4t-4s-t^2+2st-s^2 \right)
\\\\=
(t-s) \left( 2+4t-4s-t^2+2st-s^2 \right)
.\end{array}
