Answer
$-t \left( t^{3}-8t^{2}+12 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
First, get the $GCF$ of the given expression, $
-t^4+8t^3-12t
.$ 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 $\{
-1,8,-12
\}$ 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 $\{
t^4,t^3,t
\}$ is $
t
.$ Hence, the entire expression has $GCF=
t
.$
Factoring the $GCF=
t
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
t \left( \dfrac{-t^4}{t}+\dfrac{8t^3}{{t}}-\dfrac{12t}{{t}} \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 \left( -t^{4-1}+8t^{3-1}-12t^{1-1} \right)
\\\\=
t \left( -t^{3}+8t^{2}-12t^{0} \right)
\\\\=
t \left( -t^{3}+8t^{2}-12(1) \right)
\\\\=
t \left( -t^{3}+8t^{2}-12 \right)
.\end{array}
Factoring out $-1$ from the second factor results to
\begin{array}{l}\require{cancel}
-t \left( t^{3}-8t^{2}+12 \right)
.\end{array}
