## Intermediate Algebra (12th Edition)

$-t \left( t^{3}-8t^{2}+12 \right)$
$\bf{\text{Solution Outline:}}$ 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}