## Intermediate Algebra (12th Edition)

$-5a^3 \left( 1-2a+3a^{2} \right)$
$\bf{\text{Solution Outline:}}$ First, get the $GCF$ of the given expression, $-5a^3+10a^4-15a^5 .$ 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 $\{ -5,10,-15 \}$ 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 $\{ a^3,a^4,a^5 \}$ is $a^3 .$ Hence, the entire expression has $GCF= 5a^3 .$ Factoring the $GCF= 5a^3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5a^3 \left( \dfrac{-5a^3}{5a^3}+\dfrac{10a^4}{5a^3}-\dfrac{15a^5}{5a^3} \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} 5a^3 \left( -a^{3-3}+2a^{4-3}-3a^{5-3} \right) \\\\= 5a^3 \left( -a^{0}+2a^{1}-3a^{2} \right) \\\\= 5a^3 \left( -1+2a-3a^{2} \right) .\end{array} Factoring out $-1$ from the second factor results to \begin{array}{l}\require{cancel} -5a^3 \left( 1-2a+3a^{2} \right) .\end{array}