## Intermediate Algebra (12th Edition)

$m^{-5} \left( 3+m^{2} \right)$
$\bf{\text{Solution Outline:}}$ Factor the variable with the lesser exponent in the given expression, $3m^{-5}+m^{-3} .$ Then, divide the given expression and the variable with the lesser exponent. $\bf{\text{Solution Details:}}$ Factoring $m^{-5}$ (the variable with the lesser exponent), the expression above is equivalent to \begin{array}{l}\require{cancel} m^{-5} \left( \dfrac{3m^{-5}}{m^{-5}}+\dfrac{m^{-3}}{m^{-5}} \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} m^{-5} \left( 3m^{-5-(-5)}+m^{-3-(-5)} \right) \\\\= m^{-5} \left( 3m^{-5+5}+m^{-3+5} \right) \\\\= m^{-5} \left( 3m^{0}+m^{2} \right) \\\\= m^{-5} \left( 3(1)+m^{2} \right) \\\\= m^{-5} \left( 3+m^{2} \right) .\end{array}