## Intermediate Algebra (12th Edition)

$4p-8+\dfrac{6}{p}$
Dividing each of the terms of the nuemerator by the denominator, the given expression, $\dfrac{16p^3-32p^2+24p}{4p^2} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{16p^3}{4p^2}-\dfrac{32p^2}{4p^2}+\dfrac{24p}{4p^2} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the quotient of the expression above is \begin{array}{l}\require{cancel} 4p^{3-2}-8p^{2-2}+6p^{1-2} \\\\= 4p^{1}-8p^{0}+6p^{-1} \\\\= 4p-8(1)+6p^{-1} \\\\= 4p-8+6p^{-1} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4p-8+\dfrac{6}{p^1} \\\\= 4p-8+\dfrac{6}{p} .\end{array}