#### Answer

$4p-8+\dfrac{6}{p}$

#### Work Step by Step

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}