# How do you find the quotient of (p^3-4p^2+9)div(p-1) using long division?

Sep 18, 2017

${p}^{2} - 3 p + 3$ and remainder$\frac{12}{p - 1}$

#### Explanation:

$\left({p}^{3} - 4 {p}^{2} + 9\right) \div \left(p - 1\right)$

$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} {p}^{2} - 3 p + 3$
$p - 1 | \overline{{p}^{3} - 4 {p}^{2} + 0 + 9}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots} \underline{{p}^{3} - {p}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} - 3 {p}^{2} + 0$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} \underline{- 3 {p}^{2} - 3 p}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 3 p + 9$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{3 p - 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 12$

$\frac{{p}^{3} - 4 {p}^{2} + 9}{p - 1} = {p}^{2} - 3 p + 3$ and remainder$\frac{12}{p - 1}$