# How do you find the quotient of (9n^3-13n+8)div(3n-1) using long division?

Apr 11, 2017

The quotient is $= 3 {n}^{2} + n - 4$

#### Explanation:

Let's perform the long division

$\textcolor{w h i t e}{a a a a}$$9 {n}^{3}$$\textcolor{w h i t e}{a a a a a a}$$- 13 n + 8$$|$$3 n - 1$

$\textcolor{w h i t e}{a a a a}$$9 {n}^{3} - 3 {n}^{2}$$\textcolor{w h i t e}{a a a a a a a a a}$|3n^2+n-4

$\textcolor{w h i t e}{a a a a a a}$$0 + 3 {n}^{2} - 13 n$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 3 {n}^{2} - n$

$\textcolor{w h i t e}{a a a a a a a a a}$$+ 0 - 12 n + 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$- 12 n + 4$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$- 0 + 4$

So,

$\frac{9 {n}^{3} - 13 n + 8}{3 n - 1} = \left(3 {n}^{2} + n - 4\right) + \frac{4}{3 n - 1}$