# How do you find the quotient (6t^3-9t^2+6)div(2t-3) using long division?

May 31, 2017

The quotient is $= 3 {t}^{2}$ and the remainder $= 6$

#### Explanation:

We perform a long division

$\textcolor{w h i t e}{a a a a}$$2 t - 3$$\textcolor{w h i t e}{a a a a}$$|$$6 {t}^{3} - 9 {t}^{2} + 0 t + 6$$\textcolor{w h i t e}{a a a a}$$|$$3 {t}^{2}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$6 {t}^{3} - 9 {t}^{2}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a}$$0 - 0 + 0 t + 6$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a}$$+ 6$

The quotient is $= 3 {t}^{2}$ and the remainder $= 6$

$\frac{6 {t}^{3} - 9 {t}^{2} + 6}{2 t - 3} = 3 {t}^{2} + \frac{6}{2 t - 3}$