# How do you find the quotient of (t^4+6t^3-3t^2-t+8)/(t+1) using long division?

Dec 31, 2016

The quotient is $= {t}^{3} + 5 {t}^{2} - 8 t + 7$ and the remainder is $= 1$

#### Explanation:

Let's do the lomg division

$\textcolor{w h i t e}{a a a a}$${t}^{4} + 6 {t}^{3} - 3 {t}^{2} - t + 8$$\textcolor{w h i t e}{a a a a}$∣$t + 1$

$\textcolor{w h i t e}{a a a a}$${t}^{4} + {t}^{3}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$∣${t}^{3} + 5 {t}^{2} - 8 t + 7$

$\textcolor{w h i t e}{a a a a}$$0 + 5 {t}^{3} - 3 {t}^{2}$

$\textcolor{w h i t e}{a a a a a a}$$+ 5 {t}^{3} + 5 {t}^{2}$

$\textcolor{w h i t e}{a a a a a a a}$$+ 0 - 8 {t}^{2} - t$

$\textcolor{w h i t e}{a a a a a a a a a a a}$$- 8 {t}^{2} - 8 t$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$- 0 + 7 t + 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$+ 7 t + 7$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$+ 0 + 1$

The quotient is $= {t}^{3} + 5 {t}^{2} - 8 t + 7$ and the remainder is $= 1$