# How do you divide (-x^5-4x^3+x-12)/(x^2-x+3)?

Jul 10, 2018

The remainder is $= \left(8 x - 15\right)$ and the quotient is $= \left(- {x}^{3} - {x}^{2} - 2 x + 1\right)$

#### Explanation:

Perform a long division

$\textcolor{w h i t e}{a a}$$- {x}^{5} + 0 {x}^{4} - 4 {x}^{3} + 0 {x}^{2} + x - 12$$\textcolor{w h i t e}{a a}$$|$${x}^{2} - x + 3$

$\textcolor{w h i t e}{a a}$$- {x}^{5} + {x}^{4} - 3 {x}^{3}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$|$$- {x}^{3} - {x}^{2} - 2 x + 1$

$\textcolor{w h i t e}{a a a a}$$0 - 1 {x}^{4} - 1 {x}^{3} + 0 {x}^{2}$

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

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

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

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$0 + {x}^{2} + 7 x - 12$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$+ {x}^{2} - 1 x + 3$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$- 0 + 8 x - 15$

Therefore,

$\frac{- {x}^{5} + 0 {x}^{4} - 4 {x}^{3} + 0 {x}^{2} + x - 12}{{x}^{2} - x + 3}$

$= \left(- {x}^{3} - {x}^{2} - 2 x + 1\right) + \frac{8 x - 15}{{x}^{2} - x + 3}$

The remainder is $= \left(8 x - 15\right)$ and the quotient is $= \left(- {x}^{3} - {x}^{2} - 2 x + 1\right)$