# How do you divide ( -x^3 - 6x^2+3x+4 )/(2x^2 - x )?

Jul 7, 2018

$- \frac{1}{2} x - \frac{13}{4} + \frac{- \frac{x}{4} + 4}{2 {x}^{2} - x}$

#### Explanation:

Given: $\frac{- {x}^{3} - 6 {x}^{2} + 3 x + = 4}{2 {x}^{2} - x}$

Using long division:

$2 {x}^{2} - x | \overline{- {x}^{3} - 6 {x}^{2} + 3 x + 4}$

What times $2 {x}^{2} = - {x}^{3}$? $\text{ } - \frac{1}{2} x$

$\text{ } - \frac{1}{2} x$
$2 {x}^{2} - x | \overline{- {x}^{3} - 6 {x}^{2} + 3 x + 4}$
$\text{ } \underline{- {x}^{3} + \frac{1}{2} {x}^{2}}$
$\text{ } - \frac{13}{2} {x}^{2} + 3 x$

What times $2 {x}^{2} = - \frac{13}{2} {x}^{2}$? $\text{ } - \frac{13}{4}$

$\text{ } - \frac{1}{2} x - \frac{13}{4}$
$2 {x}^{2} - x | \overline{- {x}^{3} - 6 {x}^{2} + 3 x + 4 \text{ }}$
$\text{ } \underline{- {x}^{3} + \frac{1}{2} {x}^{2}}$
$\text{ } - \frac{13}{2} {x}^{2} + 3 x$
" "ul(-13/2x^2 + 13/4x
$\text{ } - \frac{1}{4} x + 4$ This is the remainder

$\frac{- {x}^{3} - 6 {x}^{2} + 3 x + = 4}{2 {x}^{2} - x} = - \frac{1}{2} x - \frac{13}{4} + \frac{- \frac{x}{4} + 4}{2 {x}^{2} - x}$

The remainder can be simplified if desired.