# How do you divide (-x^4-3x^3-2x^2+7x-7)/(x^2+3) ?

Jul 21, 2018

${x}^{2} - 3 x + 1 + \frac{16 x - 10}{{x}^{2} + 3}$

#### Explanation:

$\frac{- {x}^{4} - 3 {x}^{3} - 2 {x}^{2} + 7 x - 7}{{x}^{2} + 3}$

$\textcolor{w h i t e}{\ldots \ldots} \textcolor{w h i t e}{\ldots . .} - {x}^{2} - 3 x + 1$
${x}^{2} + 3 | \overline{- {x}^{4} - 3 {x}^{3} - 2 {x}^{2} + 7 x - 7}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} \underline{- {x}^{4} + \text{0} - 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots} - 3 {x}^{3} + {x}^{2} + 7 x$
color(white)(......................)ul(-3x^3+""0-9x)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} {x}^{2} + 16 x - 7$
color(white)(.....................................)ul(x^2+""0+3)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 16 x - 10$

$\frac{- {x}^{4} - 3 {x}^{3} - 2 {x}^{2} + 7 x - 7}{{x}^{2} + 3} = - {x}^{2} - 3 x + 1 + \frac{16 x - 10}{{x}^{2} + 3}$