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

Jun 25, 2018

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

#### Explanation:

Using place keepers of zero value. Example: $0 {x}^{3}$

$\textcolor{w h i t e}{\text{dddddddddddddd}} + {x}^{4} - 3 {x}^{3} - 2 {x}^{2} + 7 x + 3$
$\textcolor{m a \ge n t a}{- {x}^{2}} \left({x}^{2} + 3\right) \to \textcolor{w h i t e}{\text{d") ul(-x^4+0x^3-3x^2larr" Subtract}}$
$\textcolor{w h i t e}{\text{dddddddddddddddd}} 0 - 3 {x}^{3} + {x}^{2} + 7 x + 3$
$\textcolor{m a \ge n t a}{- 3 x} \left({x}^{2} + 3\right) \to \textcolor{w h i t e}{\text{dddd")ul(-3x^3+0x^2-9xlarr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddddddd}} 0 + {x}^{2} + 16 x + 3$
$\textcolor{m a \ge n t a}{+ 1} \left({x}^{2} + 3\right) \to \textcolor{w h i t e}{\text{dddddddddddd")ul(x^2+color(white)("..")0x+3larr" Subtract}}$
color(magenta)("Remainder "->color(white)("dddddddddddd.")0+16x+0)

$\textcolor{w h i t e}{}$

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