# How do you divide (-6x^4-3x^3-2x^2-4x-7)/(x^2+3) ?

$\textcolor{b l u e}{\frac{- 6 {x}^{4} - 3 {x}^{3} - 2 {x}^{2} - 4 x - 7}{{x}^{2} + 3} = - 6 {x}^{2} - 3 x + 16 + \frac{5 x - 55}{{x}^{2} + 3}}$

#### Explanation:

$\frac{- 6 {x}^{4} - 3 {x}^{3} - 2 {x}^{2} - 4 x - 7}{{x}^{2} + 3}$

The Long Division method:

 " " " " " " " " " " " "underline(-6x^2-3x+16" " " " " " " " " " " " " " " " " " ")
${x}^{2} + 0 \cdot x + 3 \lceiling - 6 {x}^{4} - 3 {x}^{3} - 2 {x}^{2} - 4 x - 7$
" " " " " " " " " " " "underline(-6x^4+0*x^3-18x^2" " " " " ")
$\text{ " " " " " " " " " " " " " " " } - 3 {x}^{3} + 16 {x}^{2} - 4 x - 7$
$\text{ " " " " " " " " " " " " " " " } \underline{- 3 {x}^{3} + 0 \cdot {x}^{2} - 9 x}$
$\text{ " " " " " " " " " " " " " " " " " " " " } 16 {x}^{2} + 5 x - 7$
$\text{ " " " " " " " " " " " " " " " " " " " " } \underline{16 {x}^{2} + 0 \cdot x + 48}$
$\text{ " " " " " " " " " " " " " " " " " " " " " " " " } + 5 x - 55 \leftarrow$Remainder

$\textcolor{b l u e}{\frac{- 6 {x}^{4} - 3 {x}^{3} - 2 {x}^{2} - 4 x - 7}{{x}^{2} + 3} = - 6 {x}^{2} - 3 x + 16 + \frac{5 x - 55}{{x}^{2} + 3}}$