# How do you divide (-3x^4+3x^2+2x+15)/(-x^3-x)?

Dec 30, 2017

$= 3 x - \frac{2}{x}$

#### Explanation:

You can do long division - but let's adjust the signs first, because it is not comfortable with the divisor starting with a negative sign

$\rightarrow \text{ }$ dividing by a negative changes the signs.

Take out a factor of $- 1$ in the numerator and denominator:

$\frac{- 3 {x}^{4} + 3 {x}^{2} + 2 x + 15}{- {x}^{3} - x}$

$= \frac{- \left(3 {x}^{4} - 3 {x}^{2} - 2 x - 15\right)}{- \left({x}^{3} + x\right)} \text{ } \leftarrow \left(\frac{-}{-}\right) = +$

$= \frac{3 {x}^{4} - 3 {x}^{2} - 2 x - 15}{{x}^{3} + x}$

Leave a space for the missing term in ${x}^{3}$

$\textcolor{w h i t e}{m m m m m m m m . m m m m} 3 x \text{ } \leftarrow \left(3 {x}^{4} \div {x}^{3} = 3 x\right)$
${x}^{3} + x | \overline{3 {x}^{4} \text{ } - 3 {x}^{2} - 2 x - 15}$

Multiply $3 x$ by both terms in the divisor:

$\textcolor{w h i t e}{m m m m m m m m . m m m m} 3 x$
${x}^{3} + x | \overline{3 {x}^{4} \text{ } - 3 {x}^{2} - 2 x - 15}$
$\textcolor{w h i t e}{m m m . m} \underline{3 {x}^{4} \text{ "" } 3 {x}^{2}} \textcolor{w h i t e}{\times \times \times \times x} \leftarrow$ subtract
$\textcolor{w h i t e}{m m m m m m . m} - 6 {x}^{2} - 2 x \text{ } \leftarrow$ bring down next term

$\left(3 {x}^{4} - 3 {x}^{2} - 2 x - 15\right) \div \left({x}^{3} + x\right)$

$= 3 x \text{ remainder } - 6 {x}^{2} - 2 x$

This can be written as:

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

$3 x + \frac{- 2 x \left(x + 1\right)}{{x}^{2} \left(x + 1\right)} \text{ } \leftarrow$ factorise top and bottom

$= 3 x - \frac{2}{x}$