# How do you divide ( -x^4 + 3x^3 + 9x^2 +4x)/(2x^2-3x)?

Dec 30, 2015

Trigonometry long division gives quotient of $\frac{- 4 {x}^{2} + 6 x + 9}{8}$ and remainder of $\frac{59}{8}$

#### Explanation:

This can be simplified a bit by first taking out common factors. After that you need to do the trigonometry version of long division.
$\frac{x \left(- {x}^{3} + 3 {x}^{2} + 9 x + 4\right)}{x \left(2 x - 3\right)} = \frac{- {x}^{3} + 3 {x}^{2} + 9 x + 4}{2 x - 3}$
Divide the first element by the first element of the divisor:
$- {x}^{3} / \left(2 x\right) = \ast - {x}^{2} / 2 \ast$

Multiply this result by the divisor:
$- {x}^{2} / 2 \cdot \left(2 x - 3\right) = - {x}^{3} + 3 {x}^{2} / 2$

Then subtract this from the dividend:
$\left(- {x}^{3} + 3 {x}^{2} + 9 x + 4\right) - \left(- {x}^{3} + 3 {x}^{2} / 2\right) = 3 {x}^{2} / 2 + 9 x + 4$
Repeat this process with the remaining dividend:
$\frac{3 {x}^{2} / 2}{2 x} = \ast 3 \frac{x}{4} \ast$

$\left(3 \frac{x}{4}\right) \cdot \left(2 x - 3\right) = \frac{6 {x}^{2} - 9 x}{4}$
$\left(\frac{3 {x}^{2}}{2} + 9 x + 4\right) - \left(\frac{6 {x}^{2} - 9 x}{4}\right) = \frac{9 x}{4} + 4$

And repeat again
$\frac{\frac{9 x}{4}}{2 x} = \ast \frac{9}{8} \ast$

$\left(\frac{9}{8}\right) \cdot \left(2 x - 3\right) = \frac{9 x}{4} - \frac{27}{8}$

$\left(\frac{9 x}{4} + 4\right) - \left(\frac{9 x}{4} - \frac{27}{8}\right) = \ast \frac{59}{8} \ast$

The quotient is then the sum of the factors in bold and the remainder is $\frac{59}{8}$
$- {x}^{2} / 2 + 3 \frac{x}{4} + \frac{9}{8} = \frac{- 4 {x}^{2} + 6 x + 9}{8}$