# How do you divide (9x^4-x^2-6x-2x)/(3x-1)?

Aug 1, 2018

$\frac{9 {x}^{4} - {x}^{2} - 6 x - 2 x}{3 x - 1} = 3 {x}^{3} + \frac{2}{3} x - \frac{\frac{22}{3} x}{3 x - 1}$

#### Explanation:

We ask how many times the leading term of the denominator goes into each part of the numerator and then factor that out. The leading term of the bottom is $3 x$, so we start with $9 {x}^{4}$
$\frac{9 {x}^{4}}{3 x} = 3 {x}^{3}$
so
$3 {x}^{3} \left(3 x - 1\right) + 3 {x}^{3} = 9 {x}^{4}$

We can repeat this and get the following:
$\frac{9 {x}^{4} - {x}^{2} - 6 x - 2 x}{3 x - 1} = \frac{3 {x}^{3} \left(3 x - 1\right) + 2 {x}^{2} - 6 x - 2 x}{3 x - 1}$

$= 3 {x}^{3} + \frac{\frac{2}{3} x \left(3 x - 1\right) + \frac{2}{3} x - 6 x - 2 x}{3 x - 1}$

$= 3 {x}^{3} + \frac{2}{3} x - \frac{\frac{22}{3} x}{3 x - 1}$