# How do you divide (x^3+7x^2-4x-1)/(3x-1) ?

Jun 5, 2017

$y = \frac{1}{3} {x}^{2} + \frac{22}{9} x - \frac{14}{27} - \frac{41}{27} \frac{1}{3 x - 1}$

#### Explanation:

Divide ${x}^{3}$ by $3 x$, the quotient is $\frac{1}{3} {x}^{2}$

$y = \frac{1}{3} {x}^{2} + \frac{\frac{1}{3} {x}^{2} + 7 {x}^{2} - 4 x - 1}{3 x - 1}$

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

Divide $\frac{22}{3} {x}^{2}$ by $3 x$, the quotient is $\frac{22}{9} x$

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

$y = \frac{1}{3} {x}^{2} + \frac{22}{9} x + \frac{- \frac{14}{9} x - 1}{3 x - 1}$

Divide $\frac{14}{9} x$ by $3 x$, the quotient is $- \frac{14}{27}$

$y = \frac{1}{3} {x}^{2} + \frac{22}{9} x - \frac{14}{27} + \frac{- \frac{14}{27} - 1}{3 x - 1}$

Degree above < degree below.