How do you divide $\frac{2 x^{2} - 3 x}{3 x + 1}$ $(2x^2-3x)/(3x + 1)$ ?

Question

2070 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-13T14:43:15

$\frac{2 x^{2} - 3 x}{3 x + 1} = \frac{2}{3} x - \frac{11}{9}$ $(2x^2-3x)/(3x+1) = 2/3x -11/9$ with remainder of $\frac{11}{9}$ $11/9$

See image below for Synthetic and Long Polynomial divisions:
enter image source here

How do you divide 2x2−3x3x+1(2x^2-3x)/(3x + 1)?