# How do you divide (4x^3-5x^2+9x-3)/(2x-3) ?

Dec 17, 2015

$y = \frac{\left({x}^{2} + 3\right) \left(4 x - 5\right) \left(3 x - 1\right)}{2 x - 3}$

#### Explanation:

$y = \frac{4 {x}^{3} - 5 {x}^{2} + 9 x - 3}{2 x - 3}$

Divide $2$ terms at once by a common factor.
$y = \frac{{x}^{2} \left(4 x - 5\right) + 3 \left(3 x - 1\right)}{2 x - 3}$

Rewrite the equation.
$y = \frac{\left({x}^{2} + 3\right) \left(4 x - 5\right) \left(3 x - 1\right)}{2 x - 3}$, $x \ne \frac{3}{2}$

Since the equation cannot be simplified any further, the final answer is $y = \frac{\left({x}^{2} + 3\right) \left(4 x - 5\right) \left(3 x - 1\right)}{2 x - 3}$ where $x \ne \frac{3}{2}$.