# How do you simplify (6x^2-4x-3)/(3x^2+x)?

Jan 8, 2017

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

#### Explanation:

Notice that:

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

Then $x$ is not a factor of the numerator, but $\left(3 x + 1\right)$ might be.

Let's try:

$6 {x}^{2} - 4 x - 3 = \left(6 {x}^{2} + 2 x\right) - \left(6 x + 2\right) - 1$

$\textcolor{w h i t e}{6 {x}^{2} - 4 x - 3} = 2 x \left(3 x + 1\right) - 2 \left(3 x + 1\right) - 1$

$\textcolor{w h i t e}{6 {x}^{2} - 4 x - 3} = \left(2 x - 2\right) \left(3 x + 1\right) - 1$

No. So there is no cancellable factor and we cannot substantially simplify the given rational expression.

About the most we can do is separate out the quotient $2$ as follows:

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

$\textcolor{w h i t e}{\frac{6 {x}^{2} - 4 x - 3}{3 {x}^{2} + x}} = \frac{2 \left(3 {x}^{2} + x\right) - \left(6 x + 3\right)}{3 {x}^{2} + x}$

$\textcolor{w h i t e}{\frac{6 {x}^{2} - 4 x - 3}{3 {x}^{2} + x}} = 2 - \frac{6 x + 3}{3 {x}^{2} + x}$