# How do you simplify (x^2 + x-12 )/( 2x^2 - 3x - 9?

Jun 14, 2018

See a solution process below:

#### Explanation:

First, factor the numerator and denominator of the expression as:

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

Next, cancel the common terms from the numerator and denominator:

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 3\right)}}} \left(x + 4\right)}{\left(2 x + 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 3\right)}}}} \implies$

$\frac{x + 4}{2 x + 3}$

However, because we cannot divide by $0$ we must ensure:

$2 x + 3 \ne 0$ and $x - 3 \ne 0$

Or

Condition 1:

$2 x + 3 \ne 0$

$2 x + 3 - \textcolor{red}{3} \ne 0 - \textcolor{red}{3}$

$2 x + 0 \ne - 3$

$2 x \ne - 3$

$\frac{2 x}{\textcolor{red}{2}} \ne - \frac{3}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} \ne - \frac{3}{2}$

$x \ne - \frac{3}{2}$

Condition 2:

$x - 3 \ne 0$

$x - 3 + \textcolor{red}{3} \ne 0 + \textcolor{red}{3}$

$x - 0 \ne 3$

$x \ne 3$

Therefore, the simplified expression is:

$\frac{x + 4}{2 x + 3}$ Where $x \ne - \frac{3}{2}$ and $x \ne 3$