# How do you solve (x^2-x-2)/(x+3)<0?

Mar 8, 2017

The solution is x in ]-oo, -3[ uu ]-1,2[

#### Explanation:

We solve this inequality with a sign chart

Let's factorise the numerator

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

The inequality is

$\frac{{x}^{2} - x - 2}{x + 3} = \frac{\left(x + 1\right) \left(x - 2\right)}{x + 3} < 0$

Let $f \left(x\right) = \frac{\left(x + 1\right) \left(x - 2\right)}{x + 3}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 3\right\}$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) < 0$ when x in ]-oo, -3[ uu ]-1,2[