# How do you solve (x+6)/(x^2+6x+8)>=0?

##### 1 Answer
Jan 11, 2017

The answer is x in [-6, -4[uu ] -2,+oo [

#### Explanation:

Let 's factorise the denominator

${x}^{2} + 6 x + 8 = \left(x + 4\right) \left(x + 2\right)$

Let $f \left(x\right) = \frac{x + 6}{{x}^{2} + 6 x + 8} = \frac{x + 6}{\left(x + 4\right) \left(x + 2\right)}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 2 , - 4\right\}$

We solve the inequality with a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a 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}$$x + 6$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{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 a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{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 a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$color(red)(∥)$\textcolor{w h i t e}{a a a}$$+$

Therefore,

$f \left(x\right) \ge 0$, when x in [-6, -4[uu ] -2,+oo [