# How do you solve and write the following in interval notation: (x^2-5x+6)/(x+3) >0?

Oct 11, 2017

The solution is $x \in \left(- 3 , 2\right) \cup \left(3 , + \infty\right)$

#### Explanation:

We cannot do crossing over

Factorise the inequality

$\frac{{x}^{2} - 5 x + 6}{x + 3} > 0$

$\frac{\left(x - 2\right) \left(x - 3\right)}{x + 3} > 0$

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

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}$$- 3$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$3$$\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 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 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}$$x - 3$$\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}$$f \left(x\right)$$\textcolor{w h i t e}{a 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}$$+$

Therefore,

$f \left(x\right) > 0$ when $x \in \left(- 3 , 2\right) \cup \left(3 , + \infty\right)$

graph{(x^2-5x+6)/(x+3) [-148.4, 152, -74.4, 76]}

graph{(x^2-5x+6)/(x+3) [-0.486, 4.993, -1.132, 1.61]}