Should $x=-2$ be included as an answer for $\frac{x^2+8x+12}{x^2+5x+6}>0$? $$\frac{x^2+8x+12}{x^2+5x+6}>0$$
First of all while solving inequalities I need to check domain so in this case $$x^2+5x+6\neq0$$
$$x\neq-2,\ x\neq-3$$ 
Later on
$$\frac{(x+6)(x+2)}{(x+3)(x+2)}>0$$
Then get critical values draw number line and get
$$x\in(-\infty;-6)\cup(-3;-2)\cup(-2;+\infty)$$
However according to wolframalpha $x=-2$ is included as an answer.

So am I wrong or wolframalpha is wrong?

Also I checked $\frac{x}{x}=1$ and wolframalpha also includes $x=0$ but once again I think it's incorrect? 
 A: As it seems you are simply solving an inequality, and entered it as such, you are correct: the expression is not defined at $-2$. 
Why Wolfram Alpha omitted $\,-3\,$ from the solutions, but included $\,-2\,$, seems inconsistent to me, as you note! It seems that if Wolfram is going to at least be consistent...either both values should be omitted, or both included.
A: Wolfram alpha probably simplified the expression before it calculated the solutions.  You are right.  It is worth mentioning, however, that there is a hole at $x=-2$, and the limit as $x\rightarrow -2$ satisfies the inequality.
A: The function as written is certainly not defined at $x= -2,$ but it does have a removable singularity there. The only way to extend the definition (note the word extend) of the function so that it becomes continuous at $x = -2$ is to define the value at $-2$ to be $4$. Then the new function is identically equal to $\frac{x+6}{x+3}$ except at $x = -3,$ where it is not defined (and worse, the singularity at $x = -3$ is not removable).
A: No, your expression is undefined at $x = -2$.  Are you taking a limit from the left or right???  
