How many of the integers that satisfy the inequality $\frac{(x+2)(x+3)}{{x-2}} \geq 0$ are less than $5$? How many of the integers that satisfy the inequality $\frac{(x+2)(x+3)}{{x-2}} \geq 0$ are less than $5$ ?

The solution is $4$ but I think there are $5$ solutions to this. 

There are $3$ solutions that satisfy $\frac{(x+2)(x+3)}{{x-2}}= 0$. These are when $x=-2$, $x =-3$, $x=2$. 
The expression  $\frac{(x+2)(x+3)}{{x-2}} > 0$ is satisfied when $x=4$, $x=3$ 

Why is $x=2$ not considered a solution?


$x=2$ is not considered a solution because the expression is undefined at $x=2$. (Pay particular note to the denominator; it effectively results in a division by zero.)
