Why can't I simply use algebra to solve this inequality? Consider the inequality:
$\frac{(x+3)(x-5)}{x(x+2)}\geq 0$
Why can't I simply multiply both sides by $x(x+2)$ and get $(x+3)(x-5)\geq 0$ ?
Which would yield: $x^2-2x-15\geq 0$ and I could then use the quadratic formula to derive the answer..?
This seems algebraically correct but I assume that is because I am misunderstanding something fundamental about inequalities.

if we assume that $x \gt 0$ then $(x+2) \gt 0$ so
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \ge 0$$
but if $x \cdot (x+2) < 0$
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \le 0$$
and if $x = 0$ or $x = -2$ then $\frac{(x+3)(x-5)}{x(x+2)}$ is undefined.
