Why do we consider the zeroes of the expression when solving rational inequalities? When finding the solution set of rational inequalities (of the form $ {x-3\over x+2} \geq 0 $ etc.) , why do we consider the "zeroes" of the expression.
So I've been taught to solve this as follows:
First,
$x-3 = 0  \implies x = 3 \quad and \quad x+2\not= 0 \implies x\not=-2 \,\text{(b/c, undefined)} $ are "points of interest".
Sketch a number line from $-∞$ to $+∞$, with the points of interest ($-2$ & $3$) labeled.
Then we check each region bound by those numbers to see if they're positive/negative and label them as such.

The thing I'm confused by is why do we check those zeroes? What tells us that the signs don't suddenly flip at some other random value? Can someone explain in detail this working, suggest a better method or point me to other resources which "prove" the things I'm talking about.
 A: For instance, consider $$\frac{x-a}{x-b} \ge 0, \ a\gt b$$ For a fraction to be non-negative, we need the numerator and the denominator to have the same parity, or that the numerator be zero.
If both are positive, then $$x-a\ge 0 \implies x\ge a$$and $$x-b \gt 0 \implies x\gt b \hspace{1 cm} \because x-b\ne 0$$ Taking the inetrsection of the two requirements, we get $$x\ge a \hspace{2 cm} (\mathbf 1)$$
If both are negative, then $$x-a \lt 0 \implies x\lt a \\ x-b \lt 0 \implies x\lt b $$ This means $$x\lt b \hspace{2 cm} (\mathbf 2)$$ Now, either of the two cases can happen, so we take the union of $(\mathbf 1)$ and $(\mathbf 2)$: $$x\in(-\infty, b) \cup [a, \infty) $$ To answer your question more directly, we consider the zeroes of each factor, because that’s precisely where the factor changes its sign.
A: As Peter Foreman stated in his comment (and taking into note my own comment), the inequalities
$$\frac{x-3}{x+2}\ge0$$ and $$(x-3)(x+2)\ge 0$$
are equivalent. It's vividly seen in the graph below:

The graph shows that $$\operatorname{sgn}\left(\frac{x-3}{x+2}\right)=\operatorname{sgn}\left((x-3)(x+2)\right), \ \ x \neq -2$$
