Why do we consider the zeroes of the expression when solving rational inequalities?

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 $$operatorname{sgn}left(frac{x-3}{x+2}right)=operatorname{sgn}left((x-3)(x+2)right), x neq -2$$

For instance, consider $$frac{x-a}{x-b} ge 0, agt b$$ 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 $$x-age 0 implies xge a$$and $$x-b gt 0 implies xgt b hspace{1 cm} because x-bne 0$$ Taking the inetrsection of the two requirements, we get $$xge a hspace{2 cm} (mathbf 1)$$

If both are negative, then $$x-a lt 0 implies xlt a \ x-b lt 0 implies xlt b $$ This means $$xlt b hspace{2 cm} (mathbf 2)$$ Now, either of the two cases can happen, so we take the union of $(mathbf 1)$ and $(mathbf 2)$: $$xin(-infty, b) cup [a, infty) $$ To answer your question more directly, we consider the zeroes of each factor, because that’s precisely where the factor changes its sign.