Q.Solve $\dfrac{x^2-4}{x^2-2x-15}\le0$.

Step 1. Factor: $x^2-4=(x-2)(x+2)$; $x^2-2x-15=(x-5)(x+3)$.

Step 2. Critical points: $x=-3,5$ (denominator zero, excluded), $x=-2,2$ (numerator zero, included since the inequality is $\le0$).

Step 3. Test each interval. $x=-4$: $\dfrac{(-6)(-2)}{(-9)(-1)}>0$. $x=-2.5$: $\dfrac{(-4.5)(-0.5)}{(-7.5)(0.5)}<0$. $x=0$: $\dfrac{(-2)(2)}{(-5)(3)}>0$. $x=3$: $\dfrac{(1)(5)}{(-2)(6)}<0$. $x=6$: $\dfrac{(4)(8)}{(1)(9)}>0$. …

Factor fully, test signs interval by interval, and include the numerator zeros …

• Including $x=-3$ or $x=5$, where the expression is undefined. …