Evaluate $\frac{x^3-x^2-5x-3}{x^3-x^2-15x} \ge 0$ I've been trying to simplify the problem but I can't. I tried long division factoring and perfect cube but I can't still solve it. My calculator shows $x=3$ and $x=-1$ but wait how? this is for the numerator

it seems you can factor  $x^3-x^2 - 5x - 3 = (x+1)^2(x-3)$ and $x^3 - x^2-15x=x(x^2-x-15)=x(x -a)(x-b)$  where $a = \dfrac{1-\sqrt{61}}2, b = \dfrac{1+\sqrt{61}}2$
we need to solve $$y = \dfrac{(x+1)^2(x-3)}{(x-a)x(x-b)} \ge 0$$ you can draw a line and put the numbers $a, -1, 0, 3$ and $b$ and do a sign analysis or you can see that $y = 1+ \cdots$ for large $x$ and switches signs across $a, 0, 3$ and $b$ but at $x = 1$ it touches the $x$-axis from below. you will see that $$\dfrac{x^3-x^2 - 5x - 3}{x^3 - x^2 - 15x} \ge 0\text{ on }-\infty < x < a, 0 < x \le 3\text{ and }b < x < \infty.$$
