jmc

By the Rational Root Theorem, any rational root $p/q$ of the given polynomial must have $p$ divide 24 and $q$ divide 1. Therefore, the rational roots of the polynomial are all integers that divide 24.

So, we check factors of 24 to see if the polynomial has any integer roots. If $x=1$, we have $$1-3-10+24=-12<0,$$so 1 is not a root. If $x=2$, we have $$8-3\cdot 4-10\cdot 2+24=0.$$So 2 is a root! By the Factor theorem, this means that $x-2$ must be a factor of $x^3-3x^2-10x+24$. Through polynomial division we get that $$x^3-3x^2-10x+24=(x-2)(x^2-x-12).$$To find the roots of $x^2-x-12$, we can factor it or use the quadratic formula. Factoring, we find that $x^2-x-12=(x+3)(x-4)$ and hence we have the roots $-3$ and $4$. Therefore our original polynomial has roots $\boxed{2,-3,4}$.