If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$
Let $a=2, b=-3$ and $c=4$
Substituting the values in 1 , we get
$x^{3}-(2-3+4) x^{2}+(-6-12+8) x-(-24)$
$\Rightarrow x^{3}-3 x^{2}-10 x+24$