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 $a=\frac{1}{2}, b=1$ and $c=-3$

Substituting the values in (1), we get

$x^{3}-\left(\frac{1}{2}+1-3\right) x^{2}+\left(\frac{1}{2}-3-\frac{3}{2}\right) x-\left(\frac{-3}{2}\right)$

$\Rightarrow x^{3}-\left(\frac{-3}{2}\right) x^{2}-4 x+\frac{3}{2}$

$\Rightarrow 2 x^{3}+3 x^{2}-8 x+3$