Let $p(x)=x+2 x^{3}-9 x^{2}+12$ and $g(x)=2 x-3$

$g(x)=2 x-3=0$ gives $x=\frac{3}{2}$

$g ( x )$ will be factor of $p ( x )$ if $p\left(\frac{3}{2}\right)=0 \quad$ (Factor theorem)

Now, $\quad p\left(\frac{3}{2}\right)=\frac{3}{2}+2\left(\frac{3}{2}\right)^{3}-9\left(\frac{3}{2}\right)^{2}+12=\frac{3}{2}+2\left(\frac{27}{8}\right)-9\left(\frac{9}{4}\right)+12$

$=\frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12=\frac{6+27-81+48}{4}=\frac{0}{4}=0$

Since, $p\left(\frac{3}{2}\right)=0,$ so, $g ( x )$ is a factor of $p ( x )$.