Show that $2x - 3$ is a factor of $x + 2x^3 - 9x^2 + 12$.

(N/A) Let $p(x) = 2x^3 - 9x^2 + x + 12$ and $g(x) = 2x - 3$.
According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$ where $x = a$ is the zero of $g(x)$.
Setting $g(x) = 0$,we get $2x - 3 = 0$,which implies $x = \frac{3}{2}$.
Now,we evaluate $p\left(\frac{3}{2}\right)$:
$p\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 9\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right) + 12$
$= 2\left(\frac{27}{8}\right) - 9\left(\frac{9}{4}\right) + \frac{3}{2} + 12$
$= \frac{27}{4} - \frac{81}{4} + \frac{6}{4} + \frac{48}{4}$
$= \frac{27 - 81 + 6 + 48}{4} = \frac{81 - 81}{4} = \frac{0}{4} = 0$.
Since $p\left(\frac{3}{2}\right) = 0$,by the Factor Theorem,$2x - 3$ is a factor of $p(x)$.