The zero of $2 x+3$ is $\left(-\frac{3}{2}\right)$

Now, $p(x)=2 x^{3}+21 x^{2}+67 x+60$

$\therefore p\left(-\frac{3}{2}\right)=2\left(-\frac{3}{2}\right)^{3}+21\left(-\frac{3}{2}\right)^{2}+67\left(-\frac{3}{2}\right)+60$

$=2\left(-\frac{27}{8}\right)+21\left(\frac{9}{4}\right)-\frac{201}{2}+60$

$=-\frac{27}{4}+\frac{189}{4}-\frac{201}{2}+60$

$=\frac{-27+189-402+240}{4}$

$=\frac{-429+429}{4}=\frac{0}{4}$

$\therefore p\left(-\frac{3}{2}\right)=0$

So, by the factor theorem, $2 x+3$ is a factor of $2 x^{3}+21 x^{2}+67 x+60$.