We have $p ( x )=3 x ^{3}+7 x$ and zero of $7+3 x$ is $\frac{-7}{3}$

$\left[\because 7+3 x=0 \Rightarrow x=\frac{-7}{3}\right]$

$\therefore $      $p \left(\frac{-7}{3}\right)=3\left(\frac{-7}{3}\right)^{3}+7\left(\frac{-7}{3}\right)=3\left(\frac{-343}{27}\right)+\left(\frac{-49}{3}\right)=-\frac{343}{9}-\frac{49}{3}=\frac{-490}{9}$

Since                      $\left(\frac{-490}{9}\right) \neq 0$

i.e.  the remainder is not $0$ .

$\therefore 3 x ^{3}-7 x$ is not divisible by $7+3 x$.

Thus, $(7+3 x)$ is not a factor of $3 x^{3}-7 x$.