If $x^{3}-2 m x^{2}+16$ is divisible by $x+2,$ then $x+2$ is a factor of $x^{3}-2 m x^{2}+16.$

Now, let $\quad p(x)=x^{3}-2 m x^{2}+16.$

As $x+2=x-(-2)$ is a factor of $x^{3}-2 m x^{2}+16.$

So $\quad p(-2)=0.$

Now, $\quad p(-2)=(-2)^{3}-2 m(-2)^{2}+16.$

$=-8-8 m+16=8-8 m$

Now, $\quad p(-2)=0$

$\Rightarrow \quad 8-8 m=0$

$\Rightarrow \quad m=8 \div 8$

$\Rightarrow \quad m=1$

Hence, for $m +1, x +2$ is a factor of $x^{3}-2 m x^{2}+16,$ so $x^{3}-2 m x^{2}+16$ is completely divisible by $x+2.$