Let $f(x)=x^{3}+2 x^{2}-11 x-12$

Since $-1$ is a zero of $f(x),(x+1)$ is a factor of $f(x)$.

On dividing $\mathrm{f}(\mathrm{x})$ by $(\mathrm{x}+1)$, we get

$$
\begin{aligned}
&f(x)=x^{3}+2 x^{2}-11 x-12 \\
&=(x+1)\left(x^{2}+x-12\right) \\
&=(x+1)\left\{x^{2}+4 x-3 x-12\right\} \\
&=(x+1)\{x(x+4)-3(x+4)\} \\
&=(x+1)(x-3)(x+4) \\
&\begin{aligned}
\therefore f(x)=0 & \Rightarrow(x+1)(x-3)(x+4)=0 \\
& \Rightarrow(x+1)=0 \text { or }(x-3)=0 \text { or }(x+4)=0 \\
& \Rightarrow x=-1 \text { or } x=3 \text { or } x=-4
\end{aligned}
\end{aligned}
$$

Thus, all the zeroes are $-1,3$ and $-4$.