It is given that −1 is one of the zeros of the polynomial

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 $f(x)$ by $(x+1)$, we get:

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

$=(x+1)\left(x^{2}+x-12\right)$

$=(x+1)\left\{x^{2}+4 x-3 x-12\right\}$

$=(x+1)\{x(x+4)-3(x+4)\}$

$=(x+1)(x-3)(x+4)$

$\therefore f(x)=0=>(x+1)(x-3)(x+4)=0$

$=>(x+1)=0$ or $(x-3)=0$ or $(x+4)=0$

$=>x=-1$ or $x=3$ or $x=-4$

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