If one zero of the polynomial p(x)

Given: $p(x)=x^{3}-6 x^{2}+11 x-6$ and its factor, $x+3$

Let us divide $p(x)$ by $(x-3)$.

Here, $x^{3}-6 x^{2}+11 x-6=(x-3)\left(x^{2}-3 x+2\right)$

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

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

$=(x-3)[x(x-2)-1(x-2)]$

$=(x-3)(x-1)(x-2)$

$\therefore$ The other two zeroes are 1 and 2 .