Order of field extension
Let $f=x^3+ax^2+bx+c$ be an irreducible polynomial over $\Bbb Z_{11}$.
The extension field $E = {\Bbb Z}_{11}[x]/\langle f\rangle$ contains a zero of $f$, namely the residue class $\alpha = \bar x + \langle f\rangle$. This gives $\alpha^3+a\alpha^2+b\alpha + c=0$, so $\alpha^3 = -a\alpha^2-b\alpha-c$, and the extension field is $E = \{u\alpha^2+v\alpha+w\mid u,v,w\in{\Bbb Z}_{11}\}$ with degree $[E:\Bbb Z_{11}]=3$. In particular, if $f$ is primitive, the powers of $\alpha$ are exactly the nonzero elements of $E$.
NB: $f(\bar x) = f(x +⟨f⟩)=f(x)+⟨f⟩=⟨f⟩=\bar 0$.