We are asked to find the solutions to the following congruence $$ x^3 + 8x^2 - x - 1 \equiv 0 \ (\text{mod } 11). $$
I know that the solution can be computed using Hensel's Lemma or by simply using brute-force. However, does there exist another approach to solving this type of problem?
On
When we have small primes $11$ for example, we can just try the first small numbers and find out the roots, let $f(x)=x^3+8x^2-x-1$ it's clear that: $$f(0)=-1,f(1)=7,f(2)=4, f(3)=7, f(4)=0 $$
and here we find the first root, and we try to factorize and obtain: $$x^3+8x^2-x-1=(x-4)(x^2+x+3)$$
and we continue testing the values $f(5)=1(25+5+3)=33=0$ and $x^2+x+3=(x-5)^2$ and we are done.
The Berlekamp algorithm gives, over $\mathbb{F}_{11}$ $$ x^3+8x^2-x-1=(x + 7)(x + 6)^2, $$ In an example with a degree $3$ polynomial one could also just try for a linear factor, i.e., looking for a zero.