How many solutions are there to the congruence
X^4 + 5X^3 + 4X^2 - 6X - 4 ≡ 0 (mod11) with 0 ≤X ≤11?
I need to find that that if there are 4 solutions or there are fewer than 4 solutions?
I saw this question while i was studying congeruences in mumber theory and i got stuck. Can someone help me?
Note that $$ X^4 + 5X^3 + 4X^2 - 6X - 4=(X^2 + 4X + 2)(X + 10)(X + 2), $$ over the field $\Bbb F_{11}$ and $X^2+4X+2$ is irreducible, i.e., has no root. So we have exactly two roots. Over any field, a polynomial of degree $n$ has at most $n$ roots.
How do we find the factorization? It is easy to see that $X=1=-10$ is a root, and that $X=-2=9$ is a root. Then we can divide by $(X-1)(X+2)$.
