multiplicity of zeros for polynom

Question

let $a \in \mathbb F_{11}[X] $ and $a \in \mathbb F_{11}:$

$$f=x^6-x^5+2x^4-2x^3+x^2+x+3$$ $$a = -5$$

When I set x=-5 to check the multiplicity of zero I get:

$$(-5)^6-(-5)^5+2(-5)^4-2(-5)^3+(-5)^2+(-5)+3 = 20273$$

so $m_{-5}(f)=0$.

Question: Is that correct? And what is the context of $\mathbb F_{11} $ here? I appreciate every hint.

Answer 1

Finite field

Your calculation is correct. Note that \begin{align}f(-5)=20273\equiv0&&[11]\end{align} hence $-5$ is a zero of $f$ (in $\Bbb F_{11}$).