$\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\F}{\mathbb{F}}$I have a function $f(x) = 2(x^2-x)+1$ and my question is, if this is irreducible in $\F_5$.
I now that $\F_5$ comes from $5=q=p^n$, $p = \Z/5\Z$, $n =\ $polynom with degree 1 (so something like $4x+11$ ...)
What happens to $x^2$ in $\F_5$? Thanks for your help.
The polynomial has roots in $\mathbb{F}_5$ and hence is reducible. We have $$ 2x^2-2x+1=2(x + 3)(x + 1) $$ over the field $\mathbb{F}_5$. Of course, $x^2=x\cdot x$ is reducible, too.