Show that $x^2 + x + 2\in \mathbb{F}_5[x]$ is irreducible

Question

Other than the brute force method of plugging in $i\in\{0, \dots, 4\}$ and showing $f(i) \not\equiv 0 \pmod 5$, what's the proper way to argue this with factorizations or gcd maybe?

Answer 1

You can simplify that quadratic using $1 \equiv -4\pmod5$.

$$ x^2+x+2 \equiv x^2-4x+2 \equiv (x-2)^2-2 \pmod5$$ So if your poly is reducible,

$$(x-2)^2\equiv 2 \pmod5$$ has a solution.

But the only quadratic residues mod 5 are $\{0,1,4\}$. Hence the poly is irreducible.

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Alternatively, $$ x^2+x+2 \equiv x^2+6x+9-7 \equiv (x+3)^2 - 2 \pmod 5$$