Rational Roots Theorem cannot be used since $\pm \frac{9,3,1}{5,1}$ are not zeroes of $f(x)$.
Eisentein's criterion cannot be used since the non-leading coefficients have no common factors.
For mod $p$ test, I tried $p=2$, but $\overline{f}(x) = x^{5} + x +1 = (x^{2}+x+1)(x^{3}+x^{2}+1)$.
For $p=3$, $\overline{f}(x)$ has a factor of $x$. $p=5$ is not possible since $deg \overline{f}=4 \neq 5 = deg f$.
I tried $p=7$ but got a linear factor.
Should I keep on looking for primes that could satisfy the mod p test or are there more efficient ways to show the desired conclusion?
The polynomial is indeed irreducible over $\Bbb F_{13}$, and $5$ is a unit in this field. We have $$ f=5(x^5 + x^4 + 9x^3 + 7x^2 + 4x + 7). $$