<s>[INST] Prove that $x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$ I wish to show that $$f(x) = x^5 + 4x^4 + 2x^3 + x^2 + 4x + 54$$ is irreducible over $\mathbb{Z} / 7\mathbb{Z}$. Simply testing every element, it is easy to see that there are no roots and thus no linear factors, but I'm not sure how to proceed from there.
 [/INST] You can try case by base: 5 = 1+4 = 2+3 and $F = \mathbb{Z}/7\mathbb{Z}$ is a field so, there are three options: $f = x^5 + 4x^4 + 2x^3 +x^2 + 4x + 54$ is irreducible, it decompose int terms of degree 1 and 4 or terms of degree 2 and 3. As you checked, the second case is impossible because it has no root in $F$. Let's try the other case:
$$
f = (x^2 + b x + c)(x^3 + dx^2 + ex + f) = x^5 + (d+b)x^4 + (e+bd+c)x^3 + (f+be+cd)x^2 + (bf+ce)x + cf
$$
So
$$
b+d = 4\\
e+bd+c = 2\\
f+be+cd = 1\\
bf+ce = 4\\
cf=54
$$
Or
$$
\left[
\begin{array}{l}
1& 0 &0\\
b &1 &0\\
c &b &1\\
0 &c &b\\
0 &0 &c
\end{array}
\right]
\left[
\begin{array}{l}
d\\
e\\
f
\end{array}
\right]
=
\left[
\begin{array}{l}
4-b\\
2-c\\
1\\
4\\
54
\end{array}
\right]
$$
So, we need to verify if this linear system have a solution;
$$
\left[
\begin{array}{l}
1& 0 &0\\
b &1 &0\\
c &b &1\\
0 &c &b\\
0 &0 &c
\end{array}
\right]
\begin{array}{l}
4-b\\
2-c\\
1\\
4\\
54
\end{array}
\\ \simeq
\left[
\begin{array}{l}
1& 0 &0\\
0 &1 &0\\
0 &0 &1\\
0 &0 &0\\
0 &0 &0
\end{array}
\right]
\begin{array}{l}
4-b\\
2-c - b(4-b)\\
1 - c(4-b) - b(2-c - b(4-b))\\
4 - c(2-c - b(4-b)) - b(1 - c(4-b) - b(2-c - b(4-b)))\\
54 - c(1 - c(4-b) - b(2-c - b(4-b)))
\end{array}
$$
We need
$$
4 - c(2-c - b(4-b)) - b(1 - c(4-b) - b(2-c - b(4-b))) = 0\\
54 - c(1 - c(4-b) - b(2-c - b(4-b))) = 0
$$
And you can use a computer to check all 49 cases in (b,c) and  see that this system has no solution in $F$.
Then, this case of decomposition is impossible too and $f$ is irreducible.
Note, this is a completely computacional way to do that. Essentially, I'm using that $F$ is a finite field and we can really check all it's elements.
</s>