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.

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.
