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.
 A: 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.
A: Showing a "standard" way of settling questions like this.
The key well known fact is that the polynomial $p(x)=x^{49}-x$ is the product of all the irreducible linear and quadratic polynomials over $\Bbb{F}_7$. This is because all those polynomials are minimal polynomials of some elements of $\Bbb{F}_{49}$. But OTOH all the elements of $\Bbb{F}_{49}$ are zeros of $p(x)$.
A consequence of this is that if $f(x)$ has no common factors with $p(x)$ then it has no linear or quadratic factors. That is enough to conclude that $f(x)$ must be irreducible in $\Bbb{F}_7[x]$.
That gcd can be efficiently calculated with the Euclidean algorithm. This is trivial for a suitable CAS, but can actually be done with paper & pencil work. In this particular case there are certain coincidences making it quite fast. Let me demonstrate!
We calculate certain remainders modulo $f(x)$. The most taxing step turns out to be (we are extraordinary lucky here):
$$
x^6-(x+3)f(x)\equiv 6+4x\pmod 7.\tag{1}
$$
The rest follows from this. Multiplying $(1)$ by $x$ gives the congruence
$$
x^7\equiv 6x+4x^2\pmod{f(x)}.\tag{2}
$$
Squaring $(2)$ gives then
$$
x^{14}\equiv (6x+4x^2)^2\equiv x^2-x^3+2x^4\pmod{f(x)}.\tag{3}
$$
On the other hand we are in characteristic seven, so we have the so called freshman's dream
$$
(a+b)^7=a^7+b^7
$$
that holds for all polynomials $a,b\in\Bbb{F}_7[x]$.
Let's raise the congruence $(2)$ to the seventh power. We first get that
(recall Little Fermat telling us that $c^7=c$ for all constants $x\in\Bbb{F}_7$)
$$
x^{49}\equiv 6x^7+4x^{14}\pmod{f(x)}.\tag{4}
$$
But we already calculated the remainders of both $x^7$ and $x^{14}$ modulo $f(x)$, so we arrive at the congruence
$$
\begin{aligned}
x^{49}&\equiv 6x^7+4x^{14}\\
&\equiv 6(6x+4x^2)+4(x^2-x^3+2x^4)\\
&\equiv x+3x^2+4x^2+3x^3+x^4\\
&\equiv x+3x^3+x^4.
\end{aligned}
$$
So our calculations have showed that
$$
x^{49}-x\equiv 3x^3+x^4=x^3(x+3)\pmod{f(x)}.\tag{5}
$$
This means that any eventual common factor of $p(x)$ and $f(x)$ must also be a factor of the remainder $r(x)=x^3(x+3)$. But $r(x)$ is manifestly a product of linear factors. OTOH you had already checked that $f(x)$ has no linear factors, so we are done!
