Question:
Check if:
$$
f(x) = x^4 + 4x^3 + 6x^2 + 2x + 1
$$
is reducible or irreducible over $\mathbb{Q}[x]$
My Answer [Edited]:
Suppose that $f(x)$ is irreducible over $Z_p[x]$ for a prime $p$, therefore it's going to be irreducible over $\mathbb{Q}[x]$. Let $p=2$. Therefore:
$$
f_2(x) = f(x) \mod{2} = x^4 + 1 = a_2(x)\cdot b_2(x) \in \mathbb{Z}_2[x]
$$
Since $x^4+1$ have roots in $\mathbb{Z}_2[x]$, there are two cases to check… First let's suppose that $a_2(x),b_2(x)$ have degree equal to two, therefore:
\begin{align*}
x^4+1&=(a_2x^2+a_1x+a_0)\cdot(b_2x^2+b_1x+b_0)=\\
&=(a_2b_2)x^4+(a_2b_1+a_1b_2)x^3+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x^+(a_0b_0)
\end{align*}
Since $a_i,b_i \in \mathbb{Z}_2 \rightarrow a_i,b_i\in \{0,1\}$, it follows that:
$$
a_0b_0=1 \longrightarrow a_0=b_0=1\\
a_1+b_1=0 \longrightarrow a_1=b_1=0\\
a_2b_2=1 \longrightarrow a_2=b_2=1
$$
Hence:
\begin{align*}
a_2(x)&=x^2+1\\
b_2(x)&=x^2+1\\
a_2(x)\cdot b_2(x) &= (x^2+1)^2 = x^4 + 2x + 1 \overbrace{\longrightarrow}^{\in \mathbb{Z}_2} x^4 + 1
\end{align*}
Now let's check WLOG the case for $a_2(x)$ having degree equal to $3$ and $b_2(x)$ having degree equal to $1$:
\begin{align*}
x^4+1 &= a_2(x)\cdot b_2(x) = (a_3x^3+a_2x^2+a_1x+a_0)\cdot(b_1x+b_0)\\
&= (b_1a_3)x^4 + (b_1a_2+b_0a_3)x^3 + (b_1a_1+b_0a_2)x^2 + (b_1a_0+b_0a_1)x + (b_0a_0)
\end{align*}
By the same reasoning as before:
\begin{align*}
b_0a_0=1 \longrightarrow b_0=a_0=1\\
b_1a_3=1 \longrightarrow b_1=a_3=1\\
a_2+b_0 = 0 \longrightarrow a_2=b_0=0\\
a_1+a_2 = 0 \longrightarrow a_1=a_2=0\\
\end{align*}
Conclusion, it is reducible over $\mathbb{Z}_2[x]$ and therefore I can't conclude anything.
Best Answer
The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise, $$f(x)=(x+1)^4-2x\ .$$ Hence $$f(x-1)=x^4-2x+2\ ;$$ this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.