I would like to check if the theorem proved and its corollaries are correct, and useful as criterions for determining if a polynomial is irreducible or not. Also, I would like to know about their extendability as criterions of irreducibility in $\mathbb{Z}\left[x\right]$, $\mathbb{Q}\left[x\right]$, $\mathbb{R}\left[x\right]$ and $\mathbb{C}\left[x\right]$.
In mathematics, an irreducible polynomial (or prime polynomial) can be defined as a non-constant polynomial that cannot be factored into the product of two non-constant polynomials.
In this context, we can set the following
Theorem. If some univariate polynomial $P_{n}\left(x\right)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ is reducible such that $P_{n}\left(x\right)=Q_{k}\left(x\right)R_{j}\left(x\right)$, where $Q_{k}\left(x\right)=b_{k}x^{k}+b_{k-1}x^{k-1}+...+b_{1}x+b_{0}$ and $R_{j}\left(x\right)=c_{j}x^{j}+c_{j-1}x^{j-1}+...+c_{1}x+c_{0}$ are two non constant polynomials, then $\sum_{i=0}^{n}a_{i}=\left(\sum_{i=0}^{k}b_{i}\right)\left(\sum_{i=0}^{j}c_{i}\right)$
Proof. As $P\left(x\right)=Q\left(x\right)R\left(x\right)$, it follows that $P\left(1\right)=Q\left(1\right)R\left(1\right)$. And expanding, we get
$P_{n}\left(1\right)=a_{n}1^{n}+a_{n-1}1^{n-1}+...+a_{1}1+a_{0}=\sum_{i=0}^{n}a_{i}$
$Q_{k}\left(1\right)=b_{k}1^{k}+b_{k-1}1^{k-1}+...+b_{1}1+b_{0}=\sum_{i=0}^{k}b_{i}$
$R_{j}\left(1\right)=c_{j}1^{j}+c_{j-1}1^{j-1}+...+c_{1}1+c_{0}=\sum_{i=0}^{j}c_{i}$
Getting the desired result.
Corollary 1. As a prime number can only be multiple of $1$ and itself, and there exist no sum of positive integers equal to $1$, it follows directly from the Theorem stated that, if $\sum_{i=0}^{n}a_{i}$ is some prime number, then $P_{n}\left(x\right)$ is irreducible in $\mathbb{Z^+}\left[x\right]$.
My guess is that, unfortunately, the criterion set in the corollary is not extendable to $\mathbb{Z}\left[x\right]$, as there do exist sums of positive and negative integers equal to $1$.
However, we could state that
Corollary 2. If $\sum_{i=0}^{n}a_{i}$ is some prime number, then $P_{n}\left(x\right)$ is reducible in $\mathbb{Z}\left[x\right]$ only if either $Q_{k}\left(1\right)$, or $R_{j}\left(1\right)$, or both, have at least one negative coefficient. And, as if $\sum_{i=0}^{n}a_{i}$ is some prime number, then $P_{n}\left(x\right)$ is primitive, by Gauss Lemma we have that $P_{n}\left(x\right)$ is reducible in $\mathbb{Q}\left[x\right]$ only if either $Q_{k}\left(1\right)$, or $R_{j}\left(1\right)$, or both, have at least one negative coefficient.
Also, if you do know / think of further corollaries from the theorem stated, it would be great to know about them.
Thanks in advance!