This question is first asked by me on MSE, but I haven't recieve a nice answer yet.
I would like to determine whether the polynomial $p(x)=x^n+5x+3$ is irreducible over $\mathbb{Q}$ when $n\ge 2$. Standard criteria (I have tried Eisenstein and $\bmod(p)$ method but both failed) do not seem to apply to this polynomial. I also looked through this work, but it seems none of the criteria apply to my situation.
Note: I used Matlab to calculate the circumstance $1<n\le 100$, the result says all of these polynomials are irreducible.
Best Answer
This should be provable using the results of On the irreducibility of the non-reciprocal part of polynomials of the form $f(x) x^n+g(x)$ by Filaseta-Li-Patane-Skabelund Acta Arithmetica 196 (2020), 187-201 (which builds on my paper "Irreducibility of polynomials with large gap" with Mark Shusterman and Michael Stoll - thanks to GH from MO for pointing out this improved version).
We certainly have irreducibility for all $n \geq 21892$ (which is close to two orders of magnitude from the checked irreducibility for $n\leq 100$.)
The polynomial $x^n+5x+3$ is of the form $f(x) x^n+g(x)$ with $f(x) = 1$ and $g(x)=5x+3$.
Some notation from the paper $\tilde{f}(x) =x^{\deg f} f(x^{-1}) $ and $\| f\|$ is the square root of the sum of the squares of the coefficients of $f$. They also use the Fibonacci sequence $F_n$.
Then $f$ and $g$ satisfy all the criteria of Theorem 2 of the linked paper:
In particular, since $f=1$ and $g$ is irreducible and primitive, $\tilde{f}$ and $g$ are "robust". For $D$ a nonnegative integer $\geq \max(\deg f, \deg g)=2$ and $\kappa$ such that for any polynomials $f_1(x), g_1(x)\in \mathbb Z[x]$ where the coefficients of degree $\leq D$ in $f_1(x)$ agree with the coefficients of degree $\leq D$ in $g(x)$ we have $\|f_1 \|^2 + \| g_1\|^2 \geq \kappa$, then if $$n > D F_{ \|f\|^2 +\|g\|^2 -\kappa + 4} -\deg f$$ and also $n > 3 \deg f + 4 \deg g$ then the non-reciprocal part of $f(x) x^n+ g(x)$ is irreducible.
I will check that (1) the reciprocal part is trivial and (2) we can take $D= 2 $ and $\kappa=17$ which gives irreducibility for $n > 2 F_{21}= 2\cdot 10946= 21892$.
(1) Any irreducible reciprocal factor of $x^n+ 5x+3$ must also divide $3 x^n + 5 x^{n-1} + 1$ and thus must divide $(3x +5 ) (x^n+ 5x +3) - x (3x^n+5 x^{n-1}+1) = 15 x^2+ 33 x + 15$ and hence must equal $5 x^2+ 11 x+ 5$ since that polynomial is irreducible. But this clearly can't be a factor for divisibility of the first and last coefficient reasons. So the reciprocal part is trivial and then the whole polynomial is irreducible.
(2) Up to sign, one of $f_1,g_1$ must have constant coefficient $1$ and one must have constant coefficient $3$. Wlog $f_1$ has constant coefficient $1$ and $g_1$ has constant coefficient $3$.
Then the coefficients of $x$ in $f_1$ and $f_2$ respectively must equal $ (1,2)$ or $(2,-1)$ - The solutions without the bound on $\|f_1\|^2+\|g_1\|^2 $ form an arithmetic progression, and the next terms $(0,5)$ and $(3,-4)$ have sum of squares at least $25$ which is too large.
In the first case, the coefficients of $x^2$ in $f_1$ and $g_1$ would satisfy $3a + b +2=0$ and in the second case they woudl satisfy $3a+b -2 =0$. In either case this forces $|a|^2+|b|^2\geq 2$. So the sums of squares of the coefficients of $f_1$ and $g_1$ are together at least
$$1^2+3^2+2^2+1^2+2=17$$
justifying $\kappa=17$.
Further work should get slightly higher values of $D$ and smaller values of $\kappa$, giving an improved bound.