Let $f(x)=x^p+px+1$, where $p$ is an odd prime. Prove that $f(x)$ is irreducible.
This is an exercise of a course (linear algebra, the first chapter focus on the polynomial rings), and the exercise trivial is well know, because $f(x-1)$ is Eisenstein.
Since I am a TA of this course. I want to know more and of course can share this with my students. So I come here.
If we replace $1$ to an arbitrary integer, say $-t$. Are there some other results about this problem? Conside the following polynomial $$g(x):=f(x+t)=(x+t)^p+p(x+t)-t=x^p+ph(x)+pt+t^p-t.$$ Here $pt+t^p-t$ is the constant term of $g(x)$ and $h(x)$ is a polynomial with integral coefficients. If we can prove that $g(x)$ is irreducible, then $f(x)$ is also irreducible. I want to use the Eisenstein's criterion. If we can show that $$p^2\nmid pt+t^p-t$$ then we are done. Use this method, we can deduce that many other polynomials like $f(x)=x^5+5x+2$ is irreducible, here $t=-2$ and the constant term of $f(x+t)$ is $-40$ and $25\nmid -40$. But this can just prove some special cases.
Are there any results about the irreducibility of the polynomials like this $x^p\pm px-t$? or some articles about the irreducility of the polynomials $x^p\pm px-t$. Thanks in advance.