Show that there do not exist polynomials $p(x)$ and $q(x)$ , each having integer coefficients and of degree greater than or equal to 1 such that: $$p(x)\cdot q(x) = x^5 + 2x + 1 $$
My approach
I found that only possible integral root of the $x^5 + 2x + 1 = 0 $ is ± 1 and these are not the zeroes of the polynomial. Hence obviously it doesn't have any linear factor. So i assumed $q(x)$ to be a third degree polynomial and $p(x)$ to be a second degree polynomial . That is ;$$p(x)=x^2+ax+b$$ $$q(x)=x^3+cx^2+dx+e$$ But i don't know how to solve it further. So please help me with this. And if there is some other approach to this question, then please share it .
Best Answer
Looking at questions like these modulo a prime is not guaranteed to work but it is definitely something that should be in your toolbox.
Let's look at your example polynomial $f(x)=x^5+2x+1$ modulo primes $p=2$ and $p=3$. Either choice of $p$ leads to a proof of irreducibility.
For comparison, the methods in other answers are all very fine. They work well here as well as in many other cases. They work in many cases where modular techniques may fail. But, modular tricks are good to know. For example, instead of your $f(x)$ consider the polynomial $$ f(x)=x^5+2x+7^n, $$ where $n$ is either a parameter or some ridiculously large natural number. AFAICT all the other answers fail or lead to uncomfortably many cases. But, the above proof using reduction modulo $3$ goes through verbatim. The modulo $2$ argument survives if you can do the part with the rational root test and show that $f(\pm 7^\ell)\neq0$ for all $\ell, 0\le\ell\le n$. One term will usually dominate the others.