Irreducibility of a polynomial over Rationals with condition given on its coefficients.

Let $$f = a_nX^n+\cdots+a_1X\pm p \in \mathbb{Z}[X]$$ with $$\sum_{i=1}^n |a_i| < p$$. Show that $$f$$ is irreducible in $$\mathbb{Q}[X]$$.

Hint: Show that every root of $$f \in \mathbb{C}$$ has modulus greater than $$1$$ and consider leading and constant terms of a factor of $$f$$.

I have been able to show that every roots has modulus greater than $$1$$. But I am not able to go any further? Please help.

If $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+(-1)^up$$ has a solution in $$\mathbb{Q}$$ which is $$r/s$$ with $$gcd(r,s)=1$$, then $$a_nr^n+a_{n-1}r^{n-1}s+\ldots+a_1rs^{n-1}+(-1)^ups^n=0\\\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+\ldots+a_1s^{n-1})=-(-1)^ups^n\\\Rightarrow r|p.$$

If $$p$$ is prime then we have a contradiction.

If $$p$$ is not prime then we have a counterexample $$f(x)=x^2+x-6$$

• How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility. – Mittal G Nov 22 '18 at 14:47