TITLE: Help proving that a polynomial has n roots and that they're all different
QUESTION [6 upvotes]: Given the nature of the exercise, I think the point is not using the Fundamental Theorem of Algebra to show there must be $n$ roots. Anyway, the polynomial is the following: $$x^n + x + 1$$
So far I've tried to reach an absurd by proposing a root that is both root of the polynomial and its derivative, but I'm stuck there. Any hints for this kind of exercise?

REPLY [7 votes]: We may safely assume $n\geq 2$. If $f(x)=x^n+x+1$ we have $f'(x)=n x^{n-1}+1$ and
$$ \gcd(f(x),f'(x)) = \gcd(f'(x), n f(x)-x f'(x))=\gcd(nx^{n-1}+1,(n-1)x+n). $$
On the other hand $(n-1)x+n$ only vanishes at $x=\frac{n}{1-n}$, while
$$ n\left(\frac{n}{1-n}\right)^{n-1}+1 $$
is positive for any odd $n$ and strictly negative for any even $n$, since $n^n>(n-1)^{n-1}$.
It follows that $f(x)$ and $f'(x)$ cannot have common roots, hence all the roots of $f(x)$ are simple.