I have encountered the polynomial equation


$$x^{n+1} = (1 - x)^n ( n + x )$$


where $n \geq 0$, and I am interested in its real roots.
The number $n$ can be an integer or, more generally, any positive real number.


The main reason I am posting this very specific question here is that this could be one of those classical equations in some subarea of analysis or algebra, so perhaps somebody recognizes it by name. Of course, somebody might happen to have a good hint. It seems that this polynomial has only one single positive root.



What can be said about the roots of the polynomial $x^{n+1} - (1 - x)^n ( n + x )$?
The positive root is indeed unique and single for any real $n>0$. Indeed,


1. for $x>1$, we have $$|(1-x)^n(n+x)|=(x-1)^n(n+x)<\left(\frac{n\cdot (x-1)+(n+x)}{n+1}\right)^{n+1}=x^{n+1}$$
by the (weighted) Arithemtic-Geometric Means inequality.
2. for $0\leqslant x\leqslant 1$ there exists a root in $(0,1)$ since the function $f(x):=(1-x)^n(n+x)-x^{n+1}$ is positive at $x=0$ and negative at $x=1$. To prove that it is unique and simple, note that
$$
(1-x)^{-n}x^{-1}f(x)=\frac n{x}+1-\left(\frac{x}{1-x}\right)^n
$$
is a strictly decreasing (and with strictly negative derivative) function on $(0,1)$.


