Consider the polynomial $p(x) = x^n(1-x) - r$, where $r$ is a real number. For sufficiently small $r$, there are two positive zeros of $x^n(1 - x) -r = 0$. (I came to this conclusion by graphing $x^n(1-x)$ and $r$ as functions of $x$ and seeing where they intersect.) My textbook concludes that if two positive roots exist, the larger of the two positive roots is dominant. That is, among all the roots, the larger of the two positive roots has the greatest magnitude.
How can this be seen? I don't know why one of the complex roots cannot have a greater magnitude.
This came from a mathematical biology textbook.
For sufficiently small $r$, at least, this follows from the fact that the roots of a polynomial vary continuously with its coefficients. For fixed $n$, when $r$ is sufficiently small, the roots of $p(x)=x^n(1-x)-r$ are close to the roots of $q(x)=x^n(1-x)$. But the roots of $q$ are just $0$ (with multiplicity $n$) and $1$ (with multiplicity $1$). This means that $p$ has $n$ roots close to $0$ and one root close to $1$. The root close to $1$ will be real (since, for instance, if it were not real then its conjugate would be a second root close to $1$) and thus is the larger positive root. All the other roots have smaller magnitude since they are close to $0$.