The root of the polynomial $f(x)=x^n-kx^{n-1}-kx^{n-2}-\cdots-kx-k$ of the largest magnitude?

Question

There is a root, say $\lambda$, of the polynomial $f(x)=x^n-kx^{n-1}-kx^{n-2}-\cdots-kx-k$ between k and k+1 (by the intermediate value theorem), where $n,k$ are integers and $n,k\geq 2$. Is $\lambda$ the root of $f(x)$ of the largest magnitude? Specifically, the numerical results show that $f(x)$ has only one positive real root $\lambda$ between $k$ and $k+1$ and zero (resp., one) negative real root in the case that $n$ is odd (resp., even), and all the other roots of $f(x)$ are complex with magnitude less than $\lambda$. The problem is how to prove this?

Answer 1

The assertions about the uniqueness of the positive root and the number of negative roots are easy to prove using Descartes' rule of signs. The sequence of coefficients of your polynomial has just one change of signs; therefore, there is exactly one positive root. Furthermore, $$ (x-1)f(x) = x^{n+1}-x^n-k(x^n-1) = x^{n+1} - (k+1)x^n + k, $$ whence $$ (-x-1)f(-x)=(-1)^{n+1}x^{n+1}+(-1)^{n+1}(k+1)x^n+k. $$ If $n$ is odd, then the sequence of the coefficients does not have any sign changes; therefore $(-x-1)f(-x)$ does not have any positive roots, meaning that $f(x)$ does not have negative roots. If $n$ is even, there is one sign change; hence, one negative root of $f(x)$.