How to prove that this polynomial has no more than $s$ repeated roots Let 
$\beta_{1},\beta_{2},\cdots,\beta_{s+1}\in R$,and $\alpha_{0},\alpha_{1},\cdots,\alpha_{s}$ be postive integers, with
$\alpha_{0}>\alpha_{1}>\cdots>\alpha_{s}$.
Show that: the polynomial $$f(x)=x^{\alpha_{0}}+\beta_{1}x^{\alpha_{1}}+\beta_{2}x^{\alpha_{2}}+\cdots+\beta_{s}x^{\alpha_{s}}+\beta_{s+1}$$ repeated (the repeated no more than $s$) roots,In other words,
if $$f(x)=(x-r_{0})^{m_{1}}\cdot (x-r_{1})^{m_{2}}\cdots (x-r_{k})^{m_{k}}$$
then 
$$m_{i}\le s,i=1,2,\cdots,k,m_{1}+m_{2}+\cdots+m_{k}=\alpha_{s}$$
My try: I think this problem might use 
Eisenstein's criterion,http://en.wikipedia.org/wiki/Eisenstein%27s_criterion,
But I can't, and I think this problem is very nice, and this problem is my friend ask me. Thank you everyone.
 A: Stated in current form, the statement can fail in two ways.


*

*If $\beta_{s+1} = 0$ then $f(x)$ has a repeat root with multiplicity $\ge \alpha_s$ at $x = 0$. It is clear we can push the multiplicity as high as we like.

*Even when $\beta_{s+1} \ne 0 $, $f(x)$ can have a repeated root with multiplicity $> s$.
The simplest example is
$$ s = 2,\quad(\beta_1,\beta_2,\beta_3) = (3,3,1)\quad\text{ and }\quad(\alpha_0,\alpha_1,\alpha_2) = (3,2,1)$$
$f(x) = (x+1)^3$ and this has a repeated root of multiplicity $3 = s + 1$ at $x = -1$.
However, the statement is true if we assume $\beta_{s+1} \ne 0$ and relax the constraint on multiplicity $m_i$ from $\le s$ to $\le s+1$.
Assume the contrary, let's say $f(x)$ has a repeated root at $x = r$ with multiplicity $m > s + 1$. Consider following $s+2$ polynomials for $0 \le k \le s + 1$.
$$\left( x \frac{d}{dx} \right)^k f(x) = \alpha_0^k x^{\alpha_0} + \alpha_1^k\beta_1 x^{\alpha_1} + \cdots \alpha_s^k\beta_s x^{\alpha_s} + \begin{cases}\beta_{s+1},&\text{ for } k = 0\\ \\0,&\text{ for } k \le 1 \le s+1\end{cases}$$
It is easy to see all of them can be expressed as linear combination of $f(x), f'(x), \ldots, f^{(s+1)}(x)$. As a result, they vanish at $x = r$. We can summarize this as a matrix equation:
$$
  \begin{pmatrix}0\\0\\ \vdots \\0\\0\end{pmatrix}
= \begin{pmatrix}
1 & 1 & \ldots & 1 & 1\\
\alpha_0 & \alpha_1 & \ldots & \alpha_s & 0\\
\vdots & \vdots & \ddots & \vdots & 0\\
\alpha_0^{s} & \alpha_1^{s} & \ldots & \alpha_s^{s} & 0\\
\alpha_0^{s+1} & \alpha_1^{s+1} & \ldots & \alpha_s^{s+1} & 0
\end{pmatrix}
= \begin{pmatrix}\;\;\;r^{\alpha_0}\\ \beta_1 r^{\alpha_1} \\ \vdots \\ \beta_s r^{\alpha_s} \\ \beta_{s+1}\;\end{pmatrix}
$$
The $(s\!+\!2) \times (s\!+\!2)$ matrix appeared above is a Vandermonde matrix. We know it is invertible
when $a_1, \ldots, \alpha_s$ and $0$ are all distinct. This implies
$$r^{\alpha_0} = \beta_1 r^{\alpha_1} = \cdots = \beta_s r^{\alpha_s} = \beta_{s+1} = 0$$
and hence contradict with our assumption that $\beta_{s+1} \ne 0$. From this, we can conclude the multiplicity of any repeated root is at most $s+1$.
