Prove: $na_0x^{n-1}+(n-1)a_1x^{n-2}+...+a_{n-1}=0$ has at least one positive root smaller than $x_0$ Let $a_0x^n+a_1x^{n-1}+...+a_{n-1}x=0$ has a positive root $x=x_0$.
Prove: $na_0x^{n-1}+(n-1)a_1x^{n-2}+...+a_{n-1}=0$ has at least one positive root smaller than $x_0$   
Vieta's theorem on $na_0x^{n-1}+(n-1)a_1x^{n-2}+...+a_{n-1}=0$,
$x_1+x_2+...+x_{n-1}=\frac{(n-1)a_1}{na_0}$
$x_1x_2...x_{n-1}=(-1)^{n-1}\frac{a_{n-1}}{na_0}$
How to evaluate Vieta's formulas for $a_0x^n+a_1x^{n-1}+...+a_{n-1}x=0$
Are Vieta's formulas useful for this problem?
 A: Note that the second polynomial is the derivative of the first and that the first polynomial has two zeros: $0$ and $x_0$. The statement follows from Rolle's Theorem which states that between two zeros of $f(x)$ there is a zero of $f^\prime(x)$.
A: Actually, we can use Rolle's Theorem:
If a real-valued function $f$ is continuous on a proper closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a)=f(b)$, then there exists at least one $c$ in the open interval (a, b) such that $f'(c)=0$.
Let $$f(x)=a_0x^n+a_1x^{n−1}+...+a_{n−1}x.$$ Obviously, we have $f(0)=0$ and $f(x_0)=0$ due to the assumption. Then, use Rolle's Theorem, here $a$ is $0$ and $b$ is $x_0$, so there must exist at least one $x_1$ in the $(0,x_0)$ such that $f'(x_1)=0.$
Firstly, we can claim that $x_1$ is a positive number that is smaller than $x_0$. Secondly, $$f'(x)=na_0x^{n−1}+(n−1)a_1x^{n−2}+...+a_{n−1},$$ and $f'(x_1)$ shows that $x_1$ is the solution of the formula $$na_0x^{n−1}+(n−1)a_1x^{n−2}+...+a_{n−1}=0.$$
This completes the proof.
If you are not familiar with Rolle's Theorem and want to take an deep look inside its proof, you can refer https://en.wikipedia.org/wiki/Rolle%27s_theorem.
