If the equation a_0x^n+a_1x^(n-1)+...+a_(n-1)x = 0 has

If the equation $a_{0}x^{n}+a_{1}x^{\,n-1}+\cdots+a_{n-1}x = 0$ has a positive root $x = x_{0}$, show that the equation $na_{0}x^{\,n-1}+\bigl(n-1\bigr)a_{1}x^{\,n-2}+\cdots+a_{n-1} = 0$ has at least one positive root smaller than $x_{0}$.

Step 1: Name the polynomial

\[ f(x) = a_{0}x^{n}+a_{1}x^{\,n-1}+\cdots+a_{n-1}x. \]

Differentiating term by term reproduces the second equation exactly: $f'(x) = na_{0}x^{\,n-1}+\bigl(n-1\bigr)a_{1}x^{\,n-2}+\cdots+a_{n-1}$.

So the claim is that $f'$ has a root in $\bigl(0,x_{0}\bigr)$.

Step 2: The hidden second root

The polynomial has no constant term — every term carries a factor $x$: $f(x) = x\Bigl(a_{0}x^{\,n-1}+a_{1}x^{\,n-2}+\cdots+a_{n-1}\Bigr) \;\Longrightarrow\; f(0) = 0$.

This is the observation on which everything turns: the hypothesis supplies one root $x_{0}$, and the shape of the equation supplies another at the origin.

Step 3: Apply Rolle's theorem on $\bigl[0,x_{0}\bigr]$

$f$ is a polynomial, hence continuous on $\bigl[0,x_{0}\bigr]$ and differentiable on $\bigl(0,x_{0}\bigr)$; and $f(0) = f(x_{0}) = 0$. All three hypotheses hold, so there exists $\xi\in\bigl(0,x_{0}\bigr)$ with $f'(\xi) = 0$.

Step 4: Verification

\[\text{Take } f(x) = x^{3}-4x:\ \text{roots } 0,\pm2, \text{ so } x_{0} = 2\]

\[f'(x) = 3x^{2}-4 = 0 \;\Longrightarrow\; \xi = \frac{2}{\sqrt3} \approx 1.1547 \in \bigl(0,2\bigr) \;\;\checkmark\]

\[\text{Take } f(x) = x^{2}-3x:\ x_{0} = 3,\ f'(x) = 2x-3,\ \xi = 1.5 \in \bigl(0,3\bigr) \;\;\checkmark\]