Question :
If $x_1, x_2, \cdots, x_n$ are the roots of the equation: $$\mathcal{f}(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_0 ~,$$ find the roots of the following equation : $$\mathcal{g}(x)=a_0x^n-a_1x^{n-1}+\cdots + (-1)^na_n$$
I've stated the question above.
I noticed that $\mathcal{g}(x)= x^n \mathcal{f}\left(-\frac{1}{x}\right)$. It is essentially same as writing $\mathcal{f}\left(-\frac{1}{x}\right)$ to avoid ambiguity of $g(0)=0$ which isn't true unless $a_n=0$.
Basically, we have to find the roots of $\mathcal{f}\left(-\frac{1}{x}\right)$. How to find that?
For $x_i\not=0$, we have $$\begin{align}g\left(-\frac{1}{x_i}\right)&=a_0\left(-\frac{1}{x_i}\right)^n-a_1\left(-\frac{1}{x_i}\right)^{n-1}+\cdots +(-1)^na_n\\\\&=\left(-\frac{1}{x_i}\right)^n\left(a_0+a_1x_i+\cdots +a_nx_i^{n}\right)\\\\&=\left(-\frac{1}{x_i}\right)^nf(x_i)\\\\&=0\end{align}$$