Let $f(x)=a_0+a_1x+...a_{n-1}+a_nx^n $ and $\bar{f}(x)=a_n+a_{n-1}x+...+a_1x^{n-1}+a_0x^n \in F[x]$. ($F$ is a field, of course.)
If $c\ne0$ is a zero of $f(x)$, prove $c^{-1}$ is a zero of $\bar{f}(x)$.
Do have an something with messy coefficients and saying c is a root and factoring out $x-c$. Wondering if there is a more elegant way?
$f(c)=0$ $$ 0=a_0+a_1c+...+a_{n-1}c^{n-1}+a_nc^n $$ Div by $c^n\neq0$ $$\begin{aligned}0&=\frac{0}{c^n}=a_0c^{-n}+a_1c^{1-n}+...+a_n1c^{-1}+a_n\\ 0&=a_0(c^{-1})^n+a_1(c^{-1})^{n-1}+...+a_{n-1}(c^{-1})^1+a_n \end{aligned}$$ $\therefore \bar {f}(c^{-1})= a_n + a_{n-1}(c^{-1})^1+...+a_1(c^{-1})^{n-1}+ a_0(c^{-1})^n=0$