If we suppose that the complex number $\alpha \not= 0$ is a root of the polynomial $P\in Q[x]$, I want to find a polynomial also in $Q[x]$ having $\alpha^{-1}$ as a root.

If we have $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$

Then $P(\frac{1}{x})=a_nx^{-n}+a_{n-1}x^{1-n}+...+a_1x^{-1}+a_0$

I am not sure how to proceed, any help is appreciated, thank you!

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    $\begingroup$ Hint: if $p(x)$ is such that $p(a)=0$ consider $x^dp(\frac 1x)$ for suitable $d$. $\endgroup$ – lulu Dec 31 '16 at 16:13
  • $\begingroup$ I have edited my question slightly, I understand up to what I've added but I am not sure why I should consider $x^d$ $\endgroup$ – Harrison W. Dec 31 '16 at 16:20
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    $\begingroup$ The function you have written, $F(x)=P(\frac 1x)$ does indeed satisfy $F(\frac 1a)=0$ but alas it is not a polynomial. But if you could only clear the denominators.... $\endgroup$ – lulu Dec 31 '16 at 16:21
  • $\begingroup$ ohhh haha, thank you, I was being silly.. $\endgroup$ – Harrison W. Dec 31 '16 at 16:22
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    $\begingroup$ No problem. Happy New Year! $\endgroup$ – lulu Dec 31 '16 at 16:22

$\alpha^{-1}$ satisfies the polynomial $(x)=a_n+a_{n-1}x+...+a_1x^{n-1}+a_0x^n$ following from what I added to the question.


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