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If $F$ is a field and $a \neq 0$ is a zero of $f(x) = a_{0} + a_{1}x + · · · + a_{n}x ^n$ in $F[x]$, Then $1/a$ is a zero of $a_{n} + a_{n−1}x + · · · + a_{0}x^{n}$.

I thought of this -

As $a$ is a zero of $f(x)$ so $f(a) = a_{0} + a_{1}a + · · · + a_{n}a ^n = 0$ Now as $a \in F$ so inverse of $a$ that is $a^{-1} = 1/a$ exists in $F$,somultiplying $a^{-1}$ to both sides of the above equation $n$ times we would get $a_{n} + a_{n−1}(a^{-1}) + · · · + a_{0} = 0$ implying $a^{-1}$ is the zero of $a_{n} + a_{n−1}x + · · · + a_{0}x^{n}$.

Is this approach correct?

BAYMAX
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    Seems correct to me – Peter Apr 20 '17 at 10:04
  • Ok,thanks ,I searched and got also another way of doing this here -https://math.stackexchange.com/questions/692191/show-that-for-a-zero-of-a-polynomial-in-a-field-that-1-a-is-a-zero-of-another-po – BAYMAX Apr 20 '17 at 10:19
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    The second polynomial is $X^nf(\frac{1}{X})$, so if $a$ is a zero of $f$ then $\frac{1}{a}$ is a zero of this polynomial – Gal Porat Apr 20 '17 at 10:23

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