Is it possible for a multiple critical point of a polynomial $p$ not to be a multiple zero of $p(z)+c$?

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Let $p$ be a univariate polynomial and $p'$ its derivative. Due to the product rule, it's clear that if some number is a multiple zero of $p$, it will be a zero of $p'$ as well. But is it possible for $p$ to have a critical point $a$ of order $m>1$ where $a$ is not a zero of multiplicity at least $m$ of $p(z) + c$, for any $c in mathbbC$? If not, how would you show it? I'm guessing there's some way with integration by parts, but it's escaping me.




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    Let $p$ be a univariate polynomial and $p'$ its derivative. Due to the product rule, it's clear that if some number is a multiple zero of $p$, it will be a zero of $p'$ as well. But is it possible for $p$ to have a critical point $a$ of order $m>1$ where $a$ is not a zero of multiplicity at least $m$ of $p(z) + c$, for any $c in mathbbC$? If not, how would you show it? I'm guessing there's some way with integration by parts, but it's escaping me.




    calculus complex-analysis




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      Let $p$ be a univariate polynomial and $p'$ its derivative. Due to the product rule, it's clear that if some number is a multiple zero of $p$, it will be a zero of $p'$ as well. But is it possible for $p$ to have a critical point $a$ of order $m>1$ where $a$ is not a zero of multiplicity at least $m$ of $p(z) + c$, for any $c in mathbbC$? If not, how would you show it? I'm guessing there's some way with integration by parts, but it's escaping me.




      calculus complex-analysis




      share|cite|improve this question











      Let $p$ be a univariate polynomial and $p'$ its derivative. Due to the product rule, it's clear that if some number is a multiple zero of $p$, it will be a zero of $p'$ as well. But is it possible for $p$ to have a critical point $a$ of order $m>1$ where $a$ is not a zero of multiplicity at least $m$ of $p(z) + c$, for any $c in mathbbC$? If not, how would you show it? I'm guessing there's some way with integration by parts, but it's escaping me.





      calculus complex-analysis





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      asked Aug 3 at 3:35









      asterac

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          Ir $a$ is a zero of order $m$ of $p’(z)$, then it is a zero of order $m+1$ of $p(z)-p(a)$.






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          • Typo; $a$ is a zero of $p’$.
            – Julián Aguirre
            Aug 3 at 7:27










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          Ir $a$ is a zero of order $m$ of $p’(z)$, then it is a zero of order $m+1$ of $p(z)-p(a)$.






          share|cite|improve this answer























          • Typo; $a$ is a zero of $p’$.
            – Julián Aguirre
            Aug 3 at 7:27














          up vote
          1
          down vote













          Ir $a$ is a zero of order $m$ of $p’(z)$, then it is a zero of order $m+1$ of $p(z)-p(a)$.






          share|cite|improve this answer























          • Typo; $a$ is a zero of $p’$.
            – Julián Aguirre
            Aug 3 at 7:27












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          1
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          up vote
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          Ir $a$ is a zero of order $m$ of $p’(z)$, then it is a zero of order $m+1$ of $p(z)-p(a)$.






          share|cite|improve this answer















          Ir $a$ is a zero of order $m$ of $p’(z)$, then it is a zero of order $m+1$ of $p(z)-p(a)$.







          share|cite|improve this answer















          share|cite|improve this answer



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          edited Aug 3 at 7:26


























          answered Aug 3 at 5:06









          Julián Aguirre

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          • Typo; $a$ is a zero of $p’$.
            – Julián Aguirre
            Aug 3 at 7:27
















          • Typo; $a$ is a zero of $p’$.
            – Julián Aguirre
            Aug 3 at 7:27















          Typo; $a$ is a zero of $p’$.
          – Julián Aguirre
          Aug 3 at 7:27




          Typo; $a$ is a zero of $p’$.
          – Julián Aguirre
          Aug 3 at 7:27












           

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