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I'm reading one theorem in the book called Contemporary Abstract Algebra by Gallian, and here's what puzzles me somewhat:

If $f(x) = (x-a)^mg(x) = (x-b)^m\phi(g(x))$ then how does the author deduce that the multiplicity of $b$ is at least $m$? Why not less than $m$ or equal to $m$?

Thank you for clarifying this for me.

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    $\begingroup$ What's the definition of multiplicity? $\endgroup$ – Swapnil Tripathi Mar 14 '17 at 20:57
  • $\begingroup$ That's how many times a zero of a polynomial appears in the multiset of zeros of the polynomial. $\endgroup$ – sequence Mar 14 '17 at 21:01
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    $\begingroup$ Yes. So $b$ appears $m$ times as a root already. If $b$ doesn't appear in $\phi(g(x))$ then multiplicity will be exactly $m$, if it appears $n$ times in $\phi(g(x))$, then the multiplicity of $b$ in $f(x)$ would be $m+n$ $\endgroup$ – Swapnil Tripathi Mar 14 '17 at 21:03
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    $\begingroup$ @sequence In other words, the multiplicity of $b$ is the largest positive integer $k$ such that there exists a polynomial $h(x)$ with the property that $f(x)=(x-b)^kh(x)$ and $x-b$ does not divide $h(x)$. Does this way of formulating the concept of multiplicity help you? $\endgroup$ – Git Gud Mar 14 '17 at 21:07
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From the comments above.


The multiplicity of a root of a polynomial is the number of times the roots appears in the multiset of all roots of the polynomial.

Here, $b$ appears $m$ times as a root of $f(x)$ already. If $b$ appears as a root of $\phi(g(x))$ with multiplicity $n$, then the multiplicity of $b$ in $f(x)$ would be $m + n$. Since $n \geq 0$, we have $m + n \geq m$.

It is useful to view the multiplicity of a root $b$ of a polynomial $f(x)$ to be the largest non-negative integer $k$ such that we can write $f(x) = (x-b)^k h(x)$, where $h(x)$ is a polynomial such that $x-b$ does not divide $h(x)$.

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