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Show if a is a root of multiplicity $n\geq 2\ $, then $f(a) = 0$ and $f'(a)=0.$

I was trying to learn root of multiplicity and saw this question. My TA did not go over it yet but I was wondering how this proof would look like.

I get root multiplicity under product solutions in $Z_n$ just from reading but this proof goes into detail.

Can someone please show me? Thanks.

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Edited tags - this definitely isn't a set theory question! – Clive Newstead Apr 24 '13 at 19:12
Dear 9959: you might want to think about accepting answers that are helpful: you can accept one answer per question, and to accept an answer, you can simply click on the $\checkmark$ to the left of the answer you'd like to accept. You receive two reputation points for each answer you accept! – amWhy Apr 24 '13 at 20:00

Suppose $f(x)=(x-a)^2g(x)$ and differentiate using the product rule.


If this is in an abstract algebra context, as I expect it is, then define the derivation $D : R[x] \to R[x]$, where $R$ is whichever ring you're working over, by $$D(\sum a_nx^n ) = \sum na_nx^{n-1}$$ where by $na_n$ I really mean $\underbrace{a_n + a_n + \cdots + a_n}_{n\ \text{times}}$.

Then prove that $D(fg)=fD(g)+gD(f)$, and proceed as if you're doing calculus. (You may have already done this somewhere in your course.)

If this is in a calculus context then ignore the previous paragraph.

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Ok. Yes I did prove D(fg) = f D(g) + g D(f). I get it now with differentiating using the product rule that you gave. – 9959 Apr 24 '13 at 19:29

Here's another proof of the double root test, which goes straight to the heart of the matter.

$$\rm\begin{eqnarray} &&\rm\!\! (x\!-\!a)^2 |\ p(x)\!\!\!\!\!\!\!\\ \iff\ &&\rm x\!-\!a\ |\ p(x)\ &\rm and\ \ &\rm x\!-\!a\ \bigg|\ \dfrac{p(x)}{x\!-\!a}\\ \\ \iff\ &&\rm p(a) = 0 &\rm and&\rm x\!-\!a\ \bigg|\ \dfrac{p(x)-p(a)}{x\!-\!a}\ \ \left[\!\iff \color{#C00}{\dfrac{p(x)-p(a)}{x\!-\!a}\Bigg|_{\large\:x\:=\:a}} \!=\: 0\ \right] \\ \\ \iff\ &&\rm p(a) = 0 &\rm and&\rm \color{#C00}{p'(a)} = 0\end{eqnarray}$$

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If $a$ is a root of multiplicity $\geq 2$, then you can write $f(x)=(x-a)^2g(x)$ (if $n>2$, then $g(x)$ will also have $(x-a)$ as a factor, but that's not really important to solving this problem).

Using the product rule then, you get $f'(x)=2(x-a)g(x)+(x-a)^2g'(x)$, and notice that both $f(a)=0$ and $f'(a)=0$.

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