Root of multiplicity? 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.  
 A: Suppose $f(x)=(x-a)^2g(x)$ and differentiate using the product rule.

Added
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.
A: Here's another proof of the double root test, which goes straight to the heart of the matter.
$$\rm\begin{eqnarray} &&\rm\!\! (x\!-\!a)^2 |\ p(x)\!\!\!\!\!\!\!\\
\iff\ &&\rm x\!-\!a\ |\ p(x)\  &\rm and\ \ &\rm  x\!-\!a\ \bigg|\ \dfrac{p(x)}{x\!-\!a}\\
\\
\iff\ &&\rm p(a) = 0 &\rm and&\rm x\!-\!a\ \bigg|\ \dfrac{p(x)-p(a)}{x\!-\!a}\ \ \left[\!\iff \color{#C00}{\dfrac{p(x)-p(a)}{x\!-\!a}\Bigg|_{\large\:x\:=\:a}} \!=\: 0\ \right] \\
\\
\iff\ &&\rm p(a) = 0 &\rm and&\rm \color{#C00}{p'(a)} = 0\end{eqnarray}$$
A: 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 $f'(a)=0$.
