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I don't know how to start the question. The title is self explanatory. How to approach and make a formal proof?

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    @$B$illDubuque I am extremely sorry.2012-04-04

2 Answers 2

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Hint $\ $ Wlog, by a shift, assume the root is $\rm\: r = 0.$

Notice $\rm\ \ x^2\: |\ f(x)$

$\rm\ \iff\ x\ |\ f(x)\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)}{x}$

$\rm\ \iff\ f(0) = 0\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)-f(0)}x\iff \dfrac{f(x)-f(0)}x\bigg|_{\:x\:=\:0} =\: 0$

$\rm\ \iff\ f(0) = 0\ $ and $\rm\ f'(0) = 0$

Remark $\ $ It's often overlooked that many divisibility properties of numbers are specializations of this double-root criterion for (polynomial) functions, e.g. see my answer here.

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Hint: (a slightly different approach).

Suppose that $f(x)$ has $r$ as a root $m$ times ($m\geq 1$). Then $f(x)=(x-r)^m g(x)$ for some polynomial $g(x)$ with $g(r)\not=0$. What is f'(x)? What has to be true about $m$ if f'(r)=0?

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    Er, my previous comment should've ended with "What is $f'(r)$?" Sorry.2012-04-09