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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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    @Kannappan I vote not to close as dup of *said* question - which is not an *exact* dup (but is a *related* question). Please be more careful about dup closing, since it seems often others vote to close without even looking at the proposed dup. There may be another exact dup though.2012-04-04
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    @BillDubuque 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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    m(x-r)^m-1 *g(x) + g'(x) (x-r)^m ? I forgot to mention integration is not allowed.2012-04-04
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    Sorry, i just came out of an exam. I meant integration2012-04-04
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    Ah, well you don't need to integrate. So, you have $f'(x)=m(x-r)^{m-1}g(x)+g'(x)(x-r)^m$. Suppose $f'(r)=0$. What has to be true about $m$? And, on the other hand, if $m\geq 2$ (ie if $r$ is a root more than once), then why is $r$ a root of $f'(x)$?2012-04-04
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    (g'(x) (x-r) )/ g (x) = m2012-04-04
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    is it because once we differentiate when is greater or equal to 2 then the root is still multiplying g(x) ?2012-04-04
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    Why are you trying to solve for $m$? No, no, no. We have $f'(x)=m(x-r)^{m-1}g(x)+g'(x)(x-r)^m$. What is $f'(0)$?2012-04-04
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    Er, my previous comment should've ended with "What is $f'(r)$?" Sorry.2012-04-09