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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?

Bill Dubuque
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Minestudent
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3 Answers3

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Wlog, by a shift, we may assume the root is $\rm\: r = 0,\,$ so we have:

$\bbox[1px,border:1px solid #c00]{\bbox[6px,border:1px solid #c00]{{\bf Double\ Root\ Test}\ \ \rm\ x^2\mid f(x) \iff f(0) = 0 = f'(0)}}$

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

$\rm\ \iff\ \color{#c00}{x\ |\ f(x)}\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)}{x}$

$\rm\ \iff\ \color{#c00}{f(0)\! =\! 0}\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)\!-\!\color{#c00}{f(0)}}x\, \left[\!\!\iff\!\! \color{#0a0}{\dfrac{f(x)\!-\!f(0)}x}\bigg|_{\,x\,=\,0} \!= 0\,\right]$

$\rm\ \iff\ f(0)\! =\! 0\ $ and $\rm\ \color{#0a0}{f'(0)}\! =\! 0$

Above we used the Polynomial $\rm\color{#c00}{Factor}$ Theorem and $\rm\color{#0a0}{Derivative}$ Formula.

Remark $ $ It is often overlooked that many divisibility properties of numbers are specializations of this double root criterion for (polynomial) functions, e.g. see my answer here, or below from here.

Hint: a conceptual way: we seek $\,\color{#c00}{47^2\mid f(47)}\,$ for $\,f(x) = (x\!+\!5)^{n+1} + ((n\!-\!1)x\!-\!5)(2x\!+\!5)^n$

$\,f(0)=0=f'(0)\,$ so by the above double root test (or first two terms of the Binomial Theorem), it follows that $\,x=0\,$ is a double root so $\,f(x) = x^2 g(x)\,$ for $g(x)$ a polynomial with integer coef's. Evaluating at $\,x=47\,$ yields $\,\color{#c00}{f(47)}= 47^2 g(47)$.

Bill Dubuque
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  • Or if Taylor's Theorem is already known then if follows immediately from that (first two terms). – Bill Dubuque Sep 29 '24 at 18:52
  • Thanks for sharing so important Theorems. But would you please tell where this example ${47^2\mid f(47)}$ comes from? What's the other applications of $\bf Double\ Root\ Test$? Btw, $n\ge1$ here. – MathArt Nov 06 '24 at 12:30
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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 $f'(r)=0$?

  • m(x-r)^m-1 *g(x) + g'(x) (x-r)^m ?

    I forgot to mention integration is not allowed.

     – Minestudent Apr 04 '12 at 00:33
  • Sorry, i just came out of an exam. I meant integration – Minestudent Apr 04 '12 at 00:38
  • 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 $f'(x)$? –  Apr 04 '12 at 00:40
  • (g'(x) (x-r) )/ g (x) = m – Minestudent Apr 04 '12 at 01:14
  • is it because once we differentiate when is greater or equal to 2 then the root is still multiplying g(x) ? – Minestudent Apr 04 '12 at 01:22
  • 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 $f'(0)$? –  Apr 04 '12 at 17:49
  • Er, my previous comment should've ended with "What is $f'(r)$?" Sorry. –  Apr 09 '12 at 10:31
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I was trying to solve this and came across this question. This is my solution. I hope it helps.

Suppose $P(x)$ is a polynomial of degree n and r is a root of $P(x)$ and its derivative.

$$P(x)=(x-r)Q(x)$$ $$P ^\prime(x)=(x-r)S(x)$$ where $Q(x)$ and $S(x)$ are also polynomials. Based on the first equation, we can also write $$P ^\prime(x)=Q(x)+(x-r)Q^\prime(x)$$ For $x \neq r$ $$S(x)=\frac{P ^\prime(x)}{(x-r)}=\frac{Q(x)+(x-r)Q^\prime(x)}{x-r}$$ $$=\frac{Q(x)}{x-r}+Q^\prime(x)$$ $$=\frac{P(x)}{(x-r)^2}+Q^\prime(x)$$ $$\Rightarrow P(x)=(x-r)^2(S(x)-Q^\prime(x))$$ So, r is a double root of $P(x)$.

If r is a double root of $P(x)$, we have $$P(x)=(x-r)^2T(x)$$ where $T(x)$ is a polynomial. We have $$P^\prime(x)=2(x-r)T(x)+(x-r)^2T^\prime(x)$$ So, r is also a root of the derivative of $P(x)$.

Nicki Bood
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