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I need to prove that

The roots of the derivative of a polynomial $f$ of degree $n$ (with real coefficients) whose roots are all real are real, too.

The hypothesis of the roots of $f$ being all real implies that we can express $f$ as a product $$a(x-x_1)\cdots (x-x_n).$$

Let's define $f_i(x):= x-x_i$. My idea was to calculate the derivative of $f$ applying the general Leibniz rule to it's decomposition above, and to somehow obtain from it an expression of $f'$ as a product of linear polynomials (concluding this way our proof).

If I'm not wrong, since all the multinomial coefficients equal 1 and the derivative of each function $f_i$ is one, we obtain that

$f'=a(f_1\cdots f_n)'=a[ f_2 \cdots f_n + f_1 f_3 \cdots f_n+ \cdots + f_1f_2 \cdots f_{n-1}]$.

Now I have no ideas of how to factorize it as a product of the $f_i's$ (or in general as a product of linear factors). Any suggestion is very appreciated.

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    $\begingroup$ Let $a,b$ be two consecutive unequal roots of $f$. From $f(a)=f(b)=0$ a standard argument shows the existence of an intermediate $c$ with $f'(c)=0$. Now the argument should be refined to see what happens with the inheritance of roots for multiple roots... $\endgroup$
    – dan_fulea
    Commented Apr 13, 2021 at 17:21
  • $\begingroup$ Yes ! I already considered the case were all the roots of $f$ are simple (because as you point out by means of the Role's theorem and the fact that $f'$ can at most have $n-1$ roots we are done), my problem is when the multiplicities of the roots of f are not all 1. $\endgroup$
    – Amelian
    Commented Apr 13, 2021 at 17:24
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    $\begingroup$ So if $f=(x-a)^rg$ with $g(a)\ne 0$, then $f'$ has the $a$ root with multiplicity $(r-1)$. This finishes the proof, where is still the problem? For instance, if the polynomial is $$f=(x-1)^3(x-3)(x-7)^5(x-8)(x-9)(x-10)^2\ ,$$ which are the roots we can insure by the above two arguments? $\endgroup$
    – dan_fulea
    Commented Apr 13, 2021 at 17:26
  • $\begingroup$ Ahh now I see it ! Thank you very much ! $\endgroup$
    – Amelian
    Commented Apr 13, 2021 at 17:29
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    $\begingroup$ Please try to answer your own question, it will be upvoted when everything is ok! $\endgroup$
    – dan_fulea
    Commented Apr 13, 2021 at 17:32

2 Answers 2

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With the useful point of view of @dan_fulea in the comments I was able to prove the statement, I'm sharing my development.

Let $f$ be a polynomial with real coefficients whose roots $x_1, \ldots , x_m$ are all real. If we denote the multiplicity of the root $x_i$ by $r_i$, then

$$f(x)=a(x-x_1)^{r_1}\cdots (x-x_m)^{r_m}.$$

Let's suppose further that just the first $l$ roots of $f$ have multiplicity greater than 1. Then, $$\delta(f)=r_1+ \cdots +r_l +(m-l).$$

Note first that, by Role's theorem, we know about the existence of $m-1$ roots $\xi_i$ of the polynomial $f'(x)$, with $\xi_i \in ]x_i, x_i+1[$ for all $i$. Now, we assert that for each $1 \leq i \leq l$, the number $x_i$ is a root of $f'$.

Indeed, expressing $f$ as the product $f(x)=(x-x_i)^{r_i}g(x)$, by the rule of the derivative of a product we get that

$$f'(x)=r_i(x-x_i)^{r_i-1}g(x)+(x-x_i)^{r_i}g'(x)=(x-x_i)^{r_i-1}P(x),$$ where $P(x)=r_ig(x)+(x-x_i)g'(x)$. Then $x_i$ actually is a root of $f'$; moreover, if $x-x_i$ happens to divide the polynomial $P(x)$, we would get that $x-x_i$ divides $g(x)$, but this contradicts the fact of $r_i$ being the multiplicity of the root $r_i$ of the polynomial $f$.

We found then the roots $\xi_i, \ldots, \xi_{m-1}$ and $x_1, \ldots, x_l$ of the polynomial $f'$. Since $$(m-1)+(r_1-1)+ \cdots +(r_l-1)=(r_1+ \cdots +r_l)+(m-l)-1= \delta(f)-1=\delta(f'),$$

we conclude that $f'(x)=(x-\xi_1)\cdots(x-\xi_{m-1})(x-x_1)^{r_1-1}\cdots(x-x_l)^{r_l-1}$

and we see that all the roots of $f'$ are real numbers.

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    $\begingroup$ Yes! Excelent presentation! Welcome on stackexchange! $\endgroup$
    – dan_fulea
    Commented Apr 13, 2021 at 19:41
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    $\begingroup$ There is an alternative simple explanation. See my answer. $\endgroup$
    – Jean Marie
    Commented May 29, 2021 at 7:56
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A different explanation.

It is an immediate consequence of the Gauss-Lucas theorem stating that the roots of $f'$ are situated in the convex hull of the roots of polynomial $f$, which is a line segment of the real axis.

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  • $\begingroup$ A beautiful result ! Thank you so much for sharing it ! $\endgroup$
    – Amelian
    Commented May 29, 2021 at 15:59

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