# Proving that, if a polynomial with real coefficients has all real roots, then the roots of its derivative are real, too.

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.

• 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... Commented Apr 13, 2021 at 17:21
• 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. Commented Apr 13, 2021 at 17:24
• 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? Commented Apr 13, 2021 at 17:26
• Ahh now I see it ! Thank you very much ! Commented Apr 13, 2021 at 17:29
• Please try to answer your own question, it will be upvoted when everything is ok! Commented Apr 13, 2021 at 17:32

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.

• Yes! Excelent presentation! Welcome on stackexchange! Commented Apr 13, 2021 at 19:41
• There is an alternative simple explanation. See my answer. Commented May 29, 2021 at 7:56

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.

• A beautiful result ! Thank you so much for sharing it ! Commented May 29, 2021 at 15:59