# Is there an algebraic way to prove this relationship between the roots of a real polynomial and the roots of its derivative?

Let $$\sum_{0 \leq i\leq n} a_ix^i$$

be a polynomial (real coefficients) with at least two real roots. Is there an algebraic way to show that for any two roots $k_1, k_2$ of this polynomial, the polynomial

$$\sum_{1 \leq i\leq n} i \cdot a_ix^{i-1}$$

admits at least one root $c$ satisfying $k_1 <c < k_2$?

Analytically, this is of course a consequence of Rolle's theorem.

Edit: "Algebra" is as broad as you want it to be. Elementary or abstract. The completeness of $\mathbb{R}$ is essential, so it won't be purely algebraic. I was mainly hoping for something without derivatives.

• What would qualify as "purely algebraic proof" for you? Something only out of ring/ideal theory without any analysis at all? It may be hard, or perhaps even impossible, to accomplish. – DonAntonio Aug 24 '16 at 10:40
• I also think that a "purey algebraic" proof would be impossible. Philosophically, my motivation is that algebra deals with equalities, while inequalities are the realm of analysis. – Daniel Robert-Nicoud Aug 24 '16 at 10:43
• This isn't true for all rings, even for ordered fields (take the rationals). So my guess would be that this has to use the topology of the real numbers. Very good question though! – Nitrogen Aug 24 '16 at 10:46
• It is true for the reals but not the rationals. The analytic completeness is difficult to define algebraically. – Arthur Aug 24 '16 at 10:47
• @Arthur If the fundamental theorem of algebra can be proved algebraically, why can't this? – MathematicsStudent1122 Aug 24 '16 at 10:50

Probably not. Note that in $\mathbb Q(\sqrt{6})$, the polynomial $x^3 - 6x$ has three roots. However, its derivative has no roots. These would be $\pm \sqrt{2}$. Thus, whatever you mean by purely algebraic needs to be stronger than the theory of ordered fields.