polynomial and derivatives

Question

given real numbers $a_0 , a_1 ,\cdots ,a_n$ such that $a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} =0$

prove that there is a root in the interval $(0,1)$ for the polynomial $p(x) = a_0 +a_1 x + a_2 x^2 +\cdots + a_n x^n$ ?!

I just don't know how to solve this question !.

Answer 1

Let $q(x) = a_0 x + a_1 \dfrac{x^2}{2} + \dots + a_k \dfrac{x^{k+1}}{k+1}.$ Then $q(0)=q(1)=0$, by the MVT there is a $ 0 < t < 1$ such that $q'(t) = p(t) = 0.$

Answer 2

If $p(x)\ne 0$ for all $x\in (0,1)$ then we have $p(x)<0$ for all $x\in (0,1)$ or $p(x)>0$ for all $x\in (0,1)$ so $$\Big|\int _0^1 p(x)dx \Big| >0$$

But $$\int _0^1 p(x)dx =a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} =0$$

A contradiction.

Answer 3

Let $$q(x)= a_0x+a_1{x^2\over 2} +...+a_{n}{x^{n+1}\over n+1}$$ so $q'(x)=p(x)$. By Lagrange theorem we have in $\zeta \in (0,1)$ such that $$p(\zeta) = q'(\zeta) = {q(1)-q(0)\over 1-0} = {0-0\over 1}=0$$