Roots of a power series in an interval Let $ a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} = 0 $
Prove that $ a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n = 0 $ has real roots into the interval $ (0,1) $
I found this problem in a real analysis course notes, but I don't even know how to attack the problem. I tried to affirm that all coefficients are zero, but that is cleary not true, we have many cases when the result is 0 but $ a_i \ne 0$ for some $i$. I have tried derive/integrate, isolate and substitute some coefficients ($ a_0 $ and $a_n $ where my favorite candidates). Work with factorials (and derivatives and factorials) but could not find a way to prove. I have many pages of useless scratches.
Any tips are welcome.

As Alexey Burdin points out, if you integrate $f(x) = a_0+a_1x+a_2x^2+\cdots+a_nx^n$ from $0$ to $1$, you get the sum
$$
\int_{x=0}^1 f(x) \, dx = \sum_{k=0}^n \frac{a_k}{k+1} = 0
$$
Since $f(0) = a_0$ and $f(x)$ is continuous over the reals, either $f(x) = 0$ over the interval $(0, 1)$, or $f(x)$ must be negative and positive over some portions of the interval $(0, 1)$; either way, there exist real roots in that interval as well.
