proof that at least one solution exist

Question

let us consider following problem:

i can take $n=1,2,...$ and try to understand basic relationship between this linear relation and relevant polynomial,for example

1.$n=1--> we have $a_0=0$

2.$n=2$ we will have

$a_0/1+a_1/2=0$

$a_0+a_1*x$

from which $a_1=-2*a_0$

put into first $-3*a_0*x=0$ from which $a_0=0$ or $x=0$,of course we can continue up to infinity times,in reality only up to $4$,for fifth polynomial we can't solve,so what should be shortest way to show that polynomial will have at least one zero?

Answer 1

Hint: For $p(x)=a_0 x + \dfrac{a_1}{2}x^2 + \ldots +\dfrac{a_n}{n+1}x^{n+1}$ we have $p(0)=p(1)=0$ and $p'(x)=a_0 + a_1x + \ldots + a_nx^n$.