Using rolle’s theorem equation $a_0{x}^n+a_1{x}^{n-1}+.............+a_n=0$  has at least one root between $0$  and $1$ , if :

Consider the function, $\varphi \left(x\right)={\displaystyle \int (a_0{x}^n+a_1{x}^{n–1}+\mathrm{...}+a_n)dx}$

                                                   $=\frac{a_0{x}^{n+1}}{n+1}+\frac{a_1{x}^n}{n}+\mathrm{...}+a_{n}x+A$ …(1)

Where $A$  is constant.

$\varphi \left(x\right)$  being algebraic polynomial is continuous in the closed interval $[0,1]$  and differentiable in the open interval $]0,1[$

If $f\left(0\right)=f\left(1\right)$ , then we have $\frac{a_0}{n+1}+\frac{a_1}{n}+\mathrm{...}+a_n=0$ …(2)

Thus $\varphi \left(x\right)$  is given in (1) by satisfying all conditions of Rolle’s theorem, so there exists atleast one $x=c$  s.t. $0<c<1$ , where $\varphi \text{'}\left(c\right)=0$

i.e., atleast one root of $\varphi \text{'}\left(x\right)=0$

i.e., $a_0{x}^n+a_1{x}^{n-1}+\mathrm{...}+a_n=0$  lie between 0 and 1 if (2) is satisfied