Using rolle’s theorem equation a0xn+a1xn−1+.............+an=0  has at least one root between 0  and 1 , if :

Consider the function, ϕ(x)=∫(a0xn+a1xn–1+...+an)dx                                                    =a0xn+1n+1+a1xnn+...+anx+A …(1)Where A  is constant.ϕ(x)  being algebraic polynomial is continuous in the closed interval [0,1]  and differentiable in the open interval ]0,1[ If f(0)=f(1) , then we have a0n+1+a1n+...+an=0 …(2)Thus ϕ(x)  is given in (1) by satisfying all conditions of Rolle’s theorem, so there exists atleast one x=c  s.t. 0