so

so

Divide f(x) by x-1: call the Quotient H(x) and the remainder r: then f(x)= (x-1)H(x)+ r. Then f(1)= (1- 1)H(1)+ r= r and we know that f(1)= 0 so r= 0:

f(x)= (x-1)H(x) for some polynomial H(x). By the quotient rule, f'(x)= H(x)+ (x-1)H'(x) so r'(1)= H(1)+ (1-1)H'(1)= H(1). So we must also have H(1)= 0.

Apply the same argument as above to H(x) to show that H(x)= (x-1)g(x). Then .