Let $\displaystyle Let f(x) = x^n - nx + n - 1 $ with integer n > 1. Considering f(x) and f'(x), show that $\displaystyle f(x) = (x - 1)^2 g(x) $ where g(x) is a polynomial with integer coefficient.
Let $\displaystyle Let f(x) = x^n - nx + n - 1 $ with integer n > 1. Considering f(x) and f'(x), show that $\displaystyle f(x) = (x - 1)^2 g(x) $ where g(x) is a polynomial with integer coefficient.
$\displaystyle f(x)= x^n- nx+ n- 1$ so $\displaystyle f(1)= 1- n+ n- 1= 0$
$\displaystyle f'(x)= nx^{n-1}- n$ so $\displaystyle f'(1)= n- n= 0$
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 $\displaystyle f(x)= (x-1)H(x)= (x-1)^2g(x)$.