I am trying to show that 1 is a zero of $P(x)=x^n - (n+1)x + n$ by dividing it through by $x-1$.
Now, It's definitely a zero because $P(1) = 0$, but from the long division I'm getting P(1) = -1
$ \hspace{3em} x^n + x^{n-1} + x^{n-2} + ... - (n+1) $ <- should be $-n$
$x-1{)}\hspace{-0.35em}\overline{\hspace{0.5em} x^{n+1} + 0x^n + 0x^{n-1} + ...-(n+1)x +n \hspace{7em}}$
$ \hspace{2.5em} x^{n+1} - x^n $
$ \hspace{5em}\overline{ x^n + 0x^{n-1} + ... -(n+1)x +n } $
$ \hspace{5em}{\underline{x^n - x^{n-1} \hspace{7em}}} $
$ \hspace{7em}{x^{n-1}+ ... -(n+1)x +n } $
$ \hspace{12em}{...} $
$ \hspace{12em}{{-(n+1)x +n \hspace{4em}}} $
$ \hspace{8em}{\underline{\hspace{4em}{-(n+1)x +n+1}}} $
$ \hspace{18em} -1 $
And the quotient is definitely correct because I have a proof by induction for $x^n - (n+1)x + n =(x-1)(x^n + x^{n-1} + ... + x -n)$, but just can't seem to confirm it through long division.
Can anyone help me out?
TIA
The last 5 rows of division should be \begin{array}{rrrrr} x^2 & + & -(n+1)x & + & n \\ x^2 &+& -x && \\ \hline &&-nx&+& n \\ &&-nx & + & n \\ \hline &&&&0. \end{array}