Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$

Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$

By factor theorem we know that $(x-a)$ is a factor of $f(x)$ if $f(a)=0$.

In this case, $f(x)=x^n -nx +n-1 \implies f(1)=0$

Hence we conclude that $(x-1)$ is a factor. From hereon, how can I say that $(x-1)^2$ is a factor?

Can we approach the problem without calculus approach? This problem was taken from a book of pre-calculus algebra.

Here is another elementary way using the binomial theorem

• $(1+y)^n = \sum_{k=0}^n\binom{n}{k}y^k$

Set $\boxed{y=x-1}$ and note that

• $(x-1)^2$ is a factor of $p(x) = x^n -nx +n-1$ if and only if $y^2$ is a factor of $p(y+1)$

Hence, \begin{eqnarray*} p(y+1) & = & (1+y)^n - (1+y)n + n-1\\ & = & 1+ny +\sum_{k=2}^n\binom{n}{k}y^k -n-ny+n-1 \\ & = & y^2\sum_{k=2}^n\binom{n}{k}y^{k-2} \end{eqnarray*} Done.

Hint

$$x^n -nx +n-1=(x^n-1) -n(x -1)=(x-1)(x^{n-1}+x^{n-2}+...+x+1-n)$$

now, write

$$p(x)=x^{n-1}+x^{n-2}+...+x+1-n$$

and check that $p(1)=0$.