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

Here's my approach:- Letting $P(x)=x^n-nx+n-1$ I tried putting $P(1)=0$ but it would only prove for $,(x-1)$

Here's what I can think of, $x^n-1-nx+n=(x^n-1)-(nx-n)=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}................+x+1)-n(x-1)$

Because $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{n+1}................+xy^{n-2}+y^{n-1})$

$=(x-1)[x^{n-1}+x^{n-2}+x^{n-3}................+x+1-n]$

But I am not able to proceed any further neither I am able to get $(x-1)^2$ as common factor.

Crocogator
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    Putting $P(1) = 0$ got you that $(x-1)$ is a factor; if you now look at $P'(1) = 0$, you'll get that it's a factor of the derivative, which will get you where you need to go. – John Hughes Oct 12 '19 at 03:28
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    Show that $P'(x)$ also has a zero at $x=1$. – Anurag A Oct 12 '19 at 03:28
  • John Hughes it may work but I'm not getting it, Please elaborate a little. Why it being a factor of the derivative prove (x-1)^2 to be a factor of P(x)? – Crocogator Oct 12 '19 at 03:33
  • @UnnayanUpadhyay One can write any polynomial as $f(x)=a+b(x-1)+(x-1)^2g(x)$; now consider what $f(1)$ and $f'(1)$ are. – Angina Seng Oct 12 '19 at 03:39

2 Answers2

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$P(x)=x^n-nx+n-1=(x^n-1)-(nx-n)$

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

$P(1)=((1)-1)((1)^{n-1}+(1)^{n-2}+...+(1)+1-n)=(0)(0)$

Geoffrey
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  • Oh, thanku. That is, we first take (x-1) as common and then put P(1) which shows that (x-1) is again a factor of the polynomial. So in the end we got two (x-2) factors which multiply to become (x-2)^2. Is this is what you have proven? – Crocogator Oct 12 '19 at 03:41
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    @UnnayanUpadhyay Two (x-1) factors, but yeah, you've got the idea. – Geoffrey Oct 12 '19 at 03:43
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Alternatively, experimenting with a few cases suggests the factorization $$x^n-nx+n-1=(x-1)^2(x^{n-2}+2x^{n-3}+3x^{n-4}+\cdots +(n-2)x+(n-1))$$ which is readily verified.

lhf
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