## remainder

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
faisalnir
Posts: 7
Joined: Thu Jan 27, 2011 12:30 am

### remainder

$$F(x)=x^{n}+x^{n-1}+.................+x^2+x+2$$ find the remainder when $$F(x^{n+1})$$ is divided by $$F(x)$$.

Mehfuj Zahir
Posts: 78
Joined: Thu Jan 20, 2011 10:46 am

### Re: remainder

u can try with geometric progression

Cryptic.shohag
Posts: 16
Joined: Fri Dec 17, 2010 11:32 pm
Contact:

### Re: remainder

faisalnir wrote:$$F(x)=x^{n}+x^{n-1}+.................+x^2+x+2$$ find the remainder when $$F(x^{n+1})$$ is divided by $$F(x)$$.
$f(x)=x^n+x^{n-1}+........+x^2+x+2$
$\therefore f(x^{n+1})=(x^{n+1})^n+(x^{n+1})^{n-1}+........+(x^{n+1})^2+x^{n+1}+2$

Now, $f(x)=\, \frac{x(x^n-1)}{x-1}+2$$=\, {P}x(x-1)+2$

And, $f(x^{n+1})=\, \frac{x^{n+1}((x^{n+1})^n-1)}{x^{n+1}-1}+2$$=\, {P}'x^{n+1}(x^{n+1}-1)+2$
If we observe it's clear that {P}' is a multiple of P

Now, $f(x^{n+1})-f(x)=\, {P}'x^{n+1}(x^{n+1}-1)+2-({P}x(x-1)+2)=\, KPx^{n+1}{P}''(x-1)-{P}x(x-1)=Px(x-1)(K{P}''x^{n+1}-1)$

If we divide $Px(x-1) \: by \: f(x), \: the \: remainder\: will \: be \: -2$. So, if we divide $(K{P}''x^{n+1}-1) by\: f(x)$ the remainder will be $-2\times (K{P}''x^{n+1}-1)$. But I can't find the value of $(K{P}''x^{n+1}-1)$. All I can guess is that the remainder could be $-2$.......
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein