## polynomial

For discussing Olympiad Level Algebra (and Inequality) problems
Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm

### polynomial

$p(7)=77,p(x)=85(x$ is greater than $7)$; what's the root of $p(n)$?

nayel
Posts: 268
Joined: Tue Dec 07, 2010 7:38 pm
Location: Dhaka, Bangladesh or Cambridge, UK

### Re: polynomial

Let $q(x)=p(x)-85$. Then $q(x)=0$ for all $x>7$, i.e. $q$ has infinitely many roots. So $q$ must be the zero polynomial, which implies $p(x)=85$ for all $x$.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm

### Re: polynomial

Bhaiya, $p(x)=85$ for a certain value greater than $7$ (not just any value greater than $7$) ...
Man himself is the master of his fate...

nayel
Posts: 268
Joined: Tue Dec 07, 2010 7:38 pm
Location: Dhaka, Bangladesh or Cambridge, UK

### Re: polynomial

$p(x)=11x$
$p(x)=10x+7$
$p(x)=12x-7$
$\dots$
All these polynomials satisfy your conditions, and have different roots. So I don't really know what you're asking for.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
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### Re: polynomial

Are you wishing to get such a polynomial, that satisfies $77$ for $7$, and $85$ for a definite integer greater than $7$ ?

Won't there be various such polynomials ?
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm
Two guys were talking... One said hey my age is the root of a polynomial.. Seeing the polynomial equation $P(n)$ the other said your age is $7$... But he figured it was wrong cause $p(7)=77$ ... The first guy said yes u r wrong... my age is greater than $7$... Then he guessed a number greater than $7$ ... That gave him $p(x)=85$...