polynomial

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Abdul Muntakim Rafi
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polynomial

Unread post by Abdul Muntakim Rafi » Sat Jul 14, 2012 10:40 pm

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

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nayel
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Re: polynomial

Unread post by nayel » Mon Jul 16, 2012 10:58 pm

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 $x$.
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Abdul Muntakim Rafi
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Re: polynomial

Unread post by Abdul Muntakim Rafi » Tue Jul 17, 2012 8:53 pm

Bhaiya, $p(x)=85$ for a certain value greater than $7$ (not just any value greater than $7$) ...
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nayel
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Re: polynomial

Unread post by nayel » Tue Jul 17, 2012 10:29 pm

$p(x)=11x$
$p(x)=10x+7$
$p(x)=12x-7$
$\dots$
All these polynomials satisfy your conditions, and have different roots. So I don't really know what you're asking for.
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nafistiham
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Re: polynomial

Unread post by nafistiham » Wed Jul 18, 2012 12:07 am

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 ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Abdul Muntakim Rafi
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Re: polynomial

Unread post by Abdul Muntakim Rafi » Thu Jul 19, 2012 1:09 am

The ques was like this-
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 $p(x)=85$...
Now what's his age?
Man himself is the master of his fate...

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