Bdmo Higher Secondary P3
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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$P(x)=x^3 +px^2 +qx+r$ is a polynomial in $x$. Given $P(1)<0$ ,$P(4)>0$, $P(6)<0$ and $P(10)>0$. all the roots of $P(x)$ are integers. Find the max value of $|p|$
Hmm..Hammer...Treat everything as nail
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Re: Bdmo Higher Secondary P3
is this is copied from aops? or somebody posted there...Asif Hossain wrote: ↑Sat Mar 27, 2021 3:16 pm$P(x)=x^3 +px^2 +qx+r$ is a polynomial in $x$. Given $P(1)<0$ ,$P(4)>0$, $P(6)<0$ and $P(10)>0$. all the roots of $P(x)$ are integers. Find the max value of $|p|$
Hmm..Hammer...Treat everything as nail
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Re: Bdmo Higher Secondary P3
OK copied from AOPS
Note that the roots of this polynomial are integers and if $A,B,C$ are the roots of this polynomial then $|A+B+C|=|p|$ . so using the condition $P(4)>0,P(6)<0$ the only root between them is $5$ so $5$ is a root
$P(1)<0,P(4)>0$ since we want to maximize $p$ one of the root is $3$
similarly for the maximality $P(6)<0,P(10)>0$ implies one of the root is $9$
SO the $max|p|=17$
Note that the roots of this polynomial are integers and if $A,B,C$ are the roots of this polynomial then $|A+B+C|=|p|$ . so using the condition $P(4)>0,P(6)<0$ the only root between them is $5$ so $5$ is a root
$P(1)<0,P(4)>0$ since we want to maximize $p$ one of the root is $3$
similarly for the maximality $P(6)<0,P(10)>0$ implies one of the root is $9$
SO the $max|p|=17$
Hmm..Hammer...Treat everything as nail
Re: Bdmo Higher Secondary P3
Would you care to explain why $|A+B+C|=|p|$?Asif Hossain wrote: ↑Sat Mar 27, 2021 10:08 pmOK copied from AOPS
Note that the roots of this polynomial are integers and if $A,B,C$ are the roots of this polynomial then $|A+B+C|=|p|$ . so using the condition $P(4)>0,P(6)<0$ the only root between them is $5$ so $5$ is a root
$P(1)<0,P(4)>0$ since we want to maximize $p$ one of the root is $3$
similarly for the maximality $P(6)<0,P(10)>0$ implies one of the root is $9$
SO the $max|p|=17$
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- Joined:Sat Jan 02, 2021 9:28 pm
Re: Bdmo Higher Secondary P3
Vieta relation..Pro_GRMR wrote: ↑Mon Mar 29, 2021 2:01 amWould you care to explain why $|A+B+C|=|p|$?Asif Hossain wrote: ↑Sat Mar 27, 2021 10:08 pmOK copied from AOPS
Note that the roots of this polynomial are integers and if $A,B,C$ are the roots of this polynomial then $|A+B+C|=|p|$ . so using the condition $P(4)>0,P(6)<0$ the only root between them is $5$ so $5$ is a root
$P(1)<0,P(4)>0$ since we want to maximize $p$ one of the root is $3$
similarly for the maximality $P(6)<0,P(10)>0$ implies one of the root is $9$
SO the $max|p|=17$
Hmm..Hammer...Treat everything as nail