Page 1 of 1
Bdmo Higher Secondary P3
Posted: Sat Mar 27, 2021 3:16 pm
by Asif Hossain
$P(x)=x^3 +px^2 +qx+r$ 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 $|p|$
Re: Bdmo Higher Secondary P3
Posted: Sat Mar 27, 2021 9:53 pm
by Asif Hossain
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 $|p|$
is this is copied from aops? or somebody posted there...
Re: Bdmo Higher Secondary P3
Posted: Sat Mar 27, 2021 10:08 pm
by Asif Hossain
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 $max|p|=17$
Re: Bdmo Higher Secondary P3
Posted: Mon Mar 29, 2021 2:01 am
by Pro_GRMR
Asif Hossain wrote: ↑Sat Mar 27, 2021 10:08 pm
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 $max|p|=17$
Would you care to explain why $|A+B+C|=|p|$?
Re: Bdmo Higher Secondary P3
Posted: Mon Mar 29, 2021 9:59 pm
by Asif Hossain
Pro_GRMR wrote: ↑Mon Mar 29, 2021 2:01 am
Asif Hossain wrote: ↑Sat Mar 27, 2021 10:08 pm
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 $max|p|=17$
Would you care to explain why $|A+B+C|=|p|$?
Vieta relation..