147. A problem with two polynominals.

by Virgil Nicula, Oct 6, 2010, 11:19 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=369458

Consider the polynominal $\boxed{p=X^4+X^3-1}$ with the roots $\{a,b,c,d\}$ . Ascertain the monic polynominal with the roots $\{ab, ac,ad,bc,bd,cd\}$ .

Proof 1. Denote $\left\{\begin{array}{c} ab+cd=m\\\ ac+bd=n\\\ ad+bc=p\end{array}\right\|$ . Observe that $\sum a^2=1$ and $\boxed{m+n+p=0}$ . Prove easily that $mn+np+pm=$ $\sum abc(a+b+c)=$

$\sum abc(-1-d)=$ $-\sum abc-4abcd$ $\implies$ $\boxed{mn+np+pm}=4$ . Remain to ascertain $mnp$ . Therefore, $mnp=\sum a^2b^2c^2+abcd\cdot\sum a^2=$

$\sum\frac {1}{a^2}-\sum a^2$ . The polynominal which has the roots $\left\{\frac 1a,\frac 1b,\frac 1c,\frac 1d\right\}$ is $X^4-X-1$ from where obtain $\sum\frac {1}{a^2}=0$ . Thus, $\boxed{mnp=-1}$ . Therefore,

the required polynominal is $q=\prod \left[\left(X^2-1\right)-mX\right]=$ $\left(X^2-1\right)^3-X\left(X^2-1\right)^2\cdot\sum m+X^2\left(X^2-1\right)\cdot\sum mn-X^3\cdot mnp=$

$\left(X^2-1\right)^3+4X^2\left(X^2-1\right)+X^3$ . In conclusion, $\boxed{q=X^6+X^4+X^3-X^2-1}$ .
This post has been edited 16 times. Last edited by Virgil Nicula, Nov 23, 2015, 6:13 AM

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