Question

Va rog sa ma ajutati cu exercitiul 2 , subpunctul c) .Problema cu polinom .

Answer 1

$\displaystyle Scriem~relatiile~lui~Viete~pentru~polinomul~f=X^3+X-2: \\ \\ \sigma_1=x_1+x_2+x_3=0 \\ \\ \sigma_2=x_1x_2+x_2x_3+x_1x_3=1 \\ \\ \sigma_3=x_1x_2x_3=2.$

$\displaystyle Polinomul~g=aX^3+bX^2+cX+d~are~radacinile~ \frac{1}{x_1},~\frac{1}{x_2}~si~\frac{1}{x_3}, \\ \\ deci~scriind~relatiile~lui~Viete~pentru~acesta,~avem: \\ \\ (1)~~~\frac{1}{x_1}+ \frac{1}{x_2}+ \frac{1}{x_3}=- \frac{b}{a} \\ \\ (2)~~~\frac{1}{x_1} \cdot \frac{1}{x_2}+ \frac{1}{x_2} \cdot \frac{1}{x_3}+ \frac{1}{x_1} \cdot \frac{1}{x_3}= \frac{c}{a} \\ \\ (3)~~~ \frac{1}{x_1} \cdot \frac{1}{x_2} \cdot \frac{1}{x_3}=- \frac{d}{a}.$

$\displaystyle (1) \Leftrightarrow -\frac{b}{a}= \frac{\sigma_2}{\sigma_3}= \frac{1}{2} \\ \\ (2) \Leftrightarrow \frac{c}{a}= \frac{\sigma_1}{\sigma_3}=0 \\ \\ (3) \Leftrightarrow -\frac{d}{a}= \frac{1}{\sigma_3}= \frac{1}{2}. \\ \\ Deci~b=- \frac{a}{2},~c=0,~d= -\frac{a}{2}.~Pentru~ca~a,b,c,d \in \mathbb{Z}~trebuie~ca~a \\ \\ sa~fie~numar~par~nenul.~Fie~a=2k,~k \in \mathbb{Z^*}. \\ \\ Atunci~avem~(a,b,c,d)=(2k,-k,0,-k),~si~deci \\ \\ g=2kX^3-kX^2-k=k(2X^3-X^2-1).$