[tex]\displaystyle\\
1)\\
\text{Daca }~x+1~\text{este divizor al polinomului }f(x),~\text{atunci:}\\
x=-1~ \text{ este solutie a ecuatiei }~f(x)=0\\
f(-1) = (-1)^3-2\cdot (-1)^2-2\cdot (-1)+1 = -1 - 2 + 2 + 1 = 0\\
\Longrightarrow~~\boxed{f(x) ~\vdots~ (x+1)}
[/tex]
[tex]\displaystyle\\
2)\\
\text{Forma generala a polinomului de gradul 3 este:}\\\\
f = ax^3 + bx^2+cx+d\\\\
\text{Vom folosi relatiile lui Viete pentru polinomul de gradul 3.}\\
\text{Acestea sunt:}\\
\begin {cases}
x_1+x_2+x_3= \dfrac{-b}{a}\\
x_1x_2+ x_1x_3+ x_2x_3= \dfrac{c}{a}\\
x_1x_2x_3 = \dfrac{-d}{a}
\end{cases}
[/tex]
[tex]\displaystyle\\
\text{Rezolvare: }\\\\
\frac{1}{x_1 x_2} + \frac{1}{x_2 x_3} + \frac{1}{x_3 x_1} = a(x_1 x_2+x_2 x_3 + x_3 x_1)\\\\
\frac{x_3}{x_1 x_2x_3} + \frac{x_1}{x_1x_2 x_3} + \frac{x_2}{x_1x_2x_3} = a(x_1 x_2+x_2 x_3 + x_3 x_1)\\\\
\frac{x_1 +x_2+x_3}{x_1 x_2x_3} = a(x_1 x_2+x_2 x_3 + x_3 x_1)\\\\
\frac{ \dfrac{-b}{a}}{\dfrac{-d}{a}} = a \cdot \dfrac{c}{a}\\\\
\text{A nu se confunda "a" din ecuatie cu "a" din relatiile lui Viete.}\\\\
\frac{ \dfrac{2}{1}}{\dfrac{-1}{1}} = a \cdot \dfrac{-2}{1}
[/tex]
[tex]\displaystyle\\
\frac{ 2}{-1} = a \cdot (-2)\\\\
-2 = -2a\\\\
a = \frac{-2}{-2}\\\\
\boxed{\bf a = 1}
[/tex]
