Question

Ex 2. Va rog mult. E urgent

Answer 1

Am atasat rezolvarea.

Answer 2

a)

[tex]\it f=x^3-4x^2-x+4 =(x^3-x) -4(x^2-1) = \\\;\\ =x(x^2-1)-4(x^2-1) =(x^2-1)(x-4) \ \ \ \ \ (*) \\\;\\ (*) \Longrightarrow f:(x^2-1) =x-4\ rest\ 0.[/tex]

b)

[tex]\it f=x^3-4x^2-x+4 \\\;\\ f(x_1) = 0 \Rightarrow x_1^3-4x_1^2-x_1+4 =0 \Rightarrow x_1^3 = 4x_1^2 + x_1 - 4 \ \ \ \ (1) \\\;\\ f(x_2) = 0 \Rightarrow x_2^3-4x_2^2-x_2+4 =0 \Rightarrow x_2^3 = 4x_2^2 + x_2 - 4 \ \ \ \ (2) [/tex]

$\it f(x_3) = 0 \Rightarrow x_3^3-4x_3^2-x_3+4 =0 \Rightarrow x_3^3 = 4x_3^2 + x_3 - 4 \ \ \ \ (3)$


$\it (1),(2), (3)\Rightarrow x_1^3+x_2^3+x_3^3 =4(x_1^2+x_2^2+x_3^2)+(x_1+x_2+x_3)-12$

c)

[tex]\it f=0 \stackrel{(*)}{\Longrightarrow}(x^2-1)(x-4)=0 \begin{cases}\it x^2-1=0\Rightarrow x^2=1 \Rightarrow x=\pm1 \\\;\\ \it x-4=0 \Rightarrow x=4\end{cases}[/tex]


$\it S=\{-1,\ 1,\ 4\}$