Aplici teorema lui Bezout,Adica f :(x-(-2))=Q(x)rest12
(x³-x+a):(x+2)=(x²-2x+3)rest a-6
a-6=-12 a= -6
f(x)=x³-x-6
f(x)=0
Se observa x=2 solutie
Imparti polinomulf prin x-2
(x³-x-6):(x-2)=x²+2x+3
x²+2x+3=0
Δ= -8 Ecuatia are 2 solutii complexe
x1= -1-√2i
x2=-1+√2i
S={2,-1-√2i,-1+√2i}
