Question

Sub ll,2.b,c acelea cu determinati nr real a si cu radacinile

Answer 1

ca sa fie divizibil restul impartirii trebuie sa fie 0
  x³-x²+ax+2÷x²-2x+2=x+1
 -x³+2x²-2x
 --------------              
       x²+(a-2)x+2
      -x²+2x      -2
----------------------
               ax=0, a=0

relatiile lui viete
x₁+x₂+x₃=1
x₁x₂+x₂x₃+x₁x3=a
x₁x₂x₃=-2

x₁³-x₁²+ax₁+2=0
x₂³-x₂²+ax₂+2=0
x₃³-x₃²+ax₃+2=0
---------------------------------
(x₁³+x₂³+x₃³)-(x₁²+x₂²+x₃²)+a(x₁+x₂+x₃)+6=0
x₁²+x₂²+x₃²=(x₁+x₂+x₃)²-2(x₁x₂+x₂x₃+x₁x₃)=1-2a
x₁³+x₂³+x₃³=1-2a-a-6=-3a-5
x₁³+x₂³+x₃³+3x₁x₂+3x₂x₃+3x₁x₃=-3a-5+3(x₁x₂+x₂x₃+x₁x₃)=-3a-5+3a=-5