1)f=[tex] x^{4}-3 x^{3}+2x^2=x^2( x^{2} -3x+2)= x^{2} (x-2)(x-1) [/tex]
2)[tex] x^{3}-3x^2+x+2=0,si, x_{1}=2, [/tex]. Scriem relati 1 si 3 a lui Vieta:
[tex] x_{1} + x_{2} + x_{3} =3,si, x_{1} x_{2} x_{3}=-2, [/tex] inlocuind radacina data =2, obtinem sistemul: [tex] x_{2} + x_{3}=1,si, x_{2} x_{3}=-1, [/tex] care na dau ecuatia: x²-x-1=0, cu radacinile: [tex]
x_{2}= \frac{1- \sqrt{5} }{2},si, x_{3}= \frac{1+ \sqrt{5} }{2} . [/tex]
