Metoda I: calcul direct
[tex] x^{2} +3x+2= x^{2} +x+2x+2=x(x+1)+2(x+1)=(x+1)(x+2)[/tex]⇒
[tex] x^{2} +3x+2=(x+1)(x+2)[/tex]
Metoda II: scriem trinomul de gradul al II-lea sub forma factorizata.
Pentru aceasta consideram ecuatia atasata:
[tex] x^{2} +3x+2=0[/tex]
Δ=[tex] 3^{2} -4*1*2=9-8=1[/tex]⇒Δ>0⇒[tex] x_{1,2} [/tex]∈R
[tex] x_{1} = \frac{-3+1}{2} = \frac{-2}{2} [/tex]⇒[tex] x_{1} =-1[/tex]
[tex] x_{2} = \frac{-3-1}{2} = \frac{-4}{2} [/tex]⇒[tex] x_{2} =-2[/tex]
Forma factorizata a trinomului: [tex]ax^{2} +bx+c, a \neq 0[/tex] este:
[tex]ax^{2} +bx+c=a(x- x_{1})(x- x_{2}) [/tex]
In cazul de fata avem ca: a=1, b=3, c=2 astfel obtinem:
[tex] x^{2} +3x+2=1*(x-(-1))(x-(-2))=(x+1)(x+2)[/tex]⇒
[tex] x^{2} +3x+2=(x+1)(x+2)[/tex]
