Question

Descompune polinomul in factori ireductibili x^3-8x^2-3x+6

Answer 1

Am atasat rezolvarea corectez 57 nu 67

Answer 2

   
[tex]\displaystyle\\ x^3-8x^2-3x+6=\\ =x^3+x^2-x^2-8x^2-3x+6=\\ =x^3+x^2-9x^2-9x+9x-3x+6=\\ =x^3+x^2-9x^2-9x+6x+6=\\ =(x^3+x^2)-(9x^2+9x)+(6x+6)=\\ =x^2(x+1)-9x(x+1)+6(x+1)=\\ =(x+1)(x^2-9x+6) [/tex]

[tex]\displaystyle\\ \text{Rezolvam ec. de gradul 2:}\\\\ x^2-9x+6=0\\\\ x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{9\pm\sqrt{81-24}}{2}=\frac{9\pm \sqrt{57}}{2}\\\\ x_1=\frac{9-\sqrt{57}}{2}\\\\ x_2=\frac{9+\sqrt{57}}{2}\\\\ \text{Rezulta descompunerea:}\\\\ x^3-8x^2-3x+6=\\\\ =(x+1)(x^2-9x+6)=\boxed{\Big(x+1\Big)\left(x-\frac{9-\sqrt{57}}{2}\right)\left(x-\frac{9+\sqrt{57}}{2}\right)}[/tex]