Question

pt f =$x^{3} +4x^{2} -20x -48$
Rezolvati ecuati f(x)=0

Answer 1

$\it f(x)=x^3+4x^2-20x-48 =\ \underline{x^3+6x^2} - \underline{\underline{2x^2-12x}} -8x-48=\\ \\ =x^2(x+6)-2x(x+6)-8(x+6)=(x+6)(x^2-2x-8)=\\ \\ =(x+6)(x^2-2x+1-9)=(x+6)[(x-1)^2-3^2]=(x+6)(x-1+3)(x-1-3)=\\ \\ =(x+6)(x+2)(x-4)$

$\it f(x)=0\ \Rightarrow \begin{cases}\it x+6=0 \Rightarrow x_1=-6\\ \\ \it x+2=0 \Rightarrow x_2=-2\\ \\ \it x-4=0\ \Rightarrow\ \ x_3=4\end{cases}$