Hi,
Solving:
[tex]x^{3} + 3x^{2} - 4x - 12 = 0 \\ {x}^{2} \times (x + 3) - 4x - 12 = 0 \\ {x}^{2} \times (x + 3) - 4(x - 3) = 0 \\ (x + 3) \times ( {x}^{2} - 4) = 0 \\ x + 3 = 0 \\ {x}^{2} - 4 = 0 \\ \\ (results) \\ x1 = - 3 \\ x2 = - 2 \\ x3 = 2[/tex]
Hope this helps.
r3t40
