Решите уравнение x^3-3x^2-4x+12=0.

\[x^{3} - 3x^{2} - 4x + 12 = 0\]

\[x^{2}(x - 3) - 4 \cdot (x - 3) = 0\]

\[(x - 3)\left( x^{2} - 4 \right) = 0\]

\[(x - 3)(x - 2)(x + 2) = 0\]

\[x = 3;\ \ \ \ x = 2;\ \ \ x = - 2\]

\[Ответ:x = \pm 2;\ \ x = 3.\]