Решите неравенство (x-2)(3x^2-5x-2)(x+4)≥0.

\[(x - 2)\left( 3x^{2} - 5x - 2 \right)(x + 4) \geq 0\]

\[3x^{2} - 5x - 2 = 3 \cdot \left( x + \frac{1}{3} \right)(x - 2)\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 + 7}{6} = 2;\ \ x_{2} = \frac{5 - 7}{6} = - \frac{1}{3}\]

\[3(x + 4)\left( x + \frac{1}{3} \right)(x - 2)(x - 2) \geq 0\]

\[Ответ:x \in ( - \infty; - 4\rbrack \cup \left\lbrack - \frac{1}{3}; + \infty \right).\]