Решите уравнение: 16x^3-32x^2-x+2=0.

\[16x^{3} - 32x^{2} - x + 2 = 0\ \]

\[16x^{2}(x - 2) - (x - 2) = 0\]

\[(x - 2)\left( 16x^{2} - 1 \right) = 0\]

\[(x - 2)(4x - 1)(4x + 1) = 0\]

\[x = 2;\ \ \ x = \frac{1}{4};\ \ \ x = - \frac{1}{4}.\]

\[Ответ:\ \ x = \pm 0,25;\ \ x = 2.\]