The zeros of the polynomial $\mathrm{x}^{2}+\frac{1}{6} \mathrm{x}-2$ are (a) $-3,4$ (b) $\frac{-3}{2}, \frac{4}{3}$ (c) $\frac{-4}{3}, \frac{3}{2}$ (d) none of these

The correct option is option (b) $\frac{-3}{2}, \frac{4}{3}$

$f(x)=x^{2}+\frac{1}{6} x-2=0$

$\Rightarrow 6 \mathrm{x}^{2}+\mathrm{x}-12=0$

$\Rightarrow 6 x^{2}+9 x-8 x-12=0$

$\Rightarrow 3 x(2 x+3)-4(2 x+3)=0$

$\Rightarrow(2 x+3)(3 x-4)=0$

$\therefore \mathrm{x}=\frac{-3}{2}$ or $\mathrm{x}=\frac{4}{3}$