Find the zeroes of the polynomial $x^{2}+\frac{1}{6} x-2,$ and verify the relation between the coefficients and the zeroes of the polynomial.

(A) To find the zeroes,set the polynomial to zero: $x^{2}+\frac{1}{6} x-2 = 0$.
Multiply by $6$ to simplify: $6x^{2}+x-12 = 0$.
Factorizing the quadratic: $6x^{2}+9x-8x-12 = 0$.
$3x(2x+3)-4(2x+3) = 0 \implies (3x-4)(2x+3) = 0$.
Thus,the zeroes are $\alpha = \frac{4}{3}$ and $\beta = -\frac{3}{2}$.
Verification:
Sum of zeroes: $\alpha + \beta = \frac{4}{3} - \frac{3}{2} = \frac{8-9}{6} = -\frac{1}{6}$.
From the polynomial $ax^2+bx+c$,sum $= -\frac{b}{a} = -\frac{1/6}{1} = -\frac{1}{6}$.
Product of zeroes: $\alpha \beta = \frac{4}{3} \times (-\frac{3}{2}) = -2$.
From the polynomial,product $= \frac{c}{a} = \frac{-2}{1} = -2$.
Both relations are verified.