Find the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial:
$7 y^{2}-\frac{11}{3} y-\frac{2}{3}$

(A) Let $f(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3}$.
To find the zeroes,set $f(y) = 0$,which implies $\frac{1}{3}(21y^2 - 11y - 2) = 0$,so $21y^2 - 11y - 2 = 0$.
Splitting the middle term: $21y^2 - 14y + 3y - 2 = 0$.
$7y(3y - 2) + 1(3y - 2) = 0$.
$(3y - 2)(7y + 1) = 0$.
Thus,the zeroes are $y = \frac{2}{3}$ and $y = -\frac{1}{7}$.
Verification:
Sum of zeroes = $\frac{2}{3} + (-\frac{1}{7}) = \frac{14 - 3}{21} = \frac{11}{21}$.
From the polynomial $7y^2 - \frac{11}{3}y - \frac{2}{3}$,the coefficient of $y$ is $-\frac{11}{3}$ and the coefficient of $y^2$ is $7$.
Sum of zeroes = $-\frac{\text{coefficient of } y}{\text{coefficient of } y^2} = -\frac{-11/3}{7} = \frac{11}{21}$.
Product of zeroes = $(\frac{2}{3})(-\frac{1}{7}) = -\frac{2}{21}$.
Constant term is $-\frac{2}{3}$,coefficient of $y^2$ is $7$.
Product of zeroes = $\frac{\text{constant term}}{\text{coefficient of } y^2} = \frac{-2/3}{7} = -\frac{2}{21}$.
Hence,the relationship is verified.