Find the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial:
$y^{2}+\frac{3}{2} \sqrt{5} y-5$

(A) Let $f(y) = y^{2} + \frac{3}{2} \sqrt{5} y - 5$.
To find the zeroes,we set $f(y) = 0$,which implies $y^{2} + \frac{3}{2} \sqrt{5} y - 5 = 0$.
Multiplying by $2$,we get $2y^{2} + 3\sqrt{5}y - 10 = 0$.
Splitting the middle term: $2y^{2} + 4\sqrt{5}y - \sqrt{5}y - 10 = 0$.
$2y(y + 2\sqrt{5}) - \sqrt{5}(y + 2\sqrt{5}) = 0$.
$(y + 2\sqrt{5})(2y - \sqrt{5}) = 0$.
Thus,the zeroes are $y = -2\sqrt{5}$ and $y = \frac{\sqrt{5}}{2}$.
Verification:
Sum of zeroes $= -2\sqrt{5} + \frac{\sqrt{5}}{2} = \frac{-4\sqrt{5} + \sqrt{5}}{2} = -\frac{3\sqrt{5}}{2}$.
From the polynomial $y^{2} + \frac{3}{2} \sqrt{5} y - 5$,the coefficient of $y$ is $\frac{3\sqrt{5}}{2}$ and the coefficient of $y^{2}$ is $1$. Thus,$-\frac{\text{coefficient of } y}{\text{coefficient of } y^{2}} = -\frac{3\sqrt{5}}{2}$.
Product of zeroes $= (-2\sqrt{5}) \times (\frac{\sqrt{5}}{2}) = -5$.
The constant term is $-5$ and the coefficient of $y^{2}$ is $1$. Thus,$\frac{\text{constant term}}{\text{coefficient of } y^{2}} = -5$.
Hence,the relationship is verified.