Find the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial: $4x^{2} + 5\sqrt{2}x - 3$

(N/A) Let $f(x) = 4x^{2} + 5\sqrt{2}x - 3$.
To factorise,we split the middle term: $4x^{2} + 6\sqrt{2}x - \sqrt{2}x - 3$.
$= 2\sqrt{2}x(\sqrt{2}x + 3) - 1(\sqrt{2}x + 3)$.
$= (\sqrt{2}x + 3)(2\sqrt{2}x - 1)$.
Setting $f(x) = 0$,we get $\sqrt{2}x + 3 = 0$ or $2\sqrt{2}x - 1 = 0$.
Thus,the zeroes are $x = -\frac{3}{\sqrt{2}}$ and $x = \frac{1}{2\sqrt{2}}$.
Verification:
Sum of zeroes $= -\frac{3}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{-6 + 1}{2\sqrt{2}} = -\frac{5}{2\sqrt{2}} = -\frac{5\sqrt{2}}{4}$.
Coefficient of $x$ is $5\sqrt{2}$ and coefficient of $x^{2}$ is $4$. So,$-\frac{\text{Coefficient of } x}{\text{Coefficient of } x^{2}} = -\frac{5\sqrt{2}}{4}$. (Verified)
Product of zeroes $= (-\frac{3}{\sqrt{2}}) \times (\frac{1}{2\sqrt{2}}) = -\frac{3}{4}$.
Constant term is $-3$ and coefficient of $x^{2}$ is $4$. So,$\frac{\text{Constant term}}{\text{Coefficient of } x^{2}} = -\frac{3}{4}$. (Verified)