Find the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial:
$2x^2 + \frac{7}{2}x + \frac{3}{4}$

(N/A) Let $f(x) = 2x^2 + \frac{7}{2}x + \frac{3}{4}$.
To find the zeroes,we set $f(x) = 0$,so $2x^2 + \frac{7}{2}x + \frac{3}{4} = 0$.
Multiplying the entire equation by $4$,we get $8x^2 + 14x + 3 = 0$.
Splitting the middle term: $8x^2 + 12x + 2x + 3 = 0$.
$4x(2x + 3) + 1(2x + 3) = 0$.
$(2x + 3)(4x + 1) = 0$.
Thus,the zeroes are $x = -\frac{3}{2}$ and $x = -\frac{1}{4}$.
Verification:
Sum of zeroes $= -\frac{3}{2} + (-\frac{1}{4}) = -\frac{6}{4} - \frac{1}{4} = -\frac{7}{4}$.
Coefficient relation: $-\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{7/2}{2} = -\frac{7}{4}$.
Product of zeroes $= (-\frac{3}{2}) \times (-\frac{1}{4}) = \frac{3}{8}$.
Coefficient relation: $\frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{3/4}{2} = \frac{3}{8}$.
Since both relations hold,the relationship is verified.