(N/A) Let $f(x) = 4x^2 - 3x - 1$.
To find the zeroes,we factorise the polynomial by splitting the middle term:
$f(x) = 4x^2 - 4x + x - 1$
$f(x) = 4x(x - 1) + 1(x - 1)$
$f(x) = (x - 1)(4x + 1)$
The zeroes are found by setting $f(x) = 0$:
$x - 1 = 0 \implies x = 1$
$4x + 1 = 0 \implies x = -\frac{1}{4}$
Thus,the zeroes are $\alpha = 1$ and $\beta = -\frac{1}{4}$.
Verification:
Sum of zeroes $= \alpha + \beta = 1 + (-\frac{1}{4}) = \frac{3}{4} = -\frac{-3}{4} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$.
Product of zeroes $= \alpha \cdot \beta = 1 \cdot (-\frac{1}{4}) = -\frac{1}{4} = \frac{-1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$.
Hence,the relationship is verified.