Find the zeroes of the quadratic polynomial $3x^{2}-x-4$ and verify the relationship between the zeroes and the coefficients.

(N/A) Given polynomial: $p(x) = 3x^{2}-x-4$.
To find the zeroes,we set $p(x) = 0$:
$3x^{2}-x-4 = 0$
$3x^{2}-4x+3x-4 = 0$
$x(3x-4)+1(3x-4) = 0$
$(3x-4)(x+1) = 0$
Thus,the zeroes are $x = \frac{4}{3}$ and $x = -1$.
Verification:
Sum of zeroes $= \frac{4}{3} + (-1) = \frac{4-3}{3} = \frac{1}{3}$.
From the polynomial,$-\frac{\text{coefficient of } x}{\text{coefficient of } x^{2}} = -\frac{-1}{3} = \frac{1}{3}$.
Since $\frac{1}{3} = \frac{1}{3}$,the sum is verified.
Product of zeroes $= \frac{4}{3} \times (-1) = -\frac{4}{3}$.
From the polynomial,$\frac{\text{constant term}}{\text{coefficient of } x^{2}} = \frac{-4}{3}$.
Since $-\frac{4}{3} = -\frac{4}{3}$,the product is verified.