(N/A) Let $f(x) = 3x^2 + 4x - 4$.
To find the zeroes,we factorize the polynomial by splitting the middle term:
$3x^2 + 6x - 2x - 4 = 3x(x + 2) - 2(x + 2) = (x + 2)(3x - 2)$.
The zeroes are found by setting $f(x) = 0$:
$(x + 2)(3x - 2) = 0$,which gives $x = -2$ or $x = \frac{2}{3}$.
Verification:
Sum of zeroes $= -2 + \frac{2}{3} = \frac{-6 + 2}{3} = -\frac{4}{3}$.
From the polynomial $ax^2 + bx + c$,where $a=3, b=4, c=-4$,the sum of zeroes is $-\frac{b}{a} = -\frac{4}{3}$.
Product of zeroes $= (-2) \times \frac{2}{3} = -\frac{4}{3}$.
From the polynomial,the product of zeroes is $\frac{c}{a} = \frac{-4}{3}$.
Since the sum and product match the coefficients,the relationship is verified.