Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also,verify the relationship between the zeroes and the coefficients in each case: $2x^3 + x^2 - 5x + 2; \frac{1}{2}, 1, -2$.

(A) Let $p(x) = 2x^3 + x^2 - 5x + 2$.
The given values are $\alpha = \frac{1}{2}, \beta = 1, \gamma = -2$.
First,we verify that these are zeroes:
$p(\frac{1}{2}) = 2(\frac{1}{8}) + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = -2 + 2 = 0$.
$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0$.
$p(-2) = 2(-8) + 4 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0$.
Since $p(\frac{1}{2}) = 0, p(1) = 0,$ and $p(-2) = 0$,these are indeed the zeroes.
Comparing $p(x)$ with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = 1, c = -5, d = 2$.
Verification of relationships:
$1$. Sum of zeroes: $\alpha + \beta + \gamma = \frac{1}{2} + 1 - 2 = -\frac{1}{2} = \frac{-b}{a}$.
$2$. Sum of product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (\frac{1}{2})(1) + (1)(-2) + (-2)(\frac{1}{2}) = \frac{1}{2} - 2 - 1 = \frac{1-6}{2} = -\frac{5}{2} = \frac{c}{a}$.
$3$. Product of zeroes: $\alpha\beta\gamma = (\frac{1}{2})(1)(-2) = -1 = \frac{-d}{a} = \frac{-2}{2} = -1$.
Thus,the relationship is verified.