Prove that $-2$,$4$,and $\frac{1}{2}$ are the zeros of the cubic polynomial $p(x) = 2x^3 - 5x^2 - 14x + 8$. Also,verify the relationship between the zeros and the coefficients.

(A) Given polynomial: $p(x) = 2x^3 - 5x^2 - 14x + 8$.
Comparing with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = -5, c = -14, d = 8$.
Step $1$: Check zeros.
$p(-2) = 2(-2)^3 - 5(-2)^2 - 14(-2) + 8 = 2(-8) - 5(4) + 28 + 8 = -16 - 20 + 28 + 8 = 0$.
$p(4) = 2(4)^3 - 5(4)^2 - 14(4) + 8 = 2(64) - 5(16) - 56 + 8 = 128 - 80 - 56 + 8 = 0$.
$p(1/2) = 2(1/8) - 5(1/4) - 14(1/2) + 8 = 1/4 - 5/4 - 7 + 8 = -4/4 + 1 = -1 + 1 = 0$.
Thus,$-2, 4, 1/2$ are zeros.
Step $2$: Verify relationships.
Sum of zeros: $\alpha + \beta + \gamma = -2 + 4 + 0.5 = 2.5 = 5/2 = -b/a$.
Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (-2)(4) + (4)(0.5) + (0.5)(-2) = -8 + 2 - 1 = -7 = -14/2 = c/a$.
Product of zeros: $\alpha\beta\gamma = (-2)(4)(0.5) = -4 = -8/2 = -d/a$.
All relationships are verified.