Prove that $1/2, 1$ and $-2$ are the zeros of the cubic polynomial $p(x) = 2x^3 + x^2 - 5x + 2$. Also,verify the relationship between the zeros and the coefficients.

(A) Given polynomial: $p(x) = 2x^3 + x^2 - 5x + 2$.
Comparing with $ax^3 + bx^2 + cx + d$,we get $a = 2, b = 1, c = -5, d = 2$.
To prove these are zeros,we substitute the values into $p(x)$:
$p(1/2) = 2(1/8) + (1/4) - 5(1/2) + 2 = 1/4 + 1/4 - 5/2 + 2 = 1/2 - 2.5 + 2 = 0$.
$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0$.
$p(-2) = 2(-8) + 4 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0$.
Since $p(1/2) = 0, p(1) = 0, p(-2) = 0$,these are zeros.
Verification:
Sum of zeros: $1/2 + 1 + (-2) = -1/2$. Formula: $-b/a = -1/2$. (Verified)
Sum of product of zeros taken two at a time: $(1/2)(1) + (1)(-2) + (-2)(1/2) = 1/2 - 2 - 1 = -2.5 = -5/2$. Formula: $c/a = -5/2$. (Verified)
Product of zeros: $(1/2)(1)(-2) = -1$. Formula: $-d/a = -2/2 = -1$. (Verified)