Answer:
i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.
ii) - the sum of the zeros and the corresponding coefficients are the same
-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.
-the product of the zeros of the polynomial is same as the corresponding coefficient
Step-by-step explanation:
We are given the cubic polynomial;
p(x) = 2x³ - 3x² - 3x + 2
For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.
Thus;
P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0
P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0
P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0
Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.
Now, let's verify the relationship between the zeros and the coefficients.
Let the zeros be as follows;
α = 2
β = -1
γ = ½
The coefficients are;
a = 2
b = -3
c = -3
d = 2
So, the relationships are;
α + β + γ = -b/a
αβ + βγ + γα = c/a
αβγ = -d/a
Thus,
First relationship α + β + γ = -b/a gives;
2 - 1 + ½ = -(-3/2)
1½ = 3/2
3/2 = 3/2
LHS = RHS; So, the sum of the zeros and the coefficients are the same
For the second relationship, αβ + βγ + γα = c/a it gives;
2(-1) + (-1)(½) + (½)(2) = -3/2
-2 - 1½ + 1 = -3/2
-1½ - 1½ = -3/2
-3/2 = - 3/2
LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient
For the third relationship, αβγ = -d/a gives;
2 * -1 * ½ = -2/2
-1 = - 1
LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient
and 
. Combining both the ranges we get the solution: 
.