Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial

p (x) = 2x3

- 3x2

- 3x + 2 and then verify the relationship between the

zeroes and the coefficients.

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