Question

Verify that 1,-1 and 3 are the zeroes of the cubic polynomial x3 + 3x2-x-3 and
check the relationship between zeroes and the coefficients​

Answer 1

1, -1 and -3 are zeroes of the given cubic polynomial....

3 is not the zero of given polynomial

The relationship between zeroes is add 2.

The relationship between coefficient is (x²-1)(x+3).

Answer 2

Correct Question: Verify that 1, -1, and -3 are the zeroes of the cubic polynomial  x³ + 3x² - x - 3 and check the relationship between zeroes and the coefficients.

Given:

Cubic polynomial f(x) = x³ + 3x² - x - 3

To Verify:

1, -1, and -3 are the zeroes of the Cubic polynomial x³ + 3x² - x - 3.

The relationship between the zeroes and coefficients of the polynomial.

Solution:

Cubic Polynomial: f(x) = x³ + 3x² - x - 3

If 1, -1, and -3 are the zeroes of the cubic polynomial then f(1), f(-1) and f(-3) must be equal to zero:

f(1) = (1³) + 3 (1)² - (1) - 3 = 1 + 3 - 1 - 3 = 0

f(-1) = (-1)³ + 3 (-1)² - (-1) - 3 = -1 + 3 + 1 - 3 = 0

f(-3) = (-3)³ + 3 (-3)² - (-3) - 3 = -27 + 27 + 3 - 3 = 0

As f(1), f(-1) and f(-3) are equal to zero, therefore 1, -1 and -3 are the zeroes of the cubic polynomial f(x) = x³ + 3x² - x - 3.

→ α = 1, β = −1 and γ = -3 are the zeroes of the polynomial: x³ + 3x² - x - 3.

→ The given polynomial is of the form: ax³ + bx² + cx + d

• where, a = 1 , b = 3 , c = -1 and d = -3.

→ The zeroes of this polynomial are: α = 1, β = −1 and γ = -3

           (1.) α + β + γ = 1 + (-1) + (-3) = -3 = (-b/a)

           (2.) αβ + βγ + γα = (1)(-1) + (-1)(-3) + (-3)(1) = -1 + 3 - 3 = -1 = (c/a)

           (3.) αβγ = (1)(-1)(-3) = 3 = (-d/a)

• (α + β + γ) comes out to be equal to (-b/a).

• (αβ + βγ + γα) comes out to be equal to (c/a).

• (αβγ) comes out to be equal to (-d/a).

Hence the relationship between the zeroes and the coefficients of this cubic polynomial gets established.

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