# Verify that 3, - 1, - ⅓ - are the zeroes of the cubic polynomial p(x) = 3x^{3} - 5x^{2} - 11x - 3, and then verify the relationship between the zeroes and the coefficients

**Solution:**

The zeroes of a polynomial are the values of x that satisfies the equation and makes the polynomial equal to zero as a whole.

Let us put the values of x in the polynomial to verify the zeroes of the polynomial.

Given, p(x) = 3x^{3} - 5x^{2} - 11x - 3

p(x = 3) = 3(3)^{3} - 5(3)^{2} - 11(3) - 3

⇒ 81 - 45- 33 - 3

⇒ 81 - 81 = 0

p(x = - 1) = 3(- 1)^{3} - 5(- 1)^{2} - 11(- 1) - 3

⇒ - 3 - 5 + 11 - 3

⇒ - 11 + 11 = 0

p(x = - ⅓ ) = 3(- ⅓ )^{3} - 5(- ⅓ )^{2} - 11(- ⅓) - 3

⇒ -1/9 + 5/9 + 11/3 - 3

⇒ -6/9 + ⅔ = - ⅔ + ⅔ = 0

Thus, 3, - 1, - ⅓ - are the zeroes of the cubic polynomial p(x) = 3x^{3} - 5x^{2} - 11x - 3.

3x^{3} - 5x^{2} - 11x - 3 is in the form of ax^{3} + bx^{2} + cx + d where a ≠ 0,

Let α, β, and γ are the three zeros of the given polynomial 3, -1 and -⅓ respectively.

The sum of zeroes

α + β + γ = -b/a

⇒ 3 + (-1) + (-⅓ ) = 5/9 = - coefficient of x2/ coefficient of x^{3}

The sum of the product of zeroes

αβ+ βγ + αγ = c/a =

⇒ 3 × (-1) + (- 1) × (-⅓ ) + (-⅓ ) × 3

= - 3 + ⅓ - 1

= - 11/3 = coefficient of x/ coefficient of x^{3}

The product of zeroes

αβγ is - d/a

⇒ 3 × (- 1) × (- ⅓ ) = - (- 3)/ 3 = - constant term/ coefficient of x^{3}

☛ Check: NCERT Solutions for Class 10 Maths Chapter 2

## Verify that 3, - 1, - ⅓ - are the zeroes of the cubic polynomial p(x) = 3x^{3} - 5x^{2} - 11x - 3, and then verify the relationship between the zeroes and the coefficients

**Summary:**

3, - 1, - ⅓ - are the zeroes of the cubic polynomial p(x) = 3x^{3} - 5x^{2} - 11x - 3 and the zeroes and the coefficients of a cubic polynomial can be expressed as the sum and product of the zeroes

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