Verify that 3, - 1, - ⅓ - are the zeroes of the cubic polynomial p(x) = 3x3 - 5x2 - 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) = 3x3 - 5x2 - 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) = 3x3 - 5x2 - 11x - 3.
3x3 - 5x2 - 11x - 3 is in the form of ax3 + bx2 + 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 x3
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 x3
The product of zeroes
αβγ is - d/a
⇒ 3 × (- 1) × (- ⅓ ) = - (- 3)/ 3 = - constant term/ coefficient of x3
☛ Check: NCERT Solutions for Class 10 Maths Chapter 2
Verify that 3, - 1, - ⅓ - are the zeroes of the cubic polynomial p(x) = 3x3 - 5x2 - 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) = 3x3 - 5x2 - 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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