Question

find cubic polynomial whose zeroes are 3,1/2,1​

Answer 1

EXPLANATION.

Cubic polynomial whose zeroes are = 3, 1/2, 1.

As we know that,

⇒ α = 3  and  β = 1/2   and  γ = 1

Sum of the zeroes of the cubic polynomial.

⇒ α + β + γ = -b/a.

⇒ 3 + 1/2 + 1.

⇒ 6 + 1 + 2/2 = 9/2.

⇒ α + β + γ = 9/2.

Products of zeroes of the cubic polynomial two at a time.

⇒ αβ + βγ + γα = c/a.

⇒ (3)(1/2) + (1/2)(1) + (1)(3).

⇒ 3/2 + 1/2 + 3.

⇒ 3 + 1 + 6/2 = 10/2 = 5.

Products of the zeroes of the cubic polynomial.

⇒ αβγ = -d/a.

⇒ (3)(1/2)(1) = 3/2.

Formula of the cubic polynomial.

⇒ x³ - (α + β + γ)x² + (αβ + βγ + γα)x - αβγ.

Put the values in the equation, we get.

⇒ x³ - (9/2)x² + (5)x - (3/2) = 0.

⇒ x³ - 9x²/2 + 5x - 3/2 = 0.

⇒ 2x³ - 9x² + 10x - 3 = 0.