Question

find cubic polynomial whose zeros are alpha =2 , bita =1÷3 , gama =1÷2

Answer 1

x^3 - (α+β+γ)x^2 + (αβ+βγ+γα)x + αβγ = 0
=> x^3 - (2+(1/3)+(1/2))x^2 + ((2/3)+(1/6)+1)x + 1/3 = 0
=> x^3 - (17/6)x^2 + (11/6)x + (1/3) = 0

Multiply both sides by 6 for simplification

=> 6x^3 - 17x^2 + 11x + 2 = 0

Answer 2

Hi there !!

α = 2
β = 1/3
γ = 1/2

Sum of zeros = α + β + γ
                       = 2 + 1/3 + 1/2
                      = 2 + 5/6 = 12 + 5/ 6
                      = 17/6

Sum of the product of  zeros taken two at a time = αβ + βγ + αγ
                                                      = 2× 1/3 + 1/3 × 1/2 + 2 × 1/2
                                                       = 2/3 + 1/6 +1
                                                       = 4+1/6 + 1
                                                       = 5/6 + 1
                                                       = 11/6

Product of zeros = αβγ
                           = 2 × 1/3 × 1/2
                           = 1/3

A cubic polynomial is given by :-

k{ x³ - [α + β + γ ]x² + [αβ + βγ + αγ ]x - [αβγ] }

here , k is a constant and can be any no:

k{x³ - 17/6x² + 11/6x - 1/3}
 k = 6 ,

6{x³ - 17/6x² + 11/6x - 1/3}
6x³ - 17x² +11x - 2        ------> Required polynomial