If  α, β, γ  are the zeros of the polynomial  x3 − 3x + 11  =  0 , then the polynomial whose zeros are  (α + β), (β + γ)  and  (γ + α)  is -Select an answerA x3 + 3x + 11  =  0 B x3 − 3x + 11  =  0 C x3 + 3x − 11  =  0 D x3 − 3x − 11  =  0

The sum of the roots of the original polynomial is -b/a = 0 (since there is no x² term), and the sum of the roots of the new polynomial is α+β + β+γ + γ+α = 2(α + β + γ) = 2*0 = 0.

The product of the roots taken two at a time for the original polynomial is c/a = -3, and for the new polynomial it is (α+β)(β+γ) + (β+γ)(γ+α) + (γ+α)(α+β) = 2(αβ + βγ + γα) = 2*-3 = -6.

The product of the roots for the original polynomial is -d/a = -11, and for the new polynomial it is (α+β)(β+γ)(γ+α) = αβγ + α²β + αβ² + β²γ + βγ² + αγ² = αβγ + (α²β + αβ² + β²γ + βγ² + αγ²) = -11 + 3 = -8.

So, the polynomial whose zeros are (α+β), (β+γ) and (γ+α) is x³ - 0x² - 6x - 8 = 0.

None of the options A, B, C, D match this result.