Find a third-degree polynomial equation with rational coefficients that has the roots 2 and - 3 + i.

A. 0 = x^3 + 8x^2 + 22x + 20

B. 0 = x^3 + 4x^2 - 2x - 20

C. 0 = x^3 + 4x^2 + 22x - 20

D. 0 = x^3 - 4x^2 - 2x + 20

Option B

Step-by-step explanation:

Complex roots occur as conjugate pairs so the third root is - 3 - i (note that the sign changes from + to - ).

So in factor form we have:-

(x - 2) (x - (-3 + i)) (x - (-3 - i)) = 0 Let's expand the last 2 factors first:-

(x - (-3 + i)) (x - (-3 - i))

= (x + 3 - i) (x + 3 + i)

= x^2 + 3x + ix + 3x + 9 + 3i - ix - 3i - i^2

= x^2 + 6x + 9 - (-1)

= x^2 + 6x + 10

Now multiplying by (x - 2) : -

(x - 2) (x^2 + 6x + 10) = 0

x^3 + 6x^2 + 10x - 2x^2 - 12x - 20 = 0

x^3 + 4x^2 - 2x - 20 = 0 (answer)

Option B