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

2

step-by-step explanation: