SOLUTION: The equation of a polynomial with real coefficients that has the roots 2 and 3(imaginary i) is? a. x^3-2x^2+9x-18 b. x^3-2x^2-12x-18 c. x^3+2x^2-9x-18 d. x^3-2x^2+9x-20

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: The equation of a polynomial with real coefficients that has the roots 2 and 3(imaginary i) is? a. x^3-2x^2+9x-18 b. x^3-2x^2-12x-18 c. x^3+2x^2-9x-18 d. x^3-2x^2+9x-20      Log On


   

Question 202835: The equation of a polynomial with real coefficients that has the roots 2 and 3(imaginary i) is?
a. x^3-2x^2+9x-18
b. x^3-2x^2-12x-18
c. x^3+2x^2-9x-18
d. x^3-2x^2+9x-20
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let's call the polynomial P(x). If 2 and 3i are roots of P(x) then P(x) must be:
P%28x%29+=+%28x+-2%29%28x%5E2+%2B+9%29
If this isn't immediately clear, think about it. If 2 is a root of P(x) then, by the very definition of a root, when x = 2, P(x) = 0. Can you see that if x = 2 in the equation above, that P(x) = 0? And similarly, can you see that if x = 3i (or -3i) that P(x) is zero (because x%5E2+%2B+9+=+0 if x = 3i (or -3i)?

Now all we need to do is multiply out P(x) to see which answer is correct. we should get:
x%5E3+-+2x%5E2+%2B+9x+-18 so (a) is correct.