Verified answer

If 3 - 2i is a root, so is 3 + 2i.

The factors are (x - (3 + 2i)) and (x - (3 - 2i)), or (x - 3 - 2i) and (x - 3 + 2i).

(x - 3 - 2i)(x - 3 + 2i) = (x^2 - 6x + 9) + 4 = x^2 - 6x + 13

P(x) = x^2 - 6x + 13

P(x) must be of the form ax^2 + bx + c (because it's degree 2), and a=1.

P(x) = x^2 +bx +c

When you set this polynomial to zero, one of the roots you get is 3-2i. This means that when you apply the quadratic equation, this is one of the answers you get in its most simplified form. I think you are then supposed to work backwards from here, by making 3-2i look like one of the quad equation solutions.

3-2i = (-b - √(b^2 - 4c))/2

I'm going to break up the fraction to get -b/2 - √(b^2 - 4c)/2. Now, set these individual fraction terms equal to the terms in 3 - 2i:

3 = -b/2

-b = 6

b = -6

and

-√(b^2 - 4c)/2 = -2i

√(b^2 - 4c)/2 = 2i

√(b^2 - 4c) = 4i

b^2 - 4c = 16i^2 = -16

(-6)^2 - 4c = -16

36 + 16 = 4c

52 = 4c

c=13

Therefore, P(x)= x^2 - 6x + 13