Question

I was wondering if anyone could please help me with an Algebra 2 problem involving roots. The roots are 1/2, 1, and 3+2i. I need help finding the equation of that and it passes through the point (-1,200).

I did (x-1/2)(x-1)(x-3+2i)(x-3-2i) as my first step
Is that right?
Then i simplified it but it does not pass through (-1,200)?

Answer 1

you forgot the fact that there could have been a constant in front of the whole mess.
I also used (2x-1) instead of (x - 1/2)

let y = a(2x-1)(x-1)(x-(x-(3+2i))(x - (3-2i))
= a(2x - 1)(x-1)(x - 3 - 2i)(x - 3 + 2i )

If you expand the last two, you get
x^2 - 6x + 13
so ...
y = a(2x-1)(x-1)(x^2 - 6x + 13)

now we "force" (-1,200) to be on it by subbing it in, and solving for a

200 = a(-3)(-2)(20)
200 = 120a
a = 200/120 = 5/3

equation is:
y = (5/3)(2x-1)(x-1)(x^2 - 6x + 13)

now plug in your point(-1,200), it will work.