Polynomial Function with solutions of 3, 2i, -2i

[FMMurphy]

How do you find the polynomial function if you have the solutions of 3, 2i, -2i ? I know that this is where it crosses the x axis but how do you go about finding the function?

 

[Denis]

(x - 2i)(x + 2i) = x^2 - 4i^2 = x^2 + 4 : OK :?:
Finish it.

 

[FMMurphy]

Polynomial function

I'm a little confused x^2-4i^2=x^2+4. I'll take your word on that but where does the 3 come in?

 

[Mrspi]

Re: Polynomial Function

How do you find the polynomial function if you have the solutions of 3, 2i, -2i ? I know that this is where it crosses the x axis but how do you go about finding the function?

If 3 is a solution, then x = 3, and x - 3 is a factor of the polynomial.

If 2i is a solution, then x = 2i, and x - 2i is a factor of the polynomial.

If -2i is a solution, then x = -2i, and x + 2i is a factor of the polynomial.

So, you can represent the polynomial f(x) as

f(x) = (x - 3)(x - 2i)(x + 2i)

Do the multiplication....

 

[FMMurphy]

Polynomial function

Thanks for your help. Would the final polynomial function be:
x^3-3x^2-4i^2+12i^2?
Thanks again!

 

[Denis]

Murph, i = sqrt(-1) ; you didn't know that?
So i^2 = -1 ; kapish?

> x^3-3x^2-4i^2+12i^2?
Btw, you're missing an x in 3rd term: -4i^2(x)