Question

Using the given zero, find all other zeros of f(x).

-2i is a zero of f(x) = x^4 - 32x^2 - 144

Answer 1

Complex roots always occur in pairs.
If -2i is a root, then its complex conjugate, +2i will also be a root.
That means, (x-2i) and (x+2i) are factors.
(x-2i)(x+2i) = x^2 - (2i)^2 = (x^2 + 4) is a factor.

Divide x^4 - 32x^2 - 144 by (x^2 + 4)
It should divide evenly without any remainder. The quotient will be a quadratic which you can solve to find the other two roots.

Answer 2

just to butt in i wouldn't actually divide using long division
i would write
\[x^4-32x-144=(x^2+4)(x^2+bx+c)\] and find \(b\) and \(c\)

Answer 3

typo there, i meant
\[x^4-32x^2-144=(x^2+4)(x^2+bx+c)\]

Answer 4

Ok for ranga I got to x^4-176 x^2 - 576

and for satelite I got b=0
x=+- 2i
c=-36

Answer 5

I'm not sure how to go forward with either process

Answer 6

Although 2i is one of the zeros in every solution I'm given!

Answer 7

Your long division is not correct. I will show you in the next reply. It is fairly straightforward here.
You got b = 0 and c = -36
So the other factor is: x^2 +0x - 36 or x^2-36
which you can factor as: (x+6)(x-6)

So the original function f(x) = (x-2i)(x+2i)(x-6)(x+6)

The roots are: -2i, 2i, -6, 6.

Answer 8

x^2 - 36
________________________
x^2 + 4 | x^4 - 32x^2 - 144
x^4 + 4x^2
--------------
-36x^2 - 144
-36x^2 - 144
----------------
0 0

Answer 9

Well now I'm looking at what you did and I have no clue how I got my answer :/
Oh and duh, I had to factor 36!
Thank you so much for your help :)