Stuck on this question

Question

h(x)=2x^4+7x^3+4x^2+28x-16

zero = -2i

I need help finding the remaining zeros with an explanation i'm stuck thanks!

Answer 1

Since -2i is a zero, so is 2i

So we have factors of (x-2i) and (x+2i)

If we multiply these factors together we get (x-2i)(x+2i) = x^2 + 4

That means we could divide our original function by x^2 + 4

We find that when we divide h(x) by (x^2 + 4) we obtain

2x^2 + 7x - 4

Now just factor

(2x - 1)(x + 4)

Therefore the other zeros are 1/2 and -4