Use the given zero to find the remaining zeros of each function f(x)=2x^4+5x^3+5x^2+20x-12 zero: -2i

1/2, 3

Step-by-step explanation:

This is a pretty involved problem, so I'm going to start by laying out two facts that our going to help us get there.

In 2., I mentioned that both -2i and 2i must be zeroes of f(x). This means that both and are factors of f(x), and furthermore, their product, , is also a factor. To see what's left after we factor out that product, we can use polynomial long division to find that

I'll go through to steps to factor that second expression below:

Solving both of the expressions when f(x) = 0 gets us our final two zeroes:

So, the remaining zeroes are 1/2 and 3.

The roots are -2i, 2i, 1/2 and -3.

Step-by-step explanation:

The complex roots exist as conjugate pairs so if one root is -2i the other is 2i. So a factor of f(x) is (x - 2i)(2 + 2i).  = x^2 + 4.

So dividing:

                           2x^2 + 5x - 3   < Quotient.

x^2 + 0x + 4  ) 2x^4 + 5x^3 + 5x^2 + 20x - 12

                       2x^4 + 0x^3 + 8x^2

                                   5x^3  - 3x^2 + 20x

                                   5x^3  + 0x^2 + 20x

                                                -3x^2 - 12

                                                  -3x^2 - 12.

Finding the other 2 zeroes:

2x^2 + 5x - 3 = 0

(2x - 1) ( x + 3) = 0

x = 1/2, -3.