Question

How do you determine the remaining zeroes for `g(x)=2x^5-3x^4-5x^3-15x^2-207x+108` if 3i is a zero?

Answer 1

Use conjugate zeros, factor theorem, division of polynomials, rational zeros theorem, division (again), then solve the resulting quadratic by you favorite method for quadratics.

Explanation:

`g(x)=2x^5-3x^4-5x^3-15x^2-207x+108`

First Two Zeros
Given that `3i` is a zero, we observe that the coefficients of `g` are real, so the Complex Conjugate Zero Theorem tells us that `-3i` is also a zero.
The Factor Theorem tells us that `x-3i` and `x+3i` are factors, so their product is also a factor:
`(x-3i)(x+3i) = x^2+9`

Divide `g(x)` by `x^2+9` to get:

`2x^5-3x^4-5x^3-15x^2-207x+108 = (x^2 +9)(2x^3-3x^2-23x+12)`

Third Zero

To find another zero, we need a zero of `2x^3-3x^2-23x+12`.

Neither of the easiest choices, `+-1`, are zeros. So use the Rational Zeros Theorem to search for another zero.

Possible rational zeros are: `+-1, +-2, +-3, +-4, +-6, +-12, +-1/2, +-3/2`
Test until you find a zero. (Synthetic division is recommended for this testing.)
Depending on the order in which you test, you'll find a rational zero.
I found `-3` is a zero first. So (or really because) `x+3` is a factor.

`2x^3-3x^2-23x+12 = (x+3)(2x^2-9x+4)`.

Last Two Zeros
Solve `2x^2-9x+4 =0` by whatever method you like.

`(2x-1)(x-4) = 0`
So the final two zeros of `g` are `1/2` and `4`.

List of Zeros

Zeros: `3i, -3i, -3, 4, 1/2`