Synthetic division to find zeroes using imaginary numbers.
Matthew Martinez 
I am given a polynomial with a zero of -i. I understand that by definition "i" must also be a zero. The problem asks me to use synthetic division to find all zeroes. My textbook only gives two examples using non-complex numbers, and after an exhaustive Google search I cannot find any calculators that show how to find a zero using only -i or i, only for complex numbers such as "2-i", etc. Can anyone please help me with the steps I should take to find any other zeroes, or if the answer is only "-i" and "i"?
Edit: Using Oscar Lanzi's suggestion, I was able to use a calculator to work out the synthetic division. I had the correct answer, but I was burning out a little since I have been doing math since late Friday night.
To give more info on the polynomial, it is $P(x)=x^3+3x^2+x+3$ and $-i$ is given as a zero. So to address the answer below, the polynomial has real coefficients. I hope this answers some questions.
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$\begingroup$I am given a polynomial with a zero of -i. I understand that by definition "i" must also be a zero.
That's true only if the given polynomial has real coefficients, and in that case it follows by the Complex conjugate root theorem (not by definition).
The following will asume that the question is indeed about a polynomial with real coefficients, but you should spell that out in the post.
The problem asks me to use synthetic division to find all zeroes.
Since $i$ and $-i$ are both roots, the original polynomial is divisible by $(x+i)(x-i) = x^2 +1$. Using Synthetic division by monic polynomials (sometimes called expanded synthetic division) you can divide the original polynomial by $x^2 + 1$. The quotient will be of degree 2 less than the original, and presumably easier to solve by other methods in order to find the rest of the roots.
[ EDIT ] Taking for example the polynomial $x^3 + 3x^2 + x + 3 = 0$ added to the original question, this one has real coefficients, so knowing that $-i$ is a root does imply that $i$ is a root as well. Dividing by $x^2+1$ using synthetic division gives:
$$ \begin{align} x^3 + 3x^2 + x + 3 & = (x^3 + x) - x + (3 x^2 + 3) - 3 + x + 3 \\ & = x(x^2+1) + 3(x^2+1) = \\ & = (x+3)(x^2+1) \end{align} $$
The quotient of the division is the first degree polynomial $x+3$ with the unique root $x=-3$.
[ EDIT #2 ] While the above answers the question as posed (using synthetic division) it may be worth pointing out that in the simple case of a 3rd degree polynomial with 2 known roots the easiest way to find the remaining root is using any of Vieta's formulas, rather than division and factorization. In the given example $x_1 = i$, $x_2 = -i$ and $x_1 + x_2 + x_3 = -3$ therefore $x_3 = -3$. $\endgroup$ 5
