Given some zeroes of a real polynomial of a given degree, how can one find the remaining zeroes? Here is what the problem says:

If $2$, $-\sqrt{5}$, and $3+i$ are three zeroes of a $5$th degree polynomial function with real coefficients, find the other zeroes of multiplicity $1$.

I don't think I can use synthetic division because the function is not given. I honestly have no idea how else to solve it. Any help would be greatly appreciated.
If I have not been clear enough or you need more detail, or I am doing something wrong by asking this question, please let me know and I will fix it ASAP. Thank you for your time.
 A: If a polynomial $p(x)=a_nx^n+...+a_1x+a_0$ has real coefficients and $r$ is a root, then $$0=\overline{0}=\overline{p(r)}=a_n\overline{r}^n+...+a_1\overline{r}+a_0$$
This means that if $r$ is a root then $\overline{r}$ is also a root. 
Since $3+i$ is a root, then $3-i$ is also a root. Therefore 
$$(x-2)(x+\sqrt{5})(x-3-i)(x-3+i)$$ divides our polynomial. Therefore the polynomial is any degree-5 polynomial divisible the this. 
If we knew that the coefficients were rational. Then the polynomial would have to be divisible by the minimal polynomial of $\sqrt{5}$. This is $(x-\sqrt{5})(x+\sqrt{5})$.
This would tell us that $$(x-2)(x+\sqrt{5})(x+\sqrt{5})(x-3-i)(x-3+i)$$ divides our polynomial. Therefore this is, up to a multiplicative constant, our degree-5 polynomial.
A: This is not enough information to determine the polynomial or the remaining zeroes of multiplicity $1$. Per Lab's comment, the nonreal roots of a real polynomial come in complex conjugate pairs; since the quintic polynomial $p$ at hand is real and $3 + i$ is a root, so is $3 - i$. Thus, the polynomial is a multiple of
$$q(x) = (x - 2)(x + \sqrt{5})(x - (3 + i))(x - (3 - i)) = (x - 2)(x + \sqrt{5})(x^2 - 6 x + 10).$$
On the other hand, any of the quintic polynomials
$$p(x) = A (x - b) q(x)$$
satisfies the given root conditions, so we can only say (1) that $3 - i$ is another root of $q$ (unless $A = 0$, so that $p = 0$, it must be a single root; if it were not, $3 + i$ would be a multiple root, but counting shows that this would imply that $\deg p > 5$), and (2) $b$ will be a single root iff it is not $2, -\sqrt{5}$.
Now, if we knew that $p$ was rational, since $-\sqrt{5}$ is not rational, $p$ would have as a factor the minimal polynomial of $-\sqrt{5}$, namely $x^2 - 5$. This would force the remaining factor to be $x - r$, where $r$ is the other real root of this minimal polynomial, namely $\sqrt{5}$ (which is hence a single root) and determine $p$ (up to an overall nonzero constant):
$$p(x) = A (x - 2) (x^2 - 5) (x^2 - 6 x + 10).$$
