How to find the equation of this polynomial How do you the simplified polynomial with integer coefficients and the following root(s): 
$\sqrt3+4i$
I think there is something related to conjugates, but I do not know how to do this.
Due to the complex conjugates theorem,
$$((x-\sqrt3)-4i)((x-\sqrt3)+4i) = (x-\sqrt3)^2+16=x^2-2\sqrt3x+19$$
Here, there is a problem since there are no integer coefficients. I tried squaring the expression $x^2-2\sqrt3+19$ as I thought maybe $\sqrt3+4i$ and $\sqrt3-4i$ were double roots, but this was useless, since it produced more irrational solutions.
Does this mean that the expression with $\sqrt3+4i$ and $\sqrt3-4i$ as its roots does not have rational solutions? If not, how would I go about approaching this problem?
 A: You are on the right track. The complex conjugate of $\sqrt3+4i$ is $\sqrt3-4i$ so we set the polynomial $$((x-\sqrt3)+4i))((x-\sqrt3)-4i))=(x-\sqrt3)^2-16i^2=x^2-2\sqrt3x+19=0$$ Now rearrange the polynomial $$2\sqrt3x=x^2+19$$ and on squaring, $$12x^2=x^4+38x^2+361$$ so the answer is $$\boxed{x^4+26x^2+361=0}$$
A: You can use also the following identity.
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4,$$ where $a=x$, $b=-\sqrt3$ and $c=-4i$.
Now, since $a+b+c=0$, we obtain:
$$2(3x^2-16x^2-48)-x^4-9-256=0$$ or
$$x^4+26x^2+361=0.$$
A: You can try to eliminate the surd and $i$ by isolating terms and squaring the equation $x=\sqrt{3}+4i$.
Method 1: eliminating the surd then the imaginary
$\begin{align}(x-4i)^2=3 &\implies x^2-8ix-19=0 \\&\implies (x^2-19)^2=-64x^2 \\&\implies x^4+26x^2+361=0\end{align}$
Method 2: eliminating the imaginary then the surd
$\begin{align}(x-\sqrt{3})^2=-16 &\implies x^2-2x\sqrt{3}+19=0 \\&\implies (x^2+19)^2=12x^2 \\&\implies x^4+26x^2+361=0\end{align}$
