Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$. 
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would be appreciated. 

$p(x)=(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=\sqrt{5}(x^3-1)+\sqrt{3}(x-2x^3)+1$
Let's set : $\begin{cases}a=x^3-1\\b=x-2x^3\\p(x)=\sqrt{5}a+\sqrt{3}b+1\end{cases}$
For $p(x)=0$ we have $5a^2+3b^2+2\sqrt{15}ab=(-1)^2=1$
And then $60a^2b^2=(1-5a^2-3b^2)^2$
Finally substitute $a,b$ with their values, I find the same polynomial as M.Rozenberg.
$49x^{12}−168x^{10}+140x^9+186x^8−240x^7−76x^6+60x^5+153x^4−80x^3−36x^2+16=0$
