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.
Start by collecting all the like-terms with a particular radical and moving them to the left-hand-side; I chose $\sqrt{3}$ first. Move all terms without $\sqrt{3}$ to the right-hand-side. Then, factor out the chosen radical. This is shown in $(1)$. Next, square both sides, as shown in $(2)$. Since all terms with $\sqrt{3}$ have been collected, we've taken $\sqrt{3}$ out of the equation. Although, we're still left with all other radicals ($\sqrt{5}$ in this case).

Next, collect all terms with the next radical ($\sqrt{5}$) on one side of the equation. As before, factor out that radical, as shown in $(3)$. And once again, square both sides, as shown in $(4)$. You're left with a polynomial with coefficients in $\mathbb{Z}$. I'll leave the simplification up to you.

$$\begin{aligned}
&amp;(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=0
\\
\\&amp;\bigg(-2x^3+x\bigg)\sqrt{3}=\bigg(1-x^3\bigg)\sqrt{5}-1&amp;(1)
\\
\\&amp;3\bigg(-2x^3+x\bigg)^2=5\left(1-x^3\right)^2-2\sqrt{5}\left(1-x^3\right)+1&amp;(2)
\\
\\&amp;3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1=-2\sqrt{5}\left(1-x^3\right)&amp;(3)
\\
\\&amp;\left(3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1\right)^2=20\left(1-x^3\right)^2&amp;(4)\end{aligned}$$
It's $$5(x^3-1)^2=(\sqrt3(2x^3-x)-1)^2$$ and from here
$$(5(x^3-1)^2-3(2x^3-x)^2-1)^2=12(2x^3-x)^2,$$
which is
$$(7x^6-12x^4+10x^3+3x^2-4)^2-12x^2(2x^2-1)^2=0$$ or
$$49x^{12}-168x^{10}+140x^9+186x^8-240x^7-76x^6+60x^5+153x^4-80x^3-36x^2+16=0.$$