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. 
 A: 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.
$$\begin{aligned}
&(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=0
\\
\\&\bigg(-2x^3+x\bigg)\sqrt{3}=\bigg(1-x^3\bigg)\sqrt{5}-1&(1)
\\
\\&3\bigg(-2x^3+x\bigg)^2=5\left(1-x^3\right)^2-2\sqrt{5}\left(1-x^3\right)+1&(2)
\\
\\&3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1=-2\sqrt{5}\left(1-x^3\right)&(3)
\\
\\&\left(3\bigg(-2x^3+x\bigg)^2-5\left(1-x^3\right)^2-1\right)^2=20\left(1-x^3\right)^2&(4)\end{aligned}$$
A: Let $u(x)=(\sqrt5+2\sqrt3)x^3-\sqrt3 x-\sqrt5+1$. Then $p(x)u(x)$
will have coefficients not involving $\sqrt3$ but still involving $\sqrt5$;
also $p(r)u(r)=0$ whenever $p(r)=0$
So halfway there! Now how to get rid of the $\sqrt5$s...? 
A: $p(x)=(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=\sqrt{5}(x^3-1)+\sqrt{3}(x-2x^3)+1$
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.
$49x^{12}−168x^{10}+140x^9+186x^8−240x^7−76x^6+60x^5+153x^4−80x^3−36x^2+16=0$
A: 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
$$49x^{12}-168x^{10}+140x^9+186x^8-240x^7-76x^6+60x^5+153x^4-80x^3-36x^2+16=0.$$
