Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putting the values wont help either but I am not able to factorize the polynomial or manipulate it to help me.
I would be grateful if anybody could help me.

Let $f(x) = x^6 -2\sqrt3 x^5 - x^4 + x^3 -4x^2 + 2x -\sqrt3$. Then, $f(2 + \sqrt3)$ is the remainder when $f(x)$ is divided by $x - (2+\sqrt3)$.(Remainder theorem )
Since, $$ f(x) = (x - (2+\sqrt3))(x^5 + (2-\sqrt3)x^4 + x^2 - (2-\sqrt3)x + 1) + 2$$
The remainder is $2$, thus $f(2+\sqrt3) = 2$.
