Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$?

Question

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.

Answer 1

Note that\begin{align}x^6-2\sqrt3x^5&=x^5\left(x-2\sqrt3\right)\\&=x^5\left(2-\sqrt3\right)\\&=x^4x\left(2-\sqrt3\right)\\&=x^4\end{align}and that therefore$$x^6-2\sqrt3x^5-x^4+x^3-4x^2+2x-\sqrt{3}=x^3-4x^2+2x-\sqrt3.$$Now, note that\begin{align}0&=\left(x-\left(2+\sqrt3\right)\right)\left(x-\left(2-\sqrt3\right)\right)\\&=x^2-4x+1.\end{align}So,\begin{align}x^3-4x^2+2x-\sqrt3&=x\overbrace{(x^2-4x+1)}^{\phantom0=0}+x-\sqrt3\\&=x-\sqrt3\\&=2.\end{align}

Answer 2

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 $f(2+\sqrt3) = 2$.

Answer 3

I calculated it longhand:

$$ \begin{align} x & = 2 + \sqrt 3 \\ x^2 & = 2^2 + 3 + 2 \cdot 2 \sqrt 3 \\ & = 7 + 4 \sqrt 3 \\ x^3 & = 2 \cdot 7 + 4 \cdot 3 + 2 \cdot 4 \sqrt 3 + 7 \sqrt 3 \\ & = 26 + 15 \sqrt 3 \\ x^4 & = 97 + 56 \sqrt 3 \\ x^5 & = 362 + 209 \sqrt 3 \\ x^6 & = 1351 + 780 \sqrt 3 \end{align} $$

We also have

$$ 2 \sqrt 3 x^5 = 209 \cdot 6 + 362 \cdot 2 \sqrt 3 = 1254 + 724 \sqrt 3 $$

Then

$$ \begin{align} & x^6 - 2 \sqrt 3 x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt 3 \\ & = 1351 - 1254 - 97 + 26 - 28 + 4 + (780 - 724 - 56 + 15 - 16 + 2 - 1) \sqrt 3 \\ & = 2 \end{align} $$

It was only after doing so that I noticed @JoséCarlosSantos's short cut.

Answer 4

As OP noted, $\,x = 2 + \sqrt{3}\,$, and also $\,\dfrac{1}{x} = 2 - \sqrt{3}\,$. Then $\,\color{blue}{x+ \dfrac{1}{x} = 4}\,$, $\,\color{green}{x - \dfrac{1}{x} = 2\sqrt{3}}\,$, and:

$$ \require{cancel} \begin{align} & \;\;x^6 - \color{green}{2\sqrt{3}}\, x^5 - x^4 + x^3 - \color{blue}{4}\,x^2 + 2x - \sqrt{3} \\ = &\; \cancel{x^6} - \left(\color{green}{\cancel{x} - \bcancel{\frac{1}{x}}}\right)x^5 - \bcancel{x^4} + \xcancel{x^3} - \left(\color{blue}{\xcancel{x}+\frac{1}{x}}\right)x^2 + 2x - \sqrt{3} \\ = &\;\; x - \sqrt{3} \\ = &\; \;2 \end{align} $$