The value of the expression $x^4-8x^3+18x^2-8x+2$ when $x=\cot\frac{\pi}{12}$

Question

The value of the expression $x^4-8x^3+18x^2-8x+2$ when $x=\cot\frac{\pi}{12}$

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I know that $\cot\frac{\pi}{12}=2+\sqrt3$,but putting this value and evaluating is a long process.Isn't there a short and intuitive method to solve this?

Answer 1

Hint:

$$\cot15^\circ+\tan15^\circ=\dfrac2{\sin(2\cdot15^\circ)}=?$$

$$\cot15^\circ\tan15^\circ=?$$

So, $\cot15^\circ,\tan15^\circ$ are the roots of $$t^2-4t+1=0$$

Now we can straightaway divide $$x^4-8x^3+18x^2-8x+2$$ by $$x^2-4x+1$$ to find the remainder.

Answer 2

The hint $$x^4-8x^3+18x^2-8x+2=(x^2-4x+1)^2+1$$ and try to solve $$x^2-4x+1=0.$$

Answer 3

I know that $\cot\frac{\pi}{12}=2+\sqrt3$

Following up on this line, let $a=2+\sqrt3\,$, then:

• $a^2=7+4\sqrt{3}$
• $\dfrac{1}{a}=2-\sqrt{3}\,$ so $\,a+\dfrac{1}{a}=4\,$, and $a^2+\dfrac{1}{a^2}=\left(a+\dfrac{1}{a}\right)^2-2=14$

Then $\;a^4-8a^3+18a^2-8a+2 = a^2 \cdot \left(a^2+\dfrac{1}{a^2}-8\left(a+\dfrac{1}{a}\right)+18\right)+1 = \cdots\,$