Hi

which specific formula/method I can use to find the complex roots of a polynomial of degree 4 on form:

$\displaystyle p(x) = x^4 + a x^3 + bx^2 + ax + a_0 = 0$

which has irrational coefficients

a = $\displaystyle -1 - \sqrt(2) $ and b = $\displaystyle 2+ \sqrt(2)$ and

where $\displaystyle a_0 = 1$

If I write the equation as $\displaystyle x^4-x^3+2x^2-x+1-\sqrt2(x^3-x^2+x)=0$, then you will notice that it factorises as $\displaystyle (x^2-x+1)(x^2-\sqrt2x+1)=0$

Then by fundamental theorem of algebra I see that

$\displaystyle p(x) = (x-(\frac{1}{2}) + \frac{\sqrt(3)}{2}i)) (x-(\frac{1}{2}) - \frac{\sqrt(3)}{2}i)) \cdot ((x - (\frac{\sqrt(2)}{2} + \frac{\sqrt(2)}{2} i)) ((x - (\frac{\sqrt(2)}{2} - \frac{\sqrt(2)}{2} i)) = (x^2-x+1)(x^2-\sqrt2x+1) = x^4-x^3+2x^2-x+1-\sqrt2(x^3-x^2+x)$

where the complex roots of the original polymial p(x) is

$\displaystyle x = ((\frac{1}{2}) \pm \frac{\sqrt(3)}{2}i), (\frac{\sqrt(2)}{2} \pm \frac{\sqrt(2)}{2} i)$

Getting back to the original polynomial

(*)$\displaystyle p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0$

where $\displaystyle a,b \in \mathbb{C}$

I would like to prove that a complex number x makes (*) true iff

$\displaystyle s = x + x^{-1}$ is a root of the $\displaystyle Q(s) = s^2 + as + (b-2) $

I see that that $\displaystyle Q(x + x^{-1}) = \frac{p(x)}{x^2}$

So the solutions of in my equation $\displaystyle s^2+as+(b-2)=0$ are $\displaystyle s=\frac{-a\pm\sqrt{a^2-4(b-2)}}{2}$.

I then need to plug each of these numbers into the equation $\displaystyle s=x+x^{-1}$, which can be written as $\displaystyle x^2-sx+1=0$. Using the quadratic formula again, you get $\displaystyle x=\frac{s\pm\sqrt{s^2-4}}{2}$.

By the way do x^-1 then exist ?

Yes, there's no problem about that. In fact since $\displaystyle s=x+x^{-1}$, you can see that $\displaystyle x^{-1}=s-x$

I get this x value for both values of s to be:

$\displaystyle x =\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4} $

This expression makes the equation

$\displaystyle p(x) = \frac{p(x)}{x^2}$ true

since $\displaystyle p(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4}) = \newline \frac{p(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4})}{(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4})^2} = 0$

This proves that there exist a number x which is both a root of the original polynomial and Q(s = x + x^{-1}), since $\displaystyle p(x) = \frac{p(x)}{x^2}$.

But one question remains on my part. Since x is supposedly a complex number. Is it a complex number in its current form? That I'm a bit unsure of. Since then it should be written in the form x = a + bi ?

Best Regards,

Billy