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I am not going to do every single step, you are a big boy you can handle it yourself.Originally Posted by harold
As you understand (I hope),
$\displaystyle x^4=3(\cos \pi +i\sin \pi)$
In order to use de Moiver's theorem you write,
$\displaystyle x^4=3[\cos (\pi +2\pi k)+i\sin (\pi +2\pi k)]$
Now you use de Moiver's theorem (actually it is not cuz his theorem if for integral exponents not rational).
Thus,
$\displaystyle x=3^{1/4}\left[\cos \left( \frac{\pi +2\pi k}{4} \right) +i\sin \left( \frac{\pi +2\pi k}{4} \right)\right]$
And each solution is for $\displaystyle k=0,1,2,3$
Ignore what I said before I made a mistake.
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This is the factorization
$\displaystyle
\left(
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)$$\displaystyle
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)$$\displaystyle
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) $$\displaystyle
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)
$
That is the factorization in $\displaystyle \mathbb{C}[x]$, good. Now find its factorization in $\displaystyle \mathbb{R}[x]$. The trick is to look for the conjugates. Remember that, $\displaystyle (x-z_1)(x-z_2)=x^2-x(z_1+z_2)+z_1z_2$, where $\displaystyle z_1+z_2$ is no longer complex and $\displaystyle z_1z_2$ is no longer complex.Originally Posted by harold
The first pair of conjugates are,
$\displaystyle
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)\cdot \left(x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)
$
The polynomial they produce is,
$\displaystyle x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$
The other pairs of conjugates are,
$\displaystyle
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)\cdot \left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)
$
The polynomial they produce is,
$\displaystyle x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$
So the factorization of,
$\displaystyle x^4+3$ in $\displaystyle \mathbb{R}[x]$ is,
$\displaystyle [x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}][x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}]$