Factorise Polynomial

**courteous** · September 14th 2008, 05:15 AM

$x^4-x^2+1$ Factorize.
I've set $x^2=t \rightarrow t^2-t+1=0$ . Using the formula, I get $x_{1,2}=\pm\sqrt\frac{1+i\sqrt3}{2} x_{3,4}=\pm\sqrt\frac{1-i\sqrt3}{2}$ What now?!

**bkarpuz** · September 14th 2008, 01:09 PM

Originally Posted by courteous

$x^4-x^2+1$ Factorize.
I've set $x^2=t \rightarrow t^2-t+1=0$ . Using the formula, I get $x_{1,2}=\pm\sqrt\frac{1+i\sqrt3}{2} x_{3,4}=\pm\sqrt\frac{1-i\sqrt3}{2}$ What now?!

$x^4-x^2+1=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})$

is what you want

**icemanfan** · September 14th 2008, 01:29 PM

$\pm\sqrt\frac{1+i\sqrt3}{2}$ can be rewritten as $\pm\sqrt{e^{i\frac{\pi}{3}}}$ and then you can determine the roots of that to be $e^{i\frac{\pi}{6}}$ and $e^{i\frac{7\pi}{6}}$ . For the other roots, the approach is similar.

**courteous** · September 14th 2008, 02:04 PM

Originally Posted by bkarpuz

$x^4-x^2+1=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})$

is what you want

Indeed, but what are $x_{1}, x_{2}, x_{3}, x_{4}$ , and most importantly, how do you get them?

Originally Posted by icemanfan

$\pm\sqrt\frac{1+i\sqrt3}{2}$ can be rewritten as $\pm\sqrt{e^{i\frac{\pi}{3}}}$

icemanfan, you are beyond me

, care to explain? Why is it, that $\frac{1+i\sqrt3}{2} = e^{i\frac{\pi}{3}}$

The solution should be $(x^2+\sqrt3x+1)(x^2-\sqrt3x+1)$ . How can you get it?

**bkarpuz** · September 14th 2008, 05:48 PM

Originally Posted by courteous

Indeed, but what are $x_{1}, x_{2}, x_{3}, x_{4}$ , and most importantly, how do you get them?

icemanfan, you are beyond me

, care to explain? Why is it, that $\frac{1+i\sqrt3}{2} = e^{i\frac{\pi}{3}}$

The solution should be $(x^2+\sqrt3x+1)(x^2-\sqrt3x+1)$ . How can you get it?

$x_{i}(i=1\ldots4)$ are the roots you have posted in your first message ( see Algebric proofs Section at Fundamental theorem of algebra - Wikipedia, the free encyclopedia )

$x+iy=r\mathrm{e}^{i\theta}$ , where $r=\sqrt{x^{2}+y^{2}}$ and $\theta=\mathrm{atan}(y/x)$ ( see Euler's formula - Wikipedia, the free encyclopedia ).

**log(xy)** · September 15th 2008, 07:21 AM

what you do is $x^4-x^2+1$

and just start doing normal factoring, dont use equations, this way will take you much shorter time...

unless it states otherwise, this would be the easiest way to do it

**courteous** · September 15th 2008, 08:41 AM

Originally Posted by log(xy)

what you do is $x^4-x^2+1$

and just start doing normal factoring, dont use equations, this way will take you much shorter time...

But what if you do not "see" what the factor is (in this case, $(x^2+\sqrt3x+1)(x^2-\sqrt3x+1)$ ; some tips to see such beast?), or if introducing "new" variable gets you nowhere (in this case, $t^2-t+1=0$ )?

What are considered best practices (in this case)?

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