# How to reduce x^4+4 into real factors

• Oct 11th 2011, 04:38 AM
terrorsquid
How to reduce x^4+4 into real factors
I am trying to factor a polynomial into irreducible factors in $\mathbb{R}$ and got it to the following stage:

$(x-1)(x+1)(x^4+4)$

I noticed that wolfram has it reduced even further by splitting $x^4+4$ into $(2-2 x+x^2) (2+2 x+x^2)$. I was just wondering how this was done? I understand how to split the $x^4+4$ using complex numbers; however, nothing occured to me in $\mathbb{R}$

Also, just out of interest, if you have an equation using the variable $m$, for example, and you want to talk about it within the set of real numbers, is it correct to say $\mathbb{R} [x]$ or should you say $\mathbb{R} [m]$?
• Oct 11th 2011, 04:48 AM
FernandoRevilla
Re: How to reduce x^4+4 into real factors
One way:

$x^4+4=x^4+4x^2-4x^2+4=(x^2+2)^2-(2x)^2=\ldots$
• Oct 11th 2011, 04:57 AM
terrorsquid
Re: How to reduce x^4+4 into real factors
:D too easy, thanks!
• Oct 11th 2011, 05:36 AM
CaptainBlack
Re: How to reduce x^4+4 into real factors
Quote:

Originally Posted by terrorsquid
I am trying to factor a polynomial into irreducible factors in $\mathbb{R}$ and got it to the following stage:

$(x-1)(x+1)(x^4+4)$

I noticed that wolfram has it reduced even further by splitting $x^4+4$ into $(2-2 x+x^2) (2+2 x+x^2)$. I was just wondering how this was done? I understand how to split the $x^4+4$ using complex numbers; however, nothing occured to me in $\mathbb{R}$

Also, just out of interest, if you have an equation using the variable $m$, for example, and you want to talk about it within the set of real numbers, is it correct to say $\mathbb{R} [x]$ or should you say $\mathbb{R} [m]$?

You should know that a real polynomial can be reduced to a product of real linear factors and real quadratic factors, then brute force will do the job:

$x^4+4=(x^2+ax+b)(x^2+cx+d)\\ \phantom{xxxx.xxx}=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc )x+bd$

gives the equations:

$ac+b+d=0$

$ad+cb=0$

$bd=4$

$a+c=0$

and the only real solutions to these equations gives the factorisation required.

CB
• Oct 12th 2011, 02:32 PM
HallsofIvy
Re: How to reduce x^4+4 into real factors
Here's how I would have done it: $x^4+ 4= x^4- (-4)= (x^2)^2- (2i)^2= (x^2- 2i)(x^2+ 2i)$. The two square roots of 2i are $\sqrt{2}+ i\sqrt{2}$ and $-\sqrt{2}- i\sqrt{2}$. The two square roots of -2i are $-\sqrt{2}+ i\sqrt{2}$ and $\sqrt{2}- i\sqrt{2}$ so linear factors are $(x- \sqrt{2}- i\sqrt{2})(x+ \sqrt{2}+ i\sqrt{2})(x+\sqrt{2}- i\sqrt{2})(x- \sqrt{2}+ i\sqrt{2})$

Since you want want real factors multiply $((x-\sqrt{2})- i\sqrt{2})((x-\sqrt{2})+ i\sqrt{2})= (x-\sqrt{2})^2- (i\sqrt{2})^2= x^2- 2\sqrt{2}x+ 2+ 2= x^2- 2\sqrt{2}x+ 4$ and $((x+\sqrt{2})+ i\sqrt{2})((x+\sqrt{2})- i\sqrt{2})$ $= (x+\sqrt{2})^2- (i\sqrt{2})^2= x^2+ 2\sqrt{2}x+ 2+ 2= x^2+ 2\sqrt{2}x+ 4)$ so that $x^4+ 4= (x^2- 2\sqrt{2}x+ 4)(x^2+ 2\sqrt{2}x+ 4)$