How to reduce x^4+4 into real factors

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

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

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

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

Re: How to reduce x^4+4 into real factors

One way:

$\displaystyle x^4+4=x^4+4x^2-4x^2+4=(x^2+2)^2-(2x)^2=\ldots$

Re: How to reduce x^4+4 into real factors

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 $\displaystyle \mathbb{R}$ and got it to the following stage:

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

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

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

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:

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

gives the equations:

$\displaystyle ac+b+d=0$

$\displaystyle ad+cb=0$

$\displaystyle bd=4$

$\displaystyle a+c=0$

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

CB

Re: How to reduce x^4+4 into real factors

Here's how I would have done it: $\displaystyle x^4+ 4= x^4- (-4)= (x^2)^2- (2i)^2= (x^2- 2i)(x^2+ 2i)$. The two square roots of 2i are $\displaystyle \sqrt{2}+ i\sqrt{2}$ and $\displaystyle -\sqrt{2}- i\sqrt{2}$. The two square roots of -2i are $\displaystyle -\sqrt{2}+ i\sqrt{2}$ and $\displaystyle \sqrt{2}- i\sqrt{2}$ so linear factors are $\displaystyle (x- \sqrt{2}- i\sqrt{2})(x+ \sqrt{2}+ i\sqrt{2})(x+\sqrt{2}- i\sqrt{2})(x- \sqrt{2}+ i\sqrt{2})$

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