Results 1 to 5 of 5

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

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    196

    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 $\displaystyle \mathbb{R} [m]$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

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

    One way:

    $\displaystyle x^4+4=x^4+4x^2-4x^2+4=(x^2+2)^2-(2x)^2=\ldots$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2011
    Posts
    196

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

    too easy, thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5

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

    Quote Originally Posted by terrorsquid View Post
    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
    Last edited by CaptainBlack; Oct 11th 2011 at 05:48 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,781
    Thanks
    3030

    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 $\displaystyle x^4+ 4= (x^2- 2\sqrt{2}x+ 4)(x^2+ 2\sqrt{2}x+ 4)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex (non-real) factors of a polynomial
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Feb 19th 2011, 02:56 PM
  2. Reduce
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Oct 25th 2010, 11:21 AM
  3. Replies: 1
    Last Post: Dec 7th 2009, 10:42 AM
  4. Is there a way to further reduce this?
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Sep 24th 2009, 04:06 AM
  5. Factors of 2310 and Factors of 1365
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 7th 2008, 06:56 PM

Search Tags


/mathhelpforum @mathhelpforum