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Math Help - How to reduce x^4+4 into real factors

  1. #1
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    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]?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    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
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  3. #3
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    Re: How to reduce x^4+4 into real factors

    too easy, thanks!
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    Grand Panjandrum
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    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 \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
    Last edited by CaptainBlack; October 11th 2011 at 05:48 AM.
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  5. #5
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    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)
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