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Factorizing X^4 + 1

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data
    I've seen a factorization of X^4 + 1 like

    [itex] x^{4} + 1 = (X^{2} + \sqrt{2}X + 1)(X^{2} - \sqrt{2}X + 1)[/itex]

    Is there an intuitive step-by-step procedure to find this factorization?

    2. Relevant equations



    3. The attempt at a solution
    NO IDEA..
     
  2. jcsd
  3. Jun 18, 2012 #2
  4. Jun 18, 2012 #3
    You do this by adding and subtracting terms.

    [tex]X^{4} + 1 + (2 X^{2} - 2 X^{2}) = (X^{4} + 2 X^{2} +1) - 2 X^{2}[/tex]
    This can be rewritten as
    [tex](X^{2} + 1)^{2} - (\sqrt{2} X)^2[/tex]
    which is in the form of a square minus a square and hence can be directly factorised to give the result you want.
     
  5. Jun 18, 2012 #4
    [itex]x^{4}+1[/itex]
    [itex]=x^{4}+2x^{2}-2x^{2}+1[/itex]
    [itex]=(x^{2}+1)^{2}-2x^{2}[/itex]
    [itex]=(x^{2}+1-\sqrt{2}x)(x^{2}+1+\sqrt{2}x)[/itex]
    [itex]=(x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)[/itex]
     
  6. Jun 18, 2012 #5
    I wouldn't have thought of it. Perhaps there is a systematic way in finding the complex roots, take [itex]x^8=1[/itex], which factors into [itex](x^4-1)(x^4+1)[/itex], which has roots [itex]\{\zeta,\zeta^2,\zeta^3,\dots,\zeta^7\}[/itex], where [itex]\zeta=e^{i2\pi/8}[/itex]. So we discard the roots to [itex]x^4-1[/itex], so the roots for [itex]x^4+1[/itex] are [itex]\{\zeta,\zeta^3,\zeta^5,\zeta^7\}.[/itex] Then notice that the conjugate of [itex]\zeta[/itex] is [itex]\zeta^7[/itex]. So [itex]\omega=\zeta+\zeta^7[/itex] is a real number. Maybe try to find a polynomial that it [itex]\omega[/itex] is a root of.

    I'm not testing any of this, just off the top of my head. But I should really ask first, what class is this for. If you're taking an abstract algebra course, than the steps I mentioned above might be closer to expected. Otherwise, I just don't think you'd be expected to find such a factorization.
     
  7. Jun 18, 2012 #6
    I think when something seems unable to be factored as it's written, the next step is to try completing the square, which is what Fightfish and I did by adding and subtracting 2x2
     
  8. Jun 18, 2012 #7
    Any expression of the form [itex]x^n+1[/itex], where n is a power of 2 can be written as:

    [tex]x^n+1=(x^{\sqrt n}+1)^2-K^2[/tex]
    [tex]K^2=(x^{\sqrt n}+1)^2-x^n-1=x^n+2x^{\sqrt n}+1-x^n-1=2x^{\sqrt n}[/tex]

    This means that

    [tex]x^n+1=(x^{\sqrt n}+1)^2-K^2=(x^{\sqrt n}+1)^2-2x^{\sqrt n}=[/tex]
    [tex]=(x^{\sqrt n}+1+\sqrt{2}x^{n^{ \frac{1}{4}}})(x^{\sqrt n}+1-\sqrt{2}x^{n^{ \frac{1}{4}}})[/tex]

    Let's assume that x was positive to keep it simple :biggrin:
     
  9. Jun 19, 2012 #8
    Okay I think I see what you are trying to say. And indeed I'm taking an algebra course [ring, field, gloais ...]. But the thing is in this course I've never learned a techniqe that is by knowing some roots to generate a polynomial the complete list of the roots of which are the roots known, nor to find a structure of a factorization, whatsoever. This will be an interesting study I guess, but seems too complex at this stage for me. I think the solutions other guys proposed are quite sufficuent.
     
  10. Jun 19, 2012 #9
    true, i guess i should have said even for algebra, you might not look at this till after rings and galois theory.

    I'm curious, what was the context in which you saw this factorization. Was it part of your current curriculum? Were you reading ahead?

    I guess I like the idea that somehow (from high school algebra) we should have thought to complete the square. I'll have to remember that. Algebra seems so immense.

    My current guess for the relevancy to you would be that it's interesting that this is the best factorization we can get in the reals, while in complex, we can completely factor it, and it appears a bit different.
     
  11. Jun 19, 2012 #10
    In reals, completeing the square provides a factorization which was already provided. In the complex numbers, we find the roots of the equation. These are the fourth roots of unity in the complex plane. We can go like this: Rewrite the expression as [itex]x^4-(-1)[/itex] and use the difference of two squares to get [itex](x^2-i)(x^2+i)[/itex]. Further factoring yields [itex](x-\sqrt{i})(x+\sqrt{i})(x-i\sqrt{i})(x+i\sqrt{i})[/itex], and this completely factors the expression.
     
  12. Jun 19, 2012 #11
    Okay. Btw, I've just finished this cousre and am preaparing the upcoming exam for this one.

    (It is just an undergraduate advanced courese, but personally I think the course is quite fluffy. I don't know... my knowledge is fluffy now, still not sure what I've learned from this course.. the main purpose was to prove that quintic polynomials or the higher do not have a radical solution. We've achieved it, but I just can't help feeling that his (lecturer) way of presenting basic materials are not that good. For further study I guess I should study by better textbooks like dummit or hungerford.)

    Anyways, so for the exam, was revising some tutorial questions. And now I remember in tutorial he just completed the squares for this one.
     
  13. Jun 19, 2012 #12
    Algebrat, I know this question was directed at julypraise, but I know that this factorization appears when integrating [itex]\int\sqrt{tanx}dx[/itex]. I think that's the first time I encountered it.
     
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