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Factorize in Q[x] , R[x] and C[X]

  1. Jul 21, 2006 #1
    Factorize in Q[x] , R[x] and C[X]

    [tex] 4x^6 + 8x^5 - 3x^4 - 19x^3 - 26x^2 - 15x - 3 [/tex]


    I couldnt finish it but this is up to where I did:

    [tex] 4(x+1/2)^2 (x^4+x^3-2x^2-3x-3) [/tex]
     
  2. jcsd
  3. Jul 21, 2006 #2

    arildno

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    Your factorization cannot be correct.
    In the original one, 1 is not a root, whereas 1 is a root in your second one.
     
  4. Jul 21, 2006 #3
    Im sorry I cant see that.
    In the second gives -6
     
  5. Jul 21, 2006 #4

    arildno

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    Sorry, my mistake.
     
  6. Jul 21, 2006 #5

    0rthodontist

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    Well, my calculator says that the remaining real solutions of your fourth-degree factor are [tex] \pm \sqrt{3}[/tex] but I don't know how you could find those by hand.
     
  7. Jul 23, 2006 #6
    Yes thats right thanks
     
  8. Jul 23, 2006 #7
    That was a problem in my exam
     
  9. Jul 23, 2006 #8

    0rthodontist

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    Well--how do you find them by hand then, seeing how they are irrational?
     
  10. Jul 23, 2006 #9

    StatusX

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    It's easy to check if a polynomial has any rational roots. To see if it has irrational roots of the form [itex]\sqrt{r}[/itex] for r rational, you can break up the polynomial into even and odd terms and substitute in [itex]\sqrt{r}[/itex], which will leave something like [itex]e(r)+o(r) \sqrt{r} [/itex]. For r rational, this will only be zero if e(r)=o(r)=0. In this case,

    [tex] x^4+x^3-2x^2-3x-3 = (x^4+2x^2-3) + (x^3-3x)=0[/tex]

    plugging in x= [itex]\sqrt{r}[/itex]:

    [tex](r^2+2r-3) + \sqrt{r} (r-3) [/tex]

    We see that r=3 is a solution to both terms, and so [itex]x=\pm \sqrt{3}[/itex] is a root of the original polynomial. This process can be generalized to cube roots, etc, but I doubt if it's very useful unless you know in advance that you'll have a root of this form.
     
    Last edited: Jul 23, 2006
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