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The hardest polynomial

  1. Sep 27, 2007 #1
    i don't know how to factorize it in reals.
  2. jcsd
  3. Sep 27, 2007 #2


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    Maybe because it can't be factored over the reals?
  4. Sep 27, 2007 #3
    but i read that all pol. can be factored in reals and the higher power of x can be 2
  5. Sep 27, 2007 #4

    c,a,d,f =/= 0
  6. Sep 27, 2007 #5
    X^2 +1=0, this polynominal can be factored over the reals?
  7. Sep 27, 2007 #6

    D H

    Staff: Mentor

    All polynomials can be factored in the complex numbers, not the reals.

    This has no real roots. Hint: The expression is quadratic in x2.
  8. Sep 27, 2007 #7


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    Isn't that statement equivalent to the Mertens conjecture?
  9. Sep 27, 2007 #8


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    Fundamental theorem of real algebra:
    Every monic polynomial can be uniquely factored into a product of monic irreducible polynomials. Any irreducible polynomial is either linear or quadratic.​
    Last edited: Sep 27, 2007
  10. Sep 27, 2007 #9
    Guys, he's saying that all polynomials with real coefficients can be factors as (at most) quadratics with real coefficients. This is true.
    Last edited by a moderator: Sep 27, 2007
  11. Sep 27, 2007 #10

    D H

    Staff: Mentor

    As the equation has no real roots, you are looking for the product of a pair of quadratics.
    You don't need a and d. Since ad=1, you can scale the two polynomials to make a and d equal to 1.

    This is the source of your problems. Try again.
  12. Sep 27, 2007 #11

    D H

    Staff: Mentor

    Did you read the guidelines? Don't post complete solutions.
  13. Sep 27, 2007 #12
    Apologies -- got lazy.
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