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Factorizing a polynomial using complex numbers

  1. Nov 23, 2008 #1

    finding 1 root and using synthetic division we can factorize to:


    using complex numbers to factorize (x2+2x+5) we have (x+1)(x-2i)(x+2i), and so our final solution is:


    is this correct?
    Last edited: Nov 23, 2008
  2. jcsd
  3. Nov 23, 2008 #2


    Staff: Mentor

    So far, so good.
    It should be obvious that x^2 + 2x + 5 doesn't factor to (x + 1)(x - 2i)(x + 2i). In any case, this is not a calculus question.
  4. Nov 23, 2008 #3
    i jus started in complex numbers today and thats what i came up with.. if someone could please move the thread to the correct place that would be appreciated..

    Mark44, Can you please advise how i should go about finding the solution to that part of the question?
  5. Nov 23, 2008 #4


    User Avatar
    Homework Helper

    For x2+2x+5, how would you find the roots? If they aren't rational, is there a formula you can use?
  6. Nov 23, 2008 #5
    there is an example in my notes where the quadratic equation has been used for polynomials with no real solutions. Using the same principal here we have:

    x = (-2 +- sqrt(-16)) / 2

    now, the square root of 16 is 4 so we have

    x = (-2 +- 4i)/2
    = -1 +- 2i

    and thats how i got my roots.
  7. Nov 23, 2008 #6
    is this not correct?
  8. Nov 23, 2008 #7
    (post corrected)
    all the roots you got are correct but the way you reconstructed the polynomial is not:
    it should be

    f(x)= (x+2)( x-(-1+2i) )( x-(-1-2i) ) (post was initially incorrect, but is now correct)

    notice that x only appears 3 times, which is the same as the degree of the polynomial, this will always be true

    lemme know if you have any other questions :smile:
    Last edited: Nov 23, 2008
  9. Nov 23, 2008 #8
    thank you logicalTime. It makes perfect sense now.
  10. Nov 23, 2008 #9
    actually.. i just noticed something when i actually went back and wrote it... shouldnt it really be (x+(1-2i))(x+(1+2i))? because x = - 1 is x + 1 = 0 for both solutions so the only thing that should change in the outcomes is the sign in front of the 2i

    why does one of the factors in your solution have x - 1?
  11. Nov 23, 2008 #10
    (post corrected)
    whoops, I messed up since the roots are -2, -1-2i, -1+2i then reconstructing the polynomial is:

    f(x)= (x+2)(x-(-1-2i))(x-(-1+2i))

    so when you plug in a root, one of the terms becomes zero making the whole thing equal to zero

    should be correct now, sry bout that
    Last edited: Nov 23, 2008
  12. Nov 23, 2008 #11
    actually i have x=-1 so should be (x+(1-2i))(x+(1+2i))... not (x-(1-2i))(x-(1+2i))
    Last edited: Nov 23, 2008
  13. Nov 23, 2008 #12
    agh, it's late and I'm being sloppy, time for bed
    I corrected my previous mistakes they are a little different format from yours, just because I like to make it obvious that plugging in the root leads to zero.
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