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Factoring polymonial with complex roots

  1. Jul 3, 2005 #1
    This may be a bit silly but i forget how to factor this into complex factors:

    s^2 + 6s + 25

    i know the answer is (s +3 - i4)(s +3 - j4)

    but how do i get that?
     
  2. jcsd
  3. Jul 3, 2005 #2

    lurflurf

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    You could use the quadratic formula.
     
  4. Jul 3, 2005 #3
    Yes, use the quadratic formula to find the roots of [itex]s^2 + 6s + 25 = 0[/itex] and then use the factor theorem: if f(a) = 0, then (x - a) is a factor of f(x). Your a here will be the complex number you get.

    Edit: no doubt dexter or someone will tell me this is wrong :rolleyes:.
     
  5. Jul 3, 2005 #4

    George Jones

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    Gold Member

    Set your expression equal to zero and the roots, i.e. find s = a and s = b such that

    0 = s^2 +6s +25.

    Then,

    0 = s^2 +6s +25
    = (s - a)(s - b).

    You could use the quadratic formula, but I think completing the square offers more insight.

    Write

    0 = s^ + 6s + c^2 - c^2 +25.

    Now find c such that

    s^ + 6s +c^2 = (s + c)^2.

    This means that 2c = 6 and c = 3. Therefore,

    0 = s^2 + 6s + 9 - 9 +25
    = (s+3)^2 +16

    So,

    (s + 3)^2 = -16.

    Regards,
    George
     
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