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Complex roots of a quartic polynomial

  1. Jan 10, 2014 #1
    The polynomial [itex]z^4 + 2z^3 + 9z^2 - 52z + 200 = 0[/itex] has a root [itex]z=-3+4i[/itex]. Find the other 3 roots.


    Since the given root is complex, one of the other roots must be the complex conjugate of the given root. So the 2nd root is [itex]z=-3-4i[/itex]. To find the other roots, I divided the polynomial by [itex]z^2 + 6z + 13[/itex] (this is the product of the 2 known roots), which gave [itex]z^2 - 4z + 20[/itex] with remainder [itex]-120z - 60[/itex]. I don't know how to proceed from here because I haven't done many examples where you get a remainder after doing algebraic division. What should I do?
     
  2. jcsd
  3. Jan 10, 2014 #2

    Student100

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    Check the product you're using from the two known roots.
     
  4. Jan 10, 2014 #3
    I meant the product of the 2 factors (not roots). So [itex](z - (-3 + 4i))(z - (-3 - 4i)) = z^2 + 6z +13[/itex], right?
     
  5. Jan 10, 2014 #4

    Student100

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    I get something different, do a quick check again.

    Maybe this will help you some more, write it like [itex]((z+3)^2 +16)[/itex] and look at why you're able to do that.
     
    Last edited: Jan 10, 2014
  6. Jan 10, 2014 #5

    epenguin

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    A remainder is a sign there is an error. If the quadratic is a factor of the quartic there will be no remainder from division.

    So since it is certain there is an error the only question remaining is whether it's yours or of the guy who set the question. :biggrin:
     
  7. Jan 10, 2014 #6

    Student100

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    Pretty sure it is his at the moment. : )
     
  8. Jan 10, 2014 #7
    Yup, I've got it now. Thanks :)
     
  9. Jan 11, 2014 #8

    HallsofIvy

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    You might find it easier to do as ((z+3)- 4i)((z+3)+4i) which is the product of a "sum and difference" and so equal to [itex](z+ 3)^2- (4i)^2[/itex].
     
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