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Find all complex roots of polynomial

  1. Sep 30, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    Given 2 + 3i is one root of the equation:
    x^4 - 6x^3 + 26x^2 -46x +65 = 0
    Find the remaining roots.

    The attempt at a solution
    I am thinking about this as a possible solution although it is too long to be plausible, unless i'm wrong.
    let x = a + bi and then replace in the polynomial. Then, expand and finally, divide it by 2+3i

    Another possibility (i'm not sure) is this:
    x (x^3 - 6x^2 + 26x -46) +65 = 0
    So, i try to solve the cubic equation: x^3 - 6x^2 + 26x -46
    However, note that one solution to x, is not equal to -65. So, this method is wrong, i believe.

    I frankly don't know how else to go about to solve this one.
     
  2. jcsd
  3. Sep 30, 2011 #2

    SteamKing

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    What other property about complex roots do you know? (Hint: they always come in pairs)
     
  4. Sep 30, 2011 #3

    sharks

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    So, 2-3i is also a root? How do i confirm that? By replacing in the polynomial, it should equal zero, right?

    But what about the other roots? Since the maximum power of x is 4, so i assume that there are 4 roots in all.
     
  5. Sep 30, 2011 #4

    sharks

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    I still can't figure it out. Can someone please give me a hint?
     
  6. Sep 30, 2011 #5

    LCKurtz

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    If you have two roots p and q, then (x-p)(x-q) is a factor. Use long division to divide out that factor and see what's left.
     
  7. Sep 30, 2011 #6

    SammyS

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    Notice that when p & q are complex conjugates, (x-p)(x-q) can be written as a sum times a difference which comes out quite nicely as a difference of squares.

    (x - (2-3i))(x - (2+3i)) = ((x-2) + 3i)((x-2) - 3i) = ...
     
  8. Sep 30, 2011 #7

    sharks

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    OK, here is what i did:
    The two known factors are: (2+3i) and (2-3i)
    So, the product is: (2+3i)(2-3i) = x^2 - 4x + 13

    Using this product as divisor and the polynomial as dividend, i get this:
    x^2 - 2x + 5
    OK, i got it. Thanks guys. :)
     
  9. Sep 30, 2011 #8

    SammyS

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    So, what are the two remaining roots?
     
  10. Sep 30, 2011 #9

    sharks

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    Initially, i did it like this:
    x(x-2) = -5
    So x = -5 and x = -3
    But then i don't know why but these answers are wrong (from the book).

    So, i used quadratic formula
    and i got these: 1-2i, 1+2i, which are correct from the book.

    But again, i don't understand why i can't use the first set of simpler roots; -5 and -3.
     
  11. Sep 30, 2011 #10

    SammyS

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    You can't use -5 & -3 because they're not solutions of x(x-2) = -5. Plug them in to see that.
     
  12. Sep 30, 2011 #11

    HallsofIvy

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    Basic rule: If ab= 0 then either a= 0 or b= 0. But that is a special property of 0. You cannot say "if ab= -5 then either a= -5 or b= -5" which is what you were trying to do.
     
  13. Oct 1, 2011 #12

    sharks

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    I think i have this one wrapped up. Thanks for the help. :)
     
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