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anonymous

  • 5 years ago

Find the equation of a polynomial function of lowest degree with real coefficients that has 6 and 3 - 2i as solutions.

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. amistre64
    • 5 years ago
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    (x-6)(x-(3-2i))

  3. amistre64
    • 5 years ago
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    (x-6)(x-3+2i) x^2 -3x +2xi -6x -12i+18 -------------------- x^2 -9x +18 +(2xi -12i)

  4. anonymous
    • 5 years ago
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    amistre: I dont think that this is with real coefficients

  5. amistre64
    • 5 years ago
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    im working on it lol

  6. anonymous
    • 5 years ago
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    I got it now i just forgot the first step

  7. anonymous
    • 5 years ago
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    ok, cool :-)

  8. amistre64
    • 5 years ago
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    cool; was I close?

  9. amistre64
    • 5 years ago
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    i think we square this part tho right? 3 - 2i

  10. amistre64
    • 5 years ago
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    its a cubic; with a bend above the line..

  11. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=%28x-6%29%28x-3%2B2i%29^2 might be useful

  12. anonymous
    • 5 years ago
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    Well when you have (3-2i) my definition you also have the conjugate (3+2i)

  13. anonymous
    • 5 years ago
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    *by definition*

  14. anonymous
    • 5 years ago
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    if your curious it was x^3-12x^2+49x-78

  15. anonymous
    • 5 years ago
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    f(x) = " "

  16. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=%28x-6%29%28x-%283-2i%29%29%28x-%283%2B2i%29%29

  17. amistre64
    • 5 years ago
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    i was curious :) but yeah, thats what I came up with eventually

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