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Math Help - Given a complex polynomial and one of its zeroes, find all zeroes of the function

  1. #1
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    Given a complex polynomial and one of its zeroes, find all zeroes of the function

    1. f(x) = x^3 - 7x^2 + 17x - 15 ; 2+i
    2. f(x) = x^3 + 6x + 20 ; 1-3i
    3. g(x) = x^4 - 6x^3 + 6x^2 + 24x - 40 ; 3+i
    4. g(x) = x^3 - 3x^2 + 9x - 7 ; 1
    Last edited by mr fantastic; October 9th 2010 at 02:36 PM. Reason: Edited title.
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  2. #2
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    I see that you have over thirty postings.
    By now you should understand that this is not a homework service
    So you need to either post some of your work on a problem or explain what you do not understand about the question.
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  3. #3
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    Apologies. While these questions are indeed a few from my homework, I did not ask this simply to find the answers and get done with it, as I am sure is the source of mathhelpforum wanting to make clear that they are not a homework service. I figure that if I understand these 4, I will be able to do all of them. As per request, I'll work through as much as I can of the first one.

    x^3 - 7x^2 + 17x - 15; 2+i
    By the Conjugate Pair Theorem, both 2+i and 2-i are zeroes of the function. The goal is to synthetically divide by the zeroes until the function is down to a quadratic.

    2+i| 1 -7 17 -15
    |
    | 1 2+i -9-3i 19+3i
    ------------------
    1 -5+i 8-3i 4+3i

    Since 2+i is given as a zero, the remainder should be zero, and not (4+3i) Using 2-i gives similar results. I'm not understanding what I'm doing wrong.
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  4. #4
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    Have you tried expanding [x-(2+i)][x-(2-i)]?
    That will tell us what the other factor is.
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  5. #5
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    That works out to x^2 - 4x - 3 which, when applying the quadratic formula, works out to be 3 and 1.
    but isn't that four answers to a function that only goes to a power of 3? That's not possible.
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  6. #6
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    No you did the algebra incorrectly.
    It should be x^2-4x+5.

    Now divide that into the given polynomial.
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  7. #7
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    [x-(2+i)][x-(2-i)]
    x^2-(2+i)x-(2-i)x-(2+i)(2-i)
    x^2-2x+ix-2x-ix-(4-i^2)
    x^2-4x-(4+1)
    x^2-4x-3
    How did I do the algebra incorrectly?
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  8. #8
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    Quote Originally Posted by Rocker1414 View Post
    [x-(2+i)][x-(2-i)]
    x^2-(2+i)x-(2-i)x-(2+i)(2-i) Mr F says: I haven't read past this line, where the red minus should be a plus.

    x^2-2x+ix-2x-ix-(4-i^2)
    x^2-4x-(4+1)
    x^2-4x-3
    How did I do the algebra incorrectly?
    Personally, I think it's much easier to expand something of the form [x - (a + ib)][x - (a - ib)] by recognising it as the factorisation of the sum of two squares:

    = [(x - a) + ib][(x - a) - ib] = (x - a)^2 + b^2 = ....
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  9. #9
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    (x-z)(z-\overline{z})=x^2-(z+\overline{z})x+z\cdot \overline{z}=x^2-(z+\overline{z})x+|z|^2
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  10. #10
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    Thanks guys. I believe I have it figured out now even by synthetically dividing by both given zeroes. Expanding the form works just as well, plus saves paper Thanks again.
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