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Math Help - Complex Zeros in polynomials

  1. #1
    Ife
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    Unhappy Complex Zeros in polynomials

    I need a little help with this please:

    Use the given zero to find the remaining zeros of the polynomial:

    h(x) = x^4 - 9x^3 + 21x^2 +21x -130; zero: 3-2i

    Now i kno all about the zeros when complex existing with their conjugates, i know all about factoring using the different rules and theorems. Had this been a real root, i would've had to divide using long division and testing for roots to factor further, but with this complex one, i am not sure how to do this. Can long division be done on a complex root??? I am not seeing the way of this one. I did one before that had a zero of -2i, but now that the 3 is there i am not sure what to do. Can i get some help please????

    Thanks a lot.
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  2. #2
    MHF Contributor red_dog's Avatar
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    If x_1=3-2i then another root is x_2=3+2i.

    Now divide the polynomial by (x-x_1)(x-x_2)=x^2-6x+13.
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  3. #3
    Ife
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    thanks, but...

    ok this i know... but how do i do this division, now? and also, how do you get those well laid out math symbols on this or on the pc on the whole? can you give me a little help on that division? i think i only know how to divide by a simple divisor, one of degree 1 (eg x+1)...
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  4. #4
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    You can do synthetic division by complex roots. It's a little tricky, but it can be done.

    First, divide x^4 - 9x^3 + 21x^2 +21x -130 by 3 - 2i. We know that the remainder will be zero:

    Code:
    3 - 2i| 1  -9      21     21     -130
    -------     3-2i  -22+6i   9+20i  130
          --------------------------------
            1  -6-2i   -1+6i  30+20i    0
    Now, let's divide the quotient by 3 + 2i, the conjugate of 3 - 2i:
    Code:
    3 + 2i| 1  -6-2i   -1+6i  30+20i
    -------     3+2i   -9-6i -30-20i
          ---------------------------
            1  -3     -10          0
    So you're left with the quadratic
    x^2 - 3x - 10 = 0
    which factors into
    (x + 2)(x - 5) = 0

    The two remaining roots are -2 and 5.


    01
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  5. #5
    MHF Contributor red_dog's Avatar
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    For long division see this: Polynomial Long Division

    To write mathematical symbols see Latex Tutorial in Latex help section on this forum.
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  6. #6
    Ife
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    Quote Originally Posted by yeongil View Post
    You can do synthetic division by complex roots. It's a little tricky, but it can be done.

    First, divide x^4 - 9x^3 + 21x^2 +21x -130 by 3 - 2i. We know that the remainder will be zero:

    Code:
    3 - 2i| 1  -9      21     21     -130
    -------     3-2i  -22+6i   9+20i  130
          --------------------------------
            1  -6-2i   -1+6i  30+20i    0
    Now, let's divide the quotient by 3 + 2i, the conjugate of 3 - 2i:
    Code:
    3 + 2i| 1  -6-2i   -1+6i  30+20i
    -------     3+2i   -9-6i -30-20i
          ---------------------------
            1  -3     -10          0
    So you're left with the quadratic
    x^2 - 3x - 10 = 0
    which factors into
    (x + 2)(x - 5) = 0

    The two remaining roots are -2 and 5.


    01
    This helps a lot, i thought it was possible, but then i have never seen it done before so i wasn't sure.. thanks a lot! i knew the solution couldn't have been that hard. THanks
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