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Math Help - Factoring Complex Polynomials

  1. #1
    Junior Member
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    Factoring Complex Polynomials

    Hi

    I am asked to show all the seventh roots of -128 in polar form, no probs. But I am then asked

    Factorise the polynomial x^7+128 into polynomial factors with
    real coefficients, where the factors are either linear or quadratic. Give exact values for the coefficients, which may involve
    trigonometric functions.

    Not sure where to begin.

    Thanks in advance
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  2. #2
    Senior Member pacman's Avatar
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    x^7 + 128 = x^7 + 2^7 = (x + 2)(x^6 - 2x^5 + 4x^4 - 8x^3 + 16x^2 - 32x + 64)
    Last edited by pacman; September 3rd 2009 at 08:54 PM.
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  3. #3
    Member alunw's Avatar
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    If you draw a picture you will see that there is one real root: -2 or and that the other roots come in pairs that are complex conjugates of each other. If you multiply the linear terms you get from each of these factors together you will get a quadratic polynomial with real coefficients. If you express the roots use a general trigonometric formula (no doubt involving various multiples of 2*pi/7 radians) it should be easy enough to get the required factorisation.
    Last edited by alunw; September 3rd 2009 at 07:10 AM. Reason: mistake corrected
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  4. #4
    Senior Member pacman's Avatar
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    Real Root: x1 = -2

    Complex Roots:

    x2 = 2(-1)^(1/7),

    x3 = -2(-1)^(2/7),

    x4 = 2(-1)^(3/7),

    x5 = -2(-1)^(4/7),

    x6 = 2(-1)^(5/7),

    x7 = -2(-1)^(6/7).

    Now, you have an idea how to convert that to polar form as suggested by the the graph
    Attached Thumbnails Attached Thumbnails Factoring Complex Polynomials-.gif  
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