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College algebra

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What is the fundamental theorem of algebra for x^6-64?

  • College algebra -

    I would understand it to say that
    x^6 - 64 = 0 has 6 roots

    Proof:
    first treat x^6 as a difference of squares
    x^6 - 64
    = (x^3 - 8)(x^3 + 8)
    now as a difference of cubes for the first factor, and a sum of cubes for the second part
    = (x-2)(x^2 + 2x + 4)(x+2)(x^2 - 2x + 4)

    so x^6 - 64 = 0
    (x-2)(x^2 + 2x + 4)(x+2)(x^2 - 2x + 4) = 0

    x-2=0 ----> x=2
    x+2=0 ----> x=-2
    x^2+2x+4)=0 ---> x = (-2 ± √-12)/2 = -1 ± √3 i
    x^2 - 2x + 4=0 --> x = (2 ± √-12)/2 = 1 ± √3 i

    6 roots!

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