Precalculus

"Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the two real roots and the four imaginary roots of this equation."

I used synthetic division to get the real roots 2 and -1, but I can't figure out how to get the imaginary roots. I checked HotMath but all it showed me was how to get the real roots and then it practically gave me the imaginary roots without showing how to work it out. Does anyone know how to do this? Any help is GREATLY appreciated!! :D

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  1. how about this
    let y = x^3
    then your equation becomes
    y^2 - 7y - 8 = 0
    (y-8)(y+1) = 0
    so y = 8 or y = -1

    then x^3 = 8 or x^3 = -1
    x^3 - 8 = 0
    (x-2)(x^2 + 2x + 4) = 0
    this will give you the real root of x=2 and 2 imaginary roots from the quadratic

    or
    x^3 + 1 = 0
    (x+1)(x^2 - x + 1) = 0
    here x = -1 plus 2 more imaginary roots

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  2. Ohhhh duh!! I can't believe I totally overlooked that. Thanks Reiny!! :D

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