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March 26, 2017

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What are the roots, real and imaginary of:

Y=x^4-6x^3+9x^2-6x+8

  • Precal - ,

    Y = X^4 - 6X^3 + 9X^2 -6X + 8 = 0,

    It was determined by trial and error
    that when X = 2, Y = 0. Therefore, 2
    is a real solution.

    X = 2,
    X - 2 = 0,
    Using synthetic division, divide the
    4th degree Eq by X - 2 and get:

    X^3 -4X^2 + X - 4.
    Now we have:
    Y = (X - 2)(X^3 - 4X^2 + X - 4 = 0,
    In the cubic Eq,when X = 4, Y = 0:
    X = 4,
    X - 4 = 0.
    We divide the cubic Eq by X - 4 and
    get:

    X^2 + 1.
    The factored form of our 4th degree Eq is:

    Y = (X - 4)(X - 2)(X^2 + 1) = 0.
    X - 4 = 0,
    X = 4.

    X - 2 = 0,
    X = 2.

    X^2 + 1 = 0,
    X^2 = -1,
    X = sqrt(-1) = i.

    Solution Set: X = 4, X = 2, X = i.
    So we have 2 real and 1 imaginary sol.

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