algerbra 2

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what are the other roots of 3x^4-12x^3+41x-8x+26 that include 2-3i

  • algerbra 2 -

    Assume you mean:
    3x^4-12x^3+41x^2-8x+26

    Well then 2+3i must be one
    multiply them
    (x -2+3i)(x - 2-3i) = x^2 -4x +13
    divide by that and get
    3x^2+2
    so the other two roots are
    x=sqrt(2/3) and x = -2/3

  • algerbra 2 -

    First of all, I will assume you have a typo and the third term should be 41x^2

    complex roots always come in conjugate pairs, that is,
    if 2-3i is a root, so is 2+3i
    sum of those roots = 4
    product of those roots = 3-9i^2 = 13
    so the quadratic that yields those two roots is
    x^2 - 4x + 13

    I then did an algebraic long division by x^2 - 4x + 13 and that came out very nicely to an answer of 3x^2 + 2 with no remainder.
    so ...

    3x^4-12x^3+41x^2-8x+26= 0
    (x^2 - 4x + 13)(3x^2 + 2) = 0
    x = 2-3i, 2+3i from the first part and

    x^2 = -2/3
    x = ± √-(2/3)
    = ± i√(2/3) or ±(1/3)i √6

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