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algebra

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According to the rational root theorem, which is not a possible rational root of x3 + 8x2 x 6 = 0?

  • algebra -

    verify that you meant
    x^3 + 8x^2 + 6 = 0

  • algebra -

    x3 + 8x2-x-6 = 0?

  • algebra -

    x^3 + 8x^2 - x - 6 = 0

    Let f(x) = x^3 + 8x^2 - x - 6
    f(1) = 1+8-1-6 ≠ 0
    f(-1) = -1 + 8 + 1 - 6 ≠ 0
    f(2) = 8 + 32 - 2 - 6 ≠ 0
    f(-2) = -8 + 32 + 2 - 6 ≠ 0
    f(3) = 27 + .... ≠ 0
    f(-3) = -27 + 72 ... ≠ 0

    numbers which are NOT possible rational roots are
    ±1 , ±2 , ± 3

    Is that what you wanted?

  • algebra -

    yeh but my homework has +-2 and +-1 as answers abd i can only choose one

  • algebra -

    I used the 1 , 2, and 3 since they were factors of the 6 at the end.
    There are an infinite number of choices of rational numbers which are NOT possible roots.
    This is a poorly worded question.

    go with the ±1 and ±2

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