Math

Find all possible rational roots using the rational root theorem.

3x^3 + 2x^2 - 1 = 0

plus/minus 1/3 and plus/minus 1?

asked by Anonymous
  1. Look at the coefficient of your leading term and your last term

    3x^3 <-- leading term

    1 <-- last term

    finding numbers that can divide into each term evenly

    Let q correspond to the leading term:

    q = +-1, +-3 (only numbers that can divide evenly with 3)

    let p correspond to the last term

    p = +1 (only numbers that can divide evenly into 1)

    now possible rational roots is just:

    q/p

    +-1/+-1, +-3/+-1

    posted by DonHo
  2. oops ignore the last part:

    It's p/q

    so roots are

    +-1/+-1 and +-1/+-3

    http://www.mathwords.com/r/rational_root_theorem.htm

    posted by DonHo
  3. let f(x) = 3x^3 + 2x^2-1
    f(1) = 3 + 2 - 1 ≠ 0
    f(-1) = -3 + 2 - 1 ≠ 0
    f(1/3) = 1/9 + 2/9 - 1 ≠ 0
    f(-1/3) ≠ 0

    So there are no "nice" roots
    looking at the graph, there appears to be a real solution at appr x = 0.5
    and two complex roots (the graph crosses the x-axis only once)

    posted by Reiny

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