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Can someone factor this for me?

i could only get up to:

  • Mica -

    That's as far as you can go using rational numbers

    (I ran x^3-2x^2-4x-8 = 0 through my "cubic equation solver and got one messy real root and 2 complex roots)

  • Mica -

    would there be another way to find the x intercepts? Cause that's what i was trying to achieve

  • Mica -

    you could sketch it and see appr. where it crosses the x-axis.
    let f(x) = x^3 - 2x^2 - 4x - 8

    If there are any rational factors, then test for
    f(x) = 0 using x = ±1, ±2,±4, and ±8

    none work, so there are no rational factors.

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