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Math Help - Factor a cubic without using CAS

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    Factor a cubic without using CAS

    How do you factor 24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1) without a computer? (Obviously, I used a computer).
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    Re: Factor a cubic without using CAS

    Quote Originally Posted by MSUMathStdnt View Post
    How do you factor 24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1) without a computer? (Obviously, I used a computer).
    Are you aware of the Rational Root Theorem?

    For convenience I'm going to divide the polynomial by 6:
    4x^3 - 16x^2 + 17x - 5 = 0

    Possible rational roots to this polynomial equation are:
    \pm \left ( 5,~\frac{5}{2},~\frac{5}{4},~1,~\frac{1}{2},~\frac  {1}{4} \right )

    That gives you 12 possible rational roots that you can check.

    If you don't know that one, then you're probably stuck with Cardano's method. (Cardano's method is about 1/2 way down the page.)

    -Dan
    Last edited by topsquark; December 10th 2012 at 11:50 AM.
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    Re: Factor a cubic without using CAS

    Quote Originally Posted by MSUMathStdnt View Post
    How do you factor 24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1) without a computer? (Obviously, I used a computer).
    If we just start with 24x^3-96x^2+102x-30, look for nice roots.

    x=1 sort of pops out. So (x-1) must be a factor.

    Once we have one factor then by division we can reduce that to a quadratic which can factored.

    I understand that it is not always possible to find a root that easily.
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