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I forgot how to factor cubic equations?

  1. Nov 12, 2011 #1
    the original equation is 4x^2-6x^2-12x+9=0

    I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
  2. jcsd
  3. Nov 13, 2011 #2
    Apply the rational root test to find a possible root.

    The possible rational roots will have the form p/q where p divides 9 and where q divides 4. So what are the possible rational roots?? Is one of these possibilities an actual root?
  4. Nov 13, 2011 #3


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    Ah, this looks like the "factoring by grouping" method, which, unfortunately, doesn't work for ALL cubic polynomials. The rational root test, as micromass mentioned, is your best bet.
  5. Nov 13, 2011 #4


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    Do you mean 4x^3 ?
  6. Nov 20, 2011 #5
    The equation you posted is not a cubic; it is a quadratic.

    4 x^2 - 6 x^2 . . .

    equals -2 x^2 . . .

    Why didn't you combine the x^2 terms?

    In any case, equations exist for solving a quadratic or cubic equation. Just do a search; I am sure you will find the appropriate formula.
  7. Nov 20, 2011 #6
    You mean 4x3-6x2-12x+9=0?

    The roots of a polynomial must be a factor of the product of the first term and the last term. Therefore, we have to try all the factors for 4*9=36. There's this really convenient method for factoring polynomials like this. It's kind of like long division, but not really. Let's try x=4 for example. Then we set up the polynomial division like this:

    2 | 4, -6, -12, 9

    First step. Drop the first term (from the left) to the right of the bar.

    2 | 4, -6, -12, 9

    Next, multiply that number by 2, and put it below the next entry.

    2 | 4, -6, -12, 9

    Add that column.

    2 | 4, -6, -12, 9


    2 | 4, -6, -12, 9

    2 | 4, -6, -12, 9

    The process I described reveals

    4x3-6x2-12x+9 = 4x2-2x-8-7/(x-2)

    The last term there, -7/(x-2) is called the remainder term. Since there is a remainder term, 2 is not a root. You just have to keep testing roots until you get one without a remainder term.
  8. Nov 20, 2011 #7

    I like Serena

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    What about just trying to substitute a possible root in the original expression, instead of doing a polynomial division?
  9. Nov 20, 2011 #8
    You could do that and then use the long division to find the coefficients of the remaining quadratic.
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