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Factoring 3rd degree polynomial

  1. Jul 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Factor 12X^3-12X^2-60X+24=0

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 8, 2009 #2

    rock.freak667

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    Homework Helper

    You forgot that part of the question.

    Do you know the rational root theorem?
     
  4. Jul 8, 2009 #3
    First I would factor 12 out of all the terms then use the rational zero test since you can't factor by grouping.
     
  5. Jul 8, 2009 #4
    so factoring out the 12 will get me a 12[x^3-X^2-5X+2]=0 and the root theorem says my it should be +-(2,1)/1 Right? So possible answers are +2, -2, or 1 and -1
     
  6. Jul 9, 2009 #5
    Yes, try those. You should only need one of them to get a quadratic you can factor.
     
  7. Jul 9, 2009 #6
    Correct. Now you can use synthetic division to test the roots. Or, you can plug and chug and then use polynomial division to figure out the quadratic remaining, which you should be able to factor.
     
  8. Jul 9, 2009 #7
    You could try something like this:

    [tex]12(x^3-x^2-5x+2)=0[/tex]

    [tex]12(x^3+2x^2-2x^2-x^2-4x-x+2)=0[/tex]

    [tex]12([x^3+2x^2] - [2x^2+4x] - [x^2+x-2])=0[/tex]

    Now factor the terms and solve the equation.

    Regards.
     
  9. Jul 9, 2009 #8

    HallsofIvy

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    How does factoring those terms help solve the equation? Boreck and Dunkle have already told him how to do this.
     
  10. Jul 9, 2009 #9
    [tex]12[x^2(x+2)-2x(x+2)-(x+2)(x-1)]=0[/tex]
    [tex]12(x+2)[x^2-2x-(x-1)]=0[/tex]

    What I did is actually, I found that -2 is the solution of the polynomial, so I found out way to factor the whole polynomial and save some time for dividing the whole polynomial with (x+2).

    Regards.
     
  11. Jul 9, 2009 #10
    I'm sure HallsofIvy saw what you were doing, but I'm guessing he thought your post was out of place because two people already gave a more straightforward and efficient way to factor it.
     
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