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How to solve a factorization

  1. Dec 18, 2004 #1


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    I know how to solve quadratics using both factorisation and the equation method ... but how can i solve :

    [tex](x-1)(2x-1)(3x-1) = 0 [/tex]

    I multiplied it all out and i got ..

    [tex]6x^3 - 2x^2 -3x -1=0[/tex]

    I just do not know where to got from here .. a little nudge in the right direction would be appreciated!

  2. jcsd
  3. Dec 18, 2004 #2
    A small hint: why do you multiply? After all, if a*b*c=0, then....
  4. Dec 18, 2004 #3


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    Why the heck did u multiply it??? :surprised It was already solved.U were being asked for the 3 possible values of "x" which cancel the expression from the LHS.I think/hope they were obvious...


    PS.But if u'd rather apply Cardano's formulae for the 3rd order algebrac equation u got,be my guest... :tongue2:
  5. Dec 18, 2004 #4


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    yes, i do understand.


    (x-1) = 0 then x =1
    (2x-1) =0 then 2x=1 so x = 1/2
    (3x-1) = 0 then 3x=1 so x=1/3

    is that right.Am i going about it right. thanks.
  6. Dec 18, 2004 #5
    Yeah, that's right.
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