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Math Help - Factoring with cubes

  1. #1
    DBA
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    Factoring with cubes

    Quote Originally Posted by DBA in orginal post
    Hi,
    - I stuck with factoring of the following:

    x^3-2x^2-4x+12

    I thought I could use grouping like

    (x^3-2x^2)- (4x-12)=

    x^2(x-2) - 4(x-3)
    +
    sorry I made a mistake...
    it should be

    - but this does not give me a common factor, so I still have 3 terms
    (x^2-4) (x-2) (x-3)
    Hi,

    sorry I made a mistake...
    it should be

    x^3-3x^2-4x+12

    where I than get different terms
    (x^3-3x^2)- (4x-12)=

    x^2(x-3) - 4(x-3)

    and then it works out fine.

    Sorry about that
    Last edited by mr fantastic; August 27th 2009 at 08:16 PM. Reason: Restored original question as a quote (at time of editing by the OP there had been no replies)
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  2. #2
    Master Of Puppets
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    Quote Originally Posted by DBA View Post
    Hi,
    I stuck with factoring of the following:

    x^3-2x^2-4x+12

    I thought I could use grouping like

    (x^3-2x^2)- (4x-12)=

    x^2(x-2) - 4(x-3)
    You can do this but it won't help. When grouping you need the terms inside the brackets to be the same. This is not the case here in what you have done.

    Quote Originally Posted by DBA View Post

    but this does not give me a common factor, so I still have 3 terms
    (x^2-4) (x-2) (x-3)
    Therefore this is not correct. The term you have here is of order x^4

    Do you know the factor theorem? It is, for P(x), if P(a) = 0 then x-a is a factor.

    You need to apply this theorem to factor a cubic of this nature.
    Last edited by pickslides; August 25th 2009 at 02:58 PM. Reason: spelling
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  3. #3
    PQR
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    x^3-3x^2-4x+12 = 0

    [(x-3)x-4]x+12 = 0<br />

    f(-2) = 0

    x^3-3x^2-4x+12 (x+2)(ax^2+bx+c)

    x^3-3x^2-4x+12 ax^3+bx^2+cx+2ax^2+2bx+2c

    x^3-3x^2-4x+12 ax^3+bx^2+2ax^2+cx+2bx+2c

    x^3-3x^2-4x+12 ax^3+(b+2a)x^2+(c+2b)x+2c

    a =1; b+2a = -3, b = -5; c+2b -4, c = 6.

    x^3-3x^2-4x+12 (x+2)(x^2-5x+6)

    x^2-5x-6 = 0

    (x-5)x-6 = 0

    f(2) = 0

    x^2-5x-4 (x-2)(ax+b)

    x^2-5x-4 ax^2+bx-2ax-2b

    x^2-5x-4 (a)x^2+(b-2a)x-2b

    a = 1; b-2a = -5, b = -3.

    x^2-5x-4 (x-2)(x-3)

    x^3-3x^2-4x+12 (x+2)(x-2)(x-3)
    Last edited by PQR; August 27th 2009 at 02:46 AM.
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