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Math Help - Factoring cubic and further polynomials

  1. #1
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    Factoring cubic and further polynomials

    Hi, just stopped lurking on mhf and made my first post.

    Ok I can factor quadratics and the differences of squares. I'm basically using the factor pairs method (finding factor pairs that add up to the b term) for quadratics as it almost always works. I can expand cubic and further polynomial brackets but I cannot do it the other way around by factorizing.

    I've hit a roadblock trying to factorize these polynomials. Could anyone please help me with them?

    x^3-3x^2-4x+12 and
    x^4+27
    3x^(3/2)-9x^(1/2)+6x^(-1/2) and finally
    x^3y-4xy
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  2. #2
    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 03:57 AM.
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  3. #3
    Senior Member pacman's Avatar
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    • grouping will do, but hard to guess....1) , the factors of 12 are many,
    (x^3 - 3x^2) - (4x -12) = x^2(x - 3) - 4(x - 3), you see (x - 3)?

    we can factor that, (x - 3)(x^2 - 4), ohh, (x^2 - 4) is the difference of 2

    squares,

    = (x - 3)(x - 2)(x + 2)


    2) , seems miscopied, it myt be x^3 + 27 = (x + 3)(x^2 - 3x + 9)
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  4. #4
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    Thanks PQR and pacman.

    I have just checked again and x^4+27 is correct and not a miscopy and the answer to it was x(x + 3)(x^2 - 3x + 9). Not sure why though.
    Last edited by Kataangel; August 31st 2009 at 03:07 PM. Reason: capslock
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  5. #5
    Member eXist's Avatar
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    It has to be a miscopy because:
    \big[x^4+27\big] \ne \big[x(x + 3)(x^2 - 3x + 9)\big]

    Are you sure the problem wasn't something like:

    x^4 + 27x?

    Because x^4 + 27x = x(x + 3)(x^2 - 3x + 9)
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  6. #6
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    I have no idea where (x+2) comes from in PQR's post:
    (x+2)(ax^2+bx+c)


    Quote Originally Posted by eXist View Post
    It has to be a miscopy because:
    \big[x^4+27\big] \ne \big[x(x + 3)(x^2 - 3x + 9)\big]

    Are you sure the problem wasn't something like:

    x^4 + 27x?

    Because x^4 + 27x = x(x + 3)(x^2 - 3x + 9)
    Yep so actually x^4 + 27x is right then.
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