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Thread: 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?

    $\displaystyle x^3-3x^2-4x+12$ and
    $\displaystyle x^4+27$
    $\displaystyle 3x^(3/2)-9x^(1/2)+6x^(-1/2)$ and finally
    $\displaystyle x^3y-4xy$
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  2. #2
    PQR
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    $\displaystyle x^3-3x^2-4x+12 = 0$

    $\displaystyle [(x-3)x-4]x+12 = 0
    $

    $\displaystyle f(-2) = 0$

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

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

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

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

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

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

    $\displaystyle x^2-5x-6 = 0$

    $\displaystyle (x-5)x-6 = 0$

    $\displaystyle f(2) = 0$

    $\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(ax+b)$

    $\displaystyle x^2-5x-4 $ ≡ $\displaystyle ax^2+bx-2ax-2b$

    $\displaystyle x^2-5x-4$ ≡ $\displaystyle (a)x^2+(b-2a)x-2b$

    $\displaystyle a = 1$; $\displaystyle b-2a$ = $\displaystyle -5$, $\displaystyle b = -3$.

    $\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(x-3)$

    $\displaystyle x^3-3x^2-4x+12$ ≡ $\displaystyle (x+2)(x-2)(x-3)$



    Last edited by PQR; Aug 27th 2009 at 02:57 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 $\displaystyle x^4+27$ is correct and not a miscopy and the answer to it was $\displaystyle x(x + 3)(x^2 - 3x + 9)$. Not sure why though.
    Last edited by Kataangel; Aug 31st 2009 at 02:07 PM. Reason: capslock
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  5. #5
    Member eXist's Avatar
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    It has to be a miscopy because:
    $\displaystyle \big[x^4+27\big] \ne \big[x(x + 3)(x^2 - 3x + 9)\big]$

    Are you sure the problem wasn't something like:

    $\displaystyle x^4 + 27x$?

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


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

    Are you sure the problem wasn't something like:

    $\displaystyle x^4 + 27x$?

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