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Math Help - Factorisation Help

  1. #1
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    Factorisation Help

    The question:
    Factorise 2(2x^3 - 3x^2 - x + 1)

    I have no idea where to start. Is there a method of factorising cubics and other high degree polynomials, preferably without remembering massive formulas?

    Thanks.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    To find the factors, it's basically trial and error.

    You should use the factor theorem, where if you a have a function:f(x), then if (x-p) is a factor of f(x), f(p) will be equal to zero.

    In your example, trying (x-1) gives:

    f(1) = 2 - 3 - 1 + 1 = -1

    So, (x-1) is not a factor. Let's try (x+1). [You normally try out the factors of the coefficient of the highest power of x and the term with the smallest power of x]

    f(-1) = -2 -3 + 1 + 1 = -3

    So, (x+1) is not a factor.

    Let's try (x-2)

    f(2) = 16 - 12 - 2 + 1 = 3

    (x-2) is not a factor. Let's try (x+2)

    f(-2) = -16 - 3 + 2 + 1 = -16

    (x+2) is not a factor either...

    Well, now, let's take (x - 0.5).

    f(0.5) = 0.25 - 0.75 - 0.5+1 = 0

    So, we get (x - 0.5) as a factor.

    Or,

    (2x - 1) as a factor.

    Do you know how to use long division to continue the factoring?
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  3. #3
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    Hmm, I just tried and I made a mess. Could you walk me though the division? Thanks.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Yep, it will be a little difficult though through the forum... and some time.

    Code:
    Long Division:
            x^2 -  x   -  1    
    
    2x - 1|2x^3 - 3x^2 -  x + 1
    
           2x^3 -  x^2
    
           0x^3 - 2x^2 -  x + 1
    
                - 2x^2 +  x
    
                  0x^2 - 2x + 1
    
                       - 2x + 1
    
                         0x + 0
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  5. #5
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    OK, I got the solution you did. However factorising the result looks different to the solution my text states, which is:

    2(x - 1)(x - 1)(2x + 1)

    When I solve x^2 - x - 1 I get:
    \frac{1\pm\sqrt{5}}{2}

    Which doesn't appear to produce the right result. What have I done wrong? Thanks.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Well, the question you posted have the solution I posted...

    See: http://www.wolframalpha.com/input/?i=(2x^3+-+3x^2+-+x+%2B+1)+%3D+0
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  7. #7
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    Right, but I want to factorise it further.
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  8. #8
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    Quote Originally Posted by Glitch View Post
    OK, I got the solution you did. However factorising the result looks different to the solution my text states, which is:

    2(x - 1)(x - 1)(2x + 1)

    When I solve x^2 - x - 1 I get:
    \frac{1\pm\sqrt{5}}{2}

    Which doesn't appear to produce the right result. What have I done wrong? Thanks.
    Since 2(x - 1)(x - 1)(2x + 1) = 4x^3 - 6x^2 + 2, either the book is wrong or you have not posted the correct question.
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  9. #9
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    Sorry guys, apparently it's supposed to be (according to my text):
    2(2x^3 - 3x^2 - x^2 + 1)

    My text has a worked solution I was trying to follow, which looks like this:
    2(2x^3 - 3x^2 - x^2 + 1)
    = 2(x - 1)( 2x^2 - x - 1)
    = 2(x - 1)(x - 1)(2x + 1)

    However, expanding it back out doesn't seem to produce the original question. I looked at it and worked out that the textbook actually meant:
    2(2x^3 - 3x^2 + 1)

    So my original question was wrong because I tried to fix it in my notes (which I wrote months ago), and that fix was wrong too! So my text and my notes were both wrong. -_-

    Anyway, thanks for your assistance. Your factorization technique is pretty darn cool, and will assist me in future. Cheers.
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