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Math Help - Factorisation

  1. #1
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    Factorisation

    Hi,

    I've a question on my assignment paper I just want to check to make sure I'm doing it right.

    Q: Factorise the numerator and denominator of each of the following and then cancel out any common factors:

    x(squared) + 2x - 3
    -------------------
    x(squared) + 5x + 6

    So what I did was get the factors of both equations -

    x(squared) + 2x - 3
    factors are (x-1)(x+3)

    x(squared) + 5x +6
    factors are (x+2)(x+3)

    so:
    (x-1)(x+3)
    -----------
    (x+2)(x+3)

    x(-1+3)
    ---------
    x(2+3)

    cancel the x's.

    -1+3
    ------
    2+3

    = 2
    ---
    5

    Is that correct?
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  2. #2
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    No, it is not. You can not bring out an x like you done in the end.

    \frac{(x-1)(x+3)}{(x+2)(x+3)}

    The x+3 cancels.
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  3. #3
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    Okay, thank you.
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  4. #4
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    I'm having a problem with another question.

    3x(squared) +6x +3
    -------------------
    9x(squared) - 27x +18

    I'm stuck with factorising the top line. I'm using the -b formula to factorise and get

    -6 +/- 0
    ----------
    6

    Is that correct? I also tried factorising the other way and got (x+3)(x+3)?
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  5. #5
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    You're going about this wrong. You should not get integers as a solution if you are merely factoring and not solving for x.

    3x^{2}+6x+3=3(x+1)^{2}

    9x^{2}-27x+18=9(x-2)(x-1)

    \frac{3(x+1)^{2}}{9(x-2)(x-1)}

    How are you getting this 6 thing?. Are you using the quadratic formula?. If so, don't. You are factoring, not solving. I least, I think.

    The solutions are 2, 1, -1 if you need those. There is no 6 involved.
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  6. #6
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    Quote Originally Posted by galactus View Post

    \frac{3(x+1)^{2}}{9(x-2)(x-1)}
    Is that the same as (3x+3)(x+1)
    ------------
    (9x-9)(x-2)

    That's what I got when I factorised again, without using the -b formula.
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  7. #7
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    Yes, to see expand it back out.

    Except, factor a 9 out of the bottom and a 3 out of the top
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  8. #8
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    Last question!!!

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

    I'm unsure how to factorise the top equation. The bottom equation is 2(x-), right?
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  9. #9
    Eater of Worlds
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    The top one does not factor nice and neat.

    It's roots are \frac{-\sqrt{5}-1}{2}, \;\ \frac{\sqrt{5}-1}{2}

    This is the related to the Golden Ratio and this is the closed form quadratic that represents it. You didn't need to know that, but.............

    The bottom is simply 2(x-1)

    Are you sure you did not mean x^{2}+2x+1 or

    x^{2}-2x+1
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  10. #10
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    No, it's x(squared) + x - 1. So I just have to leave it so.

    Thanks very much for your help.
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  11. #11
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    Are there factors for the top line? As the question asks to factorise and then cancel??
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  12. #12
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    You can always solve x^2+x-1 by the quadratic formula.
    roots are -2, 3 by it.
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