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Math Help - Please help me figure this out!

  1. #1
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    Please help me figure this out!

    It's been driving me up the wall and this is the fastest math forum I've found on google.


    Here's the problem.

    8x^2-12X-6XY+9

    How to factor???

    Thank you in advance.
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  2. #2
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    Quote Originally Posted by nikzbord View Post
    It's been driving me up the wall and this is the fastest math forum I've found on google.


    Here's the problem.

    8x^2-12X-6XY+9

    How to factor???

    Thank you in advance.
    You can't. Clearly the given expression is incorrect.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You can't. Clearly the given expression is incorrect.
    Actually you can complete the square if you notice that you can write the middle term as (-12 - 6y)x.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Actually you can complete the square if you notice that you can write the middle term as (-12 - 6y)x.
    The given expression can't be factorised: factorise 8x^2-12x-6xy+9 - Wolfram|Alpha
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  5. #5
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    8x^2 - (12 + 6y)x + 9

    = 8\left(x^2 - \frac{6 + 3y}{4}x + \frac{9}{8}\right)

     = 8\left[x^2 - \frac{6 + 3y}{4}x + \left(-\frac{6 + 3y}{8}\right)^2 - \left(-\frac{6 + 3y}{8}\right)^2 + \frac{9}{8}\right]

     = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \frac{36 + 36y + 9y^2}{64} + \frac{72}{64}\right]


     = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \frac{9y^2 + 36y -36}{64}\right]

     = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \left(\frac{3\sqrt{y^2 + 4y - 4}}{8}\right)^2\right]

     = 8\left(x - \frac{6 + 3y}{8} + \frac{3\sqrt{y^2 + 4y - 4}}{8}\right)\left(x - \frac{6 + 3y}{8} - \frac{3\sqrt{y^2 + 4y - 4}}{8}\right)

     = 8\left(\frac{8x + 3\sqrt{y^2 + 4y - 4} - 6 - 3y}{8}\right)\left(\frac{8x - 3\sqrt{y^2 + 4y - 4} - 6 - 3y}{8}\right).


    I believe this is factorised...
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  6. #6
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    Any expression can be factored if you allow any numbers (irrational and/or complex) as coefficients. Mr Fantastic is correct that this cannot be factored with integer coefficients which is what is usually meant by "factoring". Prove It is correct that what he gives is a factoring using irrational terms.
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  7. #7
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    What happened to 72/64??
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  8. #8
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    Quote Originally Posted by nikzbord View Post
    What happened to 72/64??
    I confess that I find this a very puzzling post. Is it missing?


    Okay, I went back an looked at Prove It's post. That 72/64!

    Prove it has:
    = -\frac{36+ 36y+ 9y^2}{64}+ \frac{72}{64}

    = -\frac{36+ 36y+ 9y^2}{64}- \frac{-72}{64}

    = -\frac{9y^2+ 36y+ 36- 72}{64}= -\frac{9y^2+ 36- 36}{64}
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