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Math Help - Factorisation

  1. #1
    Junior Member BobBali's Avatar
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    Factorisation

    Hello All, wishing all memebers a great start to the New-Year!
    I'm not sure if my working is entirely ok below... Tried using latex..no idea why it didnt work..?

    Factorise: [tex] 50x^2-10xy-20y^2 [\math]
    10(5x^2-xy+2y^2)
    10[(5x^2 - 2y^2) - xy]
    10[( \sqrt5x + \sqrt2y)^2 -xy )]

    10[( \sqrt5x + \sqrt2y) ( \sqrt5x + \sqrt2y) - 7.32xy ] * completing the square
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  2. #2
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    \displaystyle 50x^2 - 10xy - 20y^2 = 50\left(x^2 - \frac{1}{5}xy - \frac{2}{5}y^2\right)

    \displaystyle = 50 \left[x^2 - \frac{1}{5}xy + \left(-\frac{1}{10}y\right)^2 - \left(-\frac{1}{10}y\right)^2 - \frac{2}{5}y^2\right]

    \displaystyle = 50\left[\left(x - \frac{1}{10}y\right)^2 - \frac{1}{100}y^2 - \frac{40}{100}y^2\right]

    \displaystyle = 50\left[\left(x - \frac{1}{10}y\right)^2 - \frac{41}{100}y^2\right]

    \displaystyle = 50\left[\left(x - \frac{1}{10}y\right)^2 - \left(\frac{\sqrt{41}}{10}y\right)^2\right]

    \displaystyle = 50\left(x - \frac{1}{10}y - \frac{\sqrt{41}}{10}y\right)\left(x - \frac{1}{10}y + \frac{\sqrt{41}}{10}\right)

    \displaystyle = 50\left(x-\frac{\sqrt{41} + 1}{10}y\right)\left(x + \frac{\sqrt{41} - 1}{10}y\right)
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    We have:

    p(x,y)=10(5x^2-xy-2y^2)

    Solving on x

    5x^2-xy-2y^2=0

    we obtain

    x=\dfrac{y\pm \sqrt{41}y}{10}=\{x_1(y),x_2(y)\}

    So,

    p(x,y)=10(x-x_1(y))(x-x_2(y))=\ldots


    Fernando Revilla
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  4. #4
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    Quote Originally Posted by BobBali View Post
    Factorise: 50x^2-10xy-20y^2
    Are you sure that's not 50x^2 - 10xy - 40y^2 ?
    Factors "nicely" to: (x - 1)(5x + 4)
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  5. #5
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    Quote Originally Posted by Wilmer View Post
    Are you sure that's not 50x^2 - 10xy - 40y^2 ?
    Factors "nicely" to: (x - 1)(5x + 4)
    I'm sure you mean \displaystyle 10(x-y)(5x+4y)...
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  6. #6
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    Quote Originally Posted by Prove It View Post
    I'm sure you mean \displaystyle 10(x-y)(5x+4y)...
    Hmmm....
    I was taught (50 years ago!) to go this way:
    50x^2 - 10xy - 40y^2
    Divide by 10:
    5x^2 - xy - 4y^2
    Now factor...

    HOW serious is it to "drop" that 10 ?
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  7. #7
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    When you're asked to factorise you shouldn't "drop" terms... You can only "drop" terms when you're solving an equation.

    Besides, the bigger mistake is the fact that you left the \displaystyle y terms out...
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  8. #8
    Junior Member BobBali's Avatar
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    No Sir, unfortunately its does not end with -40y^2; it is -20y^2 and so far the replies i have gotten from some serious "super members", i mean the math they show to factorise this is just beyond me..
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  9. #9
    MHF Contributor Unknown008's Avatar
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    You can use FernandoRevilla's post which is the most elementary.

    5x^2-xy-2y^2=0

    Here, use the quadratic formula x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} to get:

    x=\dfrac{y\pm \sqrt{41}y}{10}

    You only need to substitute those now in the previous expression.

    Oh and the LaTeX works with the forward slash '/' and not backslash '\' in the math tabs.
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  10. #10
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    Quote Originally Posted by Prove It View Post
    When you're asked to factorise you shouldn't "drop" terms... You can only "drop" terms when you're solving an equation.
    Of course...my bad.
    Quote Originally Posted by Prove It View Post
    Besides, the bigger mistake is the fact that you left the \displaystyle y terms out...
    Geesh...having a real bad day...
    With those, I drop the "y"s while doing "the work", then put 'em back in
    once finished...IF I don't forget
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