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Math Help - Factorization...Again.

  1. #1
    Junior Member BobBali's Avatar
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    Factorization...Again.

    Good-day to everyone. Again, i have to factorize the given below (not solve for x), just factorize. Thanks

     <br />
36x^2 - 24xy + 48y^2<br />

    \\ 6( 6x^2 -4xy +8y^2)
     <br />
6(6x^2 + 8y^2 -4xy)<br />
     <br />
6[( \sqrt6x + \sqrt8y )^2 - 4xy)]<br />

     <br />
6[( \sqrt6x + \sqrt8y )( \sqrt6x + \sqrt8y) - 17.86xy]<br />
*Completing the square
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by BobBali View Post
    Good-day to everyone. Again, i have to factorize the given below (not solve for x), just factorize. Thanks

     <br />
36x^2 - 24xy + 48y^2<br />

    \\ 6( 6x^2 -4xy +8y^2)
     <br />
6(6x^2 + 8y^2 -4xy)<br />
     <br />
6[( \sqrt6x + \sqrt8y )^2 - 4xy)]<br />

     <br />
6[( \sqrt6x + \sqrt8y )( \sqrt6x + \sqrt8y) - 17.86xy]<br />
*Completing the square
    The HCF of 36, 24 and 48 is 12 rather than 6 (which you see in the next step because you can factor out a 2 from your second line)


    12(3x^2-2xy+4y^2)

    That quadratic term is prime so I'd just leave it at that
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  3. #3
    MHF Contributor

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    Quote Originally Posted by BobBali View Post
    Good-day to everyone. Again, i have to factorize the given below (not solve for x), just factorize. Thanks

     <br />
36x^2 - 24xy + 48y^2<br />

    \\ 6( 6x^2 -4xy +8y^2)
     <br />
6(6x^2 + 8y^2 -4xy)<br />
     <br />
6[( \sqrt6x + \sqrt8y )^2 - 4xy)]
    I believe you have made this same mistake before:
    (a+ b)^2\ne a^2+ b^2
    so that (\sqrt{6}x+ \sqrt{8}y)^2\ne 6x^2+ 8y^2


     <br />
6[( \sqrt6x + \sqrt8y )( \sqrt6x + \sqrt8y) - 17.86xy]<br />
*Completing the square
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  4. #4
    Junior Member BobBali's Avatar
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    Ok Thanks, so it cannot be factorized any further than this?? > [tex] 12(3x^2 -2xy + 4y^2) [math\]
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  5. #5
    MHF Contributor Unknown008's Avatar
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    No, this cannot be further factorised.
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  6. #6
    Junior Member BobBali's Avatar
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    Thank you ;-)
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