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Thread: More Factor By Grouping

  1. #1
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    More Factor By Grouping

    Factor By grouping.

    4a^2b^2 - 9 (ab + c)^2

    Must I FOIL (ab + c)^2 and distribute 9 before factoring?
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  2. #2
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    Re: More Factor By Grouping

    Quote Originally Posted by mathdad1965 View Post
    Factor By grouping.

    4a^2b^2 - 9 (ab + c)^2

    Must I FOIL (ab + c)^2 and distribute 9 before factoring?
    No,   \ \ \ 4a^2b^2 -  9(ab + c)^2  is the difference of two squares.

    A^2  -  B^2 \ = \ (A - B)(A + B)

    Determine what the square root of  \ \ 4a^2b^2 \ \ is \ \ and  \ \ also \ \ what \ \ the \ \ square \ \ root \ \ of \ \ 9(ab + c)^2 \ \ is.

    Those will be your A and your B, respectively.
    Last edited by greg1313; Jan 3rd 2017 at 12:30 PM.
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  3. #3
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    Re: More Factor By Grouping

    4a^2b^2 - 9 (ab + c)^2

    (2ab - 3 (ab + c))(2ab + 3 (ab + c))

    Correct?
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  4. #4
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    Re: More Factor By Grouping

    Quote Originally Posted by mathdad1965 View Post
    4a^2b^2 - 9 (ab + c)^2

    (2ab - 3 (ab + c))(2ab + 3 (ab + c))

    Correct?
    there are clearer ways to post this

    $4a^2b^2 - 9(a b+ c)^2 = x^2 - y^2$

    $x = 2ab$

    $y= 3(ab+c)$

    $x^2 - y^2 = (x+y)(x-y) = (2ab + 3(ab+c))(2ab-3(ab+c))$

    so yes, it looks like you are correct
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  5. #5
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    Re: More Factor By Grouping

    Of course, it would make sense to then "clean up" each factor:
    (2ab - 3 (ab + c))(2ab + 3 (ab + c))= (2ab- 3ab- 3c)(2ab+ 3ab+ 3c)= -(ab+ 3c)(5ab+ 3c)
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  6. #6
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    Re: More Factor By Grouping

    Quote Originally Posted by romsek View Post
    there are clearer ways to post this

    $4a^2b^2 - 9(a b+ c)^2 = x^2 - y^2$

    $x = 2ab$

    $y= 3(ab+c)$

    $x^2 - y^2 = (x+y)(x-y) = (2ab + 3(ab+c))(2ab-3(ab+c))$

    so yes, it looks like you are correct
    Good to know I got it right.
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