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Math Help - Help with factoring

  1. #1
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    Help with factoring

    Hello. I need to factor this equation and I'm a bit stumped.

    The equation is 25{x^2}{y^2}z - 9{a^2}{b^2}z

    I'm not sure how to approach it. Any help would be appreciated
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  2. #2
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    Quote Originally Posted by Thebat View Post
    Hello. I need to factor this equation and I'm a bit stumped.

    The equation is 25{x^2}{y^2}z - 9{a^2}{b^2}z

    I'm not sure how to approach it. Any help would be appreciated

    When it comes to factoring you just look at what is common to parts of the expression, in this case 25 and 9 do not have any common factors, so there is no integer value to factor out, and the variables are all unique except that both parts of the expression have a "z" so

    25{x^2}{y^2}z - 9{a^2}{b^2}z=z(25x^2y^2-9a^2b^2)

    Now though, and this takes some insight, you may realize that (25x^2y^2-9a^2b^2) is the difference of two perfect squares, and it is a well known fact that (a^2-b^2)=(a+b)(a-b)


    So (25x^2y^2-9a^2b^2)=(5xy+3ab)(5xy-3ab)

    and so the entire expression factors into z(5xy+3ab)(5xy-3ab)
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  3. #3
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    Quote Originally Posted by Thebat View Post
    Hello. I need to factor this equation and I'm a bit stumped.

    The equation is 25{x^2}{y^2}z - 9{a^2}{b^2}z

    I'm not sure how to approach it. Any help would be appreciated
    We can start by pulling out the z.

    z(25{x^2}{y^2} - 9{a^2}{b^2})

    Next, note the equation a^2-b^2=(a-b)(a+b), in your case, a=(5xy)^2 and b=(3ab)^2.

    Can you take it from here?

    EDIT - Beat to the punch :P
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  4. #4
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    Ah, I remember the difference of squares rule. The 25 and 9 should have been big clues. Thanks a lot for the help
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