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Math Help - Factor by grouping

  1. #1
    Junior Member fluffy_penguin's Avatar
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    Factor by grouping

    http://i24.tinypic.com/2hncsgh.jpg

    I don't understand how to do this...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by fluffy_penguin View Post
    http://i24.tinypic.com/2hncsgh.jpg

    I don't understand how to do this...
    Write 3a^2 + 2a - 21 = 3a^2 + (9a - 7a) - 21

    = (3a^2 + 9a) - (7a + 21)

    = 3a(a + 3) - 7(a + 3)

    = (3a - 7)(a + 3)

    -Dan
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  3. #3
    Junior Member fluffy_penguin's Avatar
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    Confused

    Quote Originally Posted by topsquark View Post
    Write 3a^2 + 2a - 21 = 3a^2 + (9a - 7a) - 21

    = (3a^2 + 9a) - (7a + 21)

    = 3a(a + 3) - 7(a + 3)

    = (3a - 7)(a + 3)

    -Dan
    How does

    3a^2 + 2a - 21 = 3a^2 + <b>(9a - 7a)</b> - 21

    I have no clue where the (9a - 7a) came from...
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fluffy_penguin View Post
    How does

    3a^2 + 2a - 21 = 3a^2 + <b>(9a - 7a)</b> - 21

    I have no clue where the (9a - 7a) came from...
    we use \bold{ } to make something bold with LaTex

    topsquark noticed that 9a - 7a = 2a. how did he do that? well, he could have done it by experience, but there is a method you can do.

    take the coefficient of a^2, that is 3 and multiply it by the constant term, that is, -21. the result is -63. now you want two numbers that when multiplied give -63 but when added give 2 (which is the coefficient of the middle term). 9 and -7 are such numbers. now we split the middle term into those two numbers and continue as topsquark did
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  5. #5
    Junior Member fluffy_penguin's Avatar
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    Okay thank you both of you. ^ ^
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fluffy_penguin View Post
    Okay thank you both of you. ^ ^
    you're welcome, come again
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