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Math Help - division

  1. #1
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    Question division

    Hi all - me again,

    I have a math exam in the morning. Lucky for me, I have some past papers, and only the numbers, not the questions, change each year.

    I was wondering if someone could help me with the solution I have here:

    1
    a) Fully factorize:

    3x ^3-7x ^2-3x+7

    The solution which I have been given is to first of all do this:

    1 I...3...............-7............-3..........7
    ....I
    ....I_________3_____-4_____7_____
    .......3............... -4............-7..........0

    Which then goes onto this:

    3x ^2 - 4x -7 = (3x-7)(x+1)

    so (x-1)(3x-7)(x+1)

    Can someone please explain all of the above to me? I have no idea what the grid thing does, how the 4 gets in there? Basically, help? Please?

    Cheers,

    Martin
    Last edited by Mathy; February 13th 2008 at 01:54 PM. Reason: To try and get the grid thing to line up
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
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    New York, USA
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    Quote Originally Posted by Mathy View Post
    Hi all - me again,

    I have a math exam in the morning. Lucky for me, I have some past papers, and only the numbers, not the questions, change each year.

    I was wondering if someone could help me with the solution I have here:

    1
    a) Fully factorize:

    3x ^3-7x ^2-3x+7

    The solution which I have been given is to first of all do this:

    1 I...3...............-7............-3..........7
    ....I
    ....I_________3_____-4_____7_____
    .......3............... -4............-7..........0

    Which then goes onto this:

    3x ^2 - 4x -7 = (3x-7)(x+1)

    so (x-1)(3x-7)(x+1)

    Can someone please explain all of the above to me? I have no idea what the grid thing does, how the 4 gets in there? Basically, help? Please?

    Cheers,

    Martin
    i don't see what you are doing here. it seems you were trying some sort of long division. that was completely unnecessary. this is a "nice" polynomial. it factors easily by grouping

    3x^3 - 7x^2 - 3x + 7

    factor out x^2 from the first two terms and -1 from the last two, we get:

    x^2 (3x - 7) - (3x - 7)

    now the 3x - 7 is common, so pull that out

    (3x - 7)(x^2 - 1)

    the last part is the difference of two squares, so we have:

    (3x - 7)(x + 1)(x - 1)
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