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Math Help - Lowest terms

  1. #1
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    Lowest terms

    Multiply or divide, write each answer in lowest terms.

    3x^3-5x-2
    ------------
    x-2

    times

    x-3
    ----
    x+1

    Where do i begin?

    (corrected, format looked odd before)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tenskypoo View Post
    Multiply or divide, write each answer in lowest terms.

    3x^3-5x-2
    ------------
    x-2

    times

    x-3
    ----
    x+1

    Where do i begin?
    begin by factoring 3x^2 - 5x - 2 and see what cancels. i suppose the power of the first x is a 2 and not 3 based on the level of questions you have asked before.
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  3. #3
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    after doing

     3x^2 - 5x - 2

    i came up with

    ( 3x + 1) ( x - 2)

    which eliminates the x-2 from the bottom of the other problem, leaving me with 3x+1?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tenskypoo View Post
    after doing

     3x^2 - 5x - 2

    i came up with

    ( 3x + 1) ( x - 2)

    which eliminates the x-2 from the bottom of the other problem, leaving me with 3x+1?
    right. so what's your final answer?
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  5. #5
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    well i have to multiply  3x + 1


    by x-3 over x+1, how do i do that?

    i want to GUESS. a heavy guess. that you rule out x+1 and you're left with x-3 over 3, or vise versa.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tenskypoo View Post
    well i have to multiply  3x + 1


    by x-3 over x+1, how do i do that?

    i want to GUESS. a heavy guess. that you rule out x+1 and you're left with x-3 over 3, or vise versa.
    that would be wrong. you cannot rule out the x + 1 from the 3x + 1. if it were 3(x + 1) you could. but that is not the case. just multiply the numerators to get the new numerator, and the denominators to get the new denominator
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  7. #7
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    I don't want to seem like a complete idiot, but could you possibly show me what the final set up before multiplying should be?
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  8. #8
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    \frac {3x^2 - 5x - 2}{x - 2} \times \frac {x - 3}{x + 1} = \frac {(3x + 1)(x - 2)}{(x - 2)} \times \frac {x - 3}{x + 1} = (3x + 1) \times \frac {x - 3}{x + 1} = \frac {(3x + 1)(x - 3)}{x + 1}
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