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Math Help - Help with simplification

  1. #1
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    Help with simplification

    I have the following problem that needs to be simplified. I already have the simplification results, but I don't understand how they got it. For example, how does there and up being 2 h's, when there is only a single h to begin with. Any help on how to simplify this problem one step at a time would be appreciated.

    Step 1: ((3-4(-2+h)^2)-(3-4(-2)^2) / h)

    Step 2: ((16h-4h^2) / h)

    Step 3: (16-4h)
    Last edited by softwareguy; March 2nd 2010 at 01:23 PM.
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  2. #2
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    You have

    \frac{16h-4h^2}{h}

    now factoring out h in the numerator

    \frac{h(16-4h)}{h}

    Which cancels with the h in the denominator

    16-4h
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  3. #3
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    But how did they get from Step 1 to Step 2? I have edited my original post to include the step numbers. I understand the last part which you explained, but I don't understand the simplification process which resulted in Step 2.
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  4. #4
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    Hello, softwareguy!

    How did they get from Step 1 to Step 2?

    Step 1: \frac{[3-4(-2+h)^2]-[3-4(-2)^2]}{h}

    Step 2: \frac{16h-4h^2}{h}

    Step 3: 16-4h

    We have: . \frac{[3-4(4-4h + h^2)] - [3-4(4)]}{h}

    . . . . . . =\;\frac{(3-16 + 16h - 4h^2) - (-13)}{h}

    . . . . . . =\;\frac{-13 + 16h - 4h^2 + 13}{h}

    . . . . . . =\; \frac{16h-4h^2}{h}

    . . . . . . =\;\frac{{\color{red}\rlap{/}}h(16-4h)}{{\color{red}\rlap{/}}h}

    . . . . . . =\;16-4h

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  5. #5
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    Thanks! I totally forgot about having to FOIL out the (-2+h)^2 equation. That was blocking my ability to proceed.
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