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Math Help - Please Help. Algebra.

  1. #1
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    Please Help. Algebra.

    For this question

    (64x^3 + 0x + 8) / (4x +2)

    I got 16x^2 -8X + 6

    Am I right
    Last edited by cleanwithit06; January 16th 2010 at 02:31 AM.
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  2. #2
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    Anybody?
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  3. #3
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    Quote Originally Posted by cleanwithit06 View Post
    For this question

    (64x^3 + 0x + 8) / (4x +2)

    I got 16x^2 -8X + 6

    Am I right
    I just did the calculations and got the answer to be 16x^2 - 8x +4. I'll post up my method, wait a few minutes.

    EDIT: I don't know how to use latex to do long division so I'll show how my answer is correct by opening up the brackets to obtain the original equation. My original method was using long division and divided 64x^3 + 8 by 4x + 2.

    (4x + 2)(16x^2 - 8x + 4)
    = 64 x^3 - 32x^2 + 16x + 32x^2 - 16x + 8
    = 64x^3 + 8
    Last edited by Simplicity; March 6th 2008 at 09:42 AM.
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  4. #4
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    Quote Originally Posted by Air View Post
    I just did the calculations and got the answer to be 16x^2 - 8x +4. I'll post up my method, wait a few minutes.
    I got that too.
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  5. #5
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    Quote Originally Posted by Air View Post
    I just did the calculations and got the answer to be 16x^2 - 8x +4. I'll post up my method, wait a few minutes.
    Thankyou.
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  6. #6
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    Ok I keep getting 16x^2 - 8x + 6

    I don't know what the crap I'm doing wrong.
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  7. #7
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    Sorry if I haven't explained clear enough.
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  8. #8
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    \frac{{64x^3  + 8}}<br />
{{4x + 2}} = \frac{{8\left( {8x^3  + 1} \right)}}<br />
{{2\left( {2x + 1} \right)}} =  \frac{{4\left( {8x^3  + 1} \right)}}<br />
{{\left( {2x + 1} \right)}} = \frac{{4\left( {2x + 1} \right)\left( {4x^2  - 2x + 1} \right)}}<br />
{{\left( {2x + 1} \right)}} = 4\left( {4x^2  - 2x + 1} \right) = 16x^2  - 8x + 4
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  9. #9
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    Quote Originally Posted by cleanwithit06
    Hey. The only step I don't get with your method is the 4th one. It says -32 -32 comes out to being 16x or something, how is that so?


    OR

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