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Math Help - Factoring quadratic polynomial fractions

  1. #1
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    Factoring quadratic polynomial fractions

    I have an additional question if someone would please help. If so, please explain how this. I simply do not understand how the numerator changes in this case.


    <br />
\frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}<br />





    Expression SOLVED
    \frac{3}{2x^2+4x-48}+\frac{7}{6x-24}+\frac{1}{4x^2+20x-24}

    =

    \frac{3}{2(x-4)(x+6)}+\frac{7}{6(x-4)}-\frac{1}{4(x+6)(x-1)}

    The author then lists the correct LCD as 12(x-4)(x+6)(x-1). I agree with everything here except the coefficient of 12 should it not be

    LCD = 2 * 6 * 4 (x-4)(x+6)(x-1) , where here I took the coeffecient factored out from each denominators expression and multiplied them by each other giving us a LCD of 48(x-4)(x+6)(x-1)
    Last edited by allyourbass2212; July 8th 2009 at 10:10 AM.
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  2. #2
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    answer

    it's 6+4+2 not 6*2*4
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  3. #3
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    My memory must be foggy, are coefficients for the LCD always added in this manner?
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  4. #4
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    Quote Originally Posted by allyourbass2212 View Post
    In this particular question the author wants me to find the LCD for the fractions.


    Expression
    \frac{3}{2x^2+4x-48}+\frac{7}{6x-24}+\frac{1}{4x^2+20x-24}

    =

    \frac{3}{2(x-4)(x+6)}+\frac{7}{6(x-4)}-\frac{1}{4(x+6)(x-1)}

    The author then lists the correct LCD as 12(x-4)(x+6)(x-1). I agree with everything here except the coefficient of 12 should it not be

    LCD = 2 * 6 * 4 (x-4)(x+6)(x-1) , where here I took the coeffecient factored out from each denominators expression and multiplied them by each other giving us a LCD of 48(x-4)(x+6)(x-1)
    In general you are right, but with this example the number 6 can be split into 2*3, that means the factor 4 is already present at:

    \underbrace{2\cdot 2}_{yields\ 4}\cdot 3 = 12
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  5. #5
    Senior Member Stroodle's Avatar
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    I think it's easiest to think of it as 12 is the lowest number that 2, 6 and 4 are all factors of.

    So you would multiply the top and bottom of the first fraction by 6(x-1) the second by 2(x+6)(x-1) and the third one by 3(x-4)
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  6. #6
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    Ah yes thank you very much earboth, small step I missed.

    2=2*1
    6=3*2
    4= 2^2

    = 2^2*3=12
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  7. #7
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    I have an additional question if someone would please help. If so, please explain this \frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}, I simply do not understand how the numerator changes in this case.
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  8. #8
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    Quote Originally Posted by allyourbass2212 View Post
    I have an additional question if someone would please help. If so, please explain how this \frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}, I simply do not understand how the numerator changes in this case.
    do the two distributions in the numerator and combine like terms ...

    1(x-1) + 1(x+2) = x - 1 + x + 2 = 2x + 1<br />
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  9. #9
    Senior Member Stroodle's Avatar
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    1(x-1)+1(x+2)=x-1+x+2=2x+1

    *edit - Skeeter beat me to it
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