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    another fraction decomposition

    x^2 / (x^3-8)
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    Quote Originally Posted by cityismine View Post
    x^2 / (x^3-8)
    Can you factor the denominator? Hint: it's a difference of cubes.
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    Still can't solve it, stuck!
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    Jen
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    Quote Originally Posted by cityismine View Post
    x^2 / (x^3-8)
    Quote Originally Posted by Reckoner View Post
    Can you factor the denominator? Hint: it's a difference of cubes.
    As Reckoner said, use the difference of cubes

    The difference of cubes is,

    a^3-b^3=(a-b)(a^2+ab+b^2)

    and 8=2^3

    Hope that helps.
    Last edited by Jen; May 28th 2008 at 02:54 PM.
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    I understand the difference of cubes, but I'm having trouble decomposing the fraction.
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    Quote Originally Posted by cityismine View Post
    I understand the difference of cubes, but I'm having trouble decomposing the fraction.
    Show us what you have done so far so that we can see what, exactly, you are having trouble with.

    Once you have factored the denominator completely, set up your new fractions:

    For every linear factor (ax + b) include the term \frac{A}{ax + b}. If you have powers of the same factor, (ax + b)^n, include all of the terms \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}.

    For each quadratic factor (or power of a quadratic factor), (ax^2 + bx + c)^n, include all of the terms \frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n}.

    So, in this case we have

    \frac{x^2}{x^3 - 8} = \frac{x^2}{(x - 2)(x^2 + 2x + 4)}

    =\frac{x^2}{{\color{red}(x - 2)}{\color{blue}(x^2 + 2x + 4)}}

    ={\color{red}\frac A{x - 2}} + {\color{blue}\frac{Bx + C}{x^2 + 2x + 4}}

    So, we have

    \frac{x^2}{x^3 - 8}=\frac A{x - 2} + \frac{Bx + C}{x^2 + 2x + 4}

    Multiply through by x^3 - 8:

    x^2 = A(x^2 + 2x + 4) + (Bx + C)(x - 2)

    Collect terms of the same degree, and equate the coefficients of both sides. Can you finish?
    Last edited by Reckoner; May 27th 2008 at 09:19 PM. Reason: Error correction
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Reckoner View Post
    Show us what you have done so far so that we can see what, exactly, you are having trouble with.

    Once you have factored the denominator completely, set up your new fractions:

    For every linear factor (ax + b) include the term \frac{A}{ax + b}. If you have powers of the same factor, (ax + b)^n, include all of the terms \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}.

    For each quadratic factor (or power of a quadratic factor), (ax^2 + bx + c)^n, include all of the terms \frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n}.

    So, in this case we have

    \frac{x^2}{x^3 - 8} = \frac{x^2}{(x - 2)(x^2 + 2x + 4)}

    =\frac{x^2}{{\color{red}(x - 2)}{\color{blue}(x^2 + 2x + 4)}}

    ={\color{red}\frac A{x - 2}} + \boxed{{\color{blue}\frac B{x^2 + 2x + 4}}}

    So, we have

    \frac{x^2}{x^3 - 8}=\frac A{x - 2} + \boxed{\frac B{x^2 + 2x + 4}}

    Multiply through by x^3 - 8:

    x^2 = A(x^2 + 2x + 4) + B(x - 2)

    Collect terms of the same degree, and equate the coefficients of both sides. Can you finish?

    The degree of the numerator is one less than the denominator. It should be \frac{Bx+C}{x^2+2x+4}, not \frac{B}{x^2+2x+4}
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    Quote Originally Posted by Chris L T521 View Post
    The degree of the numerator is one less than the denominator. It should be \frac{Bx+C}{x^2+2x+4}, not \frac{B}{x^2+2x+4}
    You're right. I even mentioned as much a few lines above. I should get some rest!
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