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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

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  2. #2
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    I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.
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  3. #3
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    My answer is the part in red (Part 3).
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    Quote Originally Posted by r_maths View Post
    My answer is the part in red (Part 3).
    Hmm, I get A = 1, B = 2, which leads to a final answer of: x+\frac{1}{x+1}+\frac{2}{x+2}. I do not see how the x term is not supposed to be there.
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  5. #5
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    I did include the x

    I got the same answer as you, but the answer from the book gives a different answer...

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    Quote Originally Posted by r_maths View Post
    I did include the x

    I got the same answer as you, but the answer from the book gives a different answer...

    I do not see how B equals one, so the book is probably wrong unless both of us missed something.
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  7. #7
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    Hello, r_maths!

    Express \frac{x^3+3x^2+5x+4}{(x+1)(x+2)} in partial fractions.

    \frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \;=\;x + \frac{3x+4}{(x+1)(x+2)} \;=\;x + \frac{1}{x+1} + \frac{2}{x+2}\quad{\color{red}Right!}

    Book answer: . x + \frac{1}{x+1} + \frac{1}{x+2}\quad{\color{red}Wrong!}
    \text{Their answer adds up to: }\;\frac{x^3+3x^2+4x+3}{(x+1)(x+2)} . . . . obviously not the original fraction.


    But your answer checks out!

    \frac{x}{1}\cdot{\color{blue}\frac{(x+1)(x+2)}{(x+  1)(x+2)}} + \frac{1}{x+1}\cdot{\color{blue}\frac{x+2}{x+2}} + \frac{2}{x+2}\cdot{\color{blue}\frac{x+1}{x+1}}

    . . = \;\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)}

    . . = \;\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)}

    . . = \;\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} . . . . See?

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  8. #8
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    \frac{{x^3  + 3x^2  + 5x + 4}}<br />
{{(x + 1)(x + 2)}} = \frac{{x^3  + 3x^2  + 2x + 3x + 4}}<br />
{{(x + 1)(x + 2)}} = \frac{{x(x + 1)(x + 2) + (3x + 4)}}<br />
{{(x + 1)(x + 2)}}.

    Now

    x + \frac{{3x + 4}}<br />
{{(x + 1)(x + 2)}} = x + \frac{{(x + 2) + 2(x + 1)}}<br />
{{(x + 1)(x + 2)}} = x + \frac{1}<br />
{{x + 1}} + \frac{2}<br />
{{x + 2}}.

    Hence \frac{{x^3  + 3x^2  + 5x + 4}}<br />
{{(x + 1)(x + 2)}} = x + \frac{1}<br />
{{x + 1}} + \frac{2}<br />
{{x + 2}}.
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