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Thread: decompose fraction question

  1. #1
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    decompose fraction question

    (3x^2+7) / ((x-1)(x^2+2x+5))
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    Member Jonboy's Avatar
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    what's stumpin' ya?
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  3. #3
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    I can decompose fractions of the type A/(x-1)+ B/(x^2+2x+5) but I'm having trouble with fractions of the type A/(x-1)+(Bx+c)(x^2+2x+5). Can someone show me a step by step method of decomposing fractions of the (Bx+C) type?
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  4. #4
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    Finally figured it out. Answer is 5/(4(x-1)) + (7x-3)/(4(x^2+2x+5)). Hope I'm right.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cityismine View Post
    (3x^2+7) / ((x-1)(x^2+2x+5))
    Quote Originally Posted by cityismine View Post
    Finally figured it out. Answer is 5/(4(x-1)) + (7x-3)/(4(x^2+2x+5)). Hope I'm right.
    I'm going to verify your answer:


    $\displaystyle \frac{3x^2+7}{(x-1)(x^2+2x+5)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2x+5}$

    Multiply both sides by the denominator of the left term:

    $\displaystyle 3x^2+7=A(x^2+2x+5)+(Bx+C)(x-1)$

    Now, we get:

    $\displaystyle 3x^2+7=Ax^2+2Ax+5A+Bx^2-Bx+Cx-C$

    $\displaystyle 3x^2+7=(A+B)x^2+(2A-B+C)x+5A-C$

    We now compare the coefficients:

    $\displaystyle \begin{aligned}
    A+B &=3 \\
    2A-B+C &=0 \\
    5A-C &=7
    \end{aligned}
    $

    From the first equation, we see that $\displaystyle B=3-A$. From the third equation, we see that $\displaystyle C=5A-7$. Now substitute these values for B and C into the second equation:

    $\displaystyle 2A-(3-A)+(5A-7)=0 \implies 8A-10=0\implies \color{red}\boxed{A=\frac{5}{4}}$

    Thus, $\displaystyle \color{red}\boxed{C=-\frac{3}{4}}$ and $\displaystyle \color{red}\boxed{B=\frac{7}{4}}$

    Thus the PFD is:

    $\displaystyle \frac{\frac{5}{4}}{x-1}+\frac{-\frac{3}{4}x+\frac{7}{4}}{x^2+2x+5}\implies \color{red}\boxed{\frac{5}{4(x-1)}+\frac{7-3x}{4(x^2+2x+5)}}$.

    It's Good!!
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