Decomposing Polynomial Fractions

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Vertciel

Junior Member
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May 13, 2007
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Hello everyone,

I am not getting the answer listed in my book for the following two problems. My work is shown below. Thank you for your help.

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1. \(\displaystyle \frac{2x + 9}{x^3 - 1}\)

= \(\displaystyle \frac{2x + 9}{(x - 1)(x^2 +x + 1)}\)

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

\(\displaystyle 2x + 9 = A(x^2 + x + 1) + (Bx - C)(x - 1)\)

\(\displaystyle = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C\)

\(\displaystyle = (A + B)x^2 + (A - B + C)x + A - C\)

\(\displaystyle A + B = 0\)

\(\displaystyle A - B + C = 2\)

\(\displaystyle A - C = 9\)

From the system of equations:

\(\displaystyle A = 11/3\)
\(\displaystyle B = -11/3\)
\(\displaystyle C = -16/3\)

Partial fraction: \(\displaystyle \frac{11}{3(x - 1)}\ + \frac{-11x - 16}{3(x^2 + x + 1)}\)

Book's answer = \(\displaystyle \frac{11}{2(x - 1)}\ + \frac{-11x + 9}{2(x^2 + x + 1)}\)




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

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

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

= \(\displaystyle \frac{A}{x}\ + \frac{B}{x-2}\ + \frac{C}{(x-2)^2}\\)

\(\displaystyle x^2 + 2x + 8 = A(x - 2)^2 + B[x(x - 2)] + Cx\)

\(\displaystyle = (A + B)x^2 + (-4A - 2B + C)x + 4A\)

From the system of equations:

\(\displaystyle A + B = 1\)
\(\displaystyle -4A - 2B + C = 2\)
\(\displaystyle 4A = 8\)

\(\displaystyle A = 2\)
\(\displaystyle B = -1\)
\(\displaystyle C = 8\)

Partial fraction: \(\displaystyle \frac{2}{x}\ - \frac{1}{x - 2}\ + \frac{8}{(x - 2)^2}\\)

Book's answer: \(\displaystyle \frac{2}{x}\ - \frac{2}{x - 2}\ + \frac{x + 6}{(x - 2)^2}\\)
 
You missed two signs.
\(\displaystyle \L 2x + 9 = A(x^2 + x + 1) + (Bx - C)(x - 1)\)
\(\displaystyle \L 2x + 9 = A(x^2 + x + 1) + (Bx^2 - Bx )+ ( - Cx + C)\)
 
1) I agree with your answer. And the book's answer does not "compress" back to the original expression, so it's not just an alternate form or something.

2) I agree with your answer, and am not sure what method the solver used to arrive at the provided answer...? But both expansions "compress" to the original expression.

(Note: I think you typoed some signs in your work, but the following steps are correct, so I think whatever you did on your scratch paper is probably correct. The errors are just transcription problems.)

Eliz.
 
Thank you very much pka and stapel for your helpful input.

I just have one more question. In the following problem, could you please explain how my answer can be simplified into the book's answer?

3. \(\displaystyle \frac{x^2 + 1}{2x^3 + 5x^2 - 3x}\)

My answer: \(\displaystyle \frac{-1/3}{x}\ + \frac{5/7}{2x - 1}\ + \frac{10/21}{x + 3}\) = \(\displaystyle \frac{-1}{3x}\ + \frac{5}{7(2x - 1)}\ + \frac{10}{21(x + 3)}\)

Answer in book: \(\displaystyle \frac{-1}{x}\ + \frac{5}{21(2x - 1)}\ + \frac{20}{21(x + 3)}\)

Thanks again.
 
I don't believe they're the same. Your answer is correct, though.

The books answer is equivalent to:

\(\displaystyle \L\\\frac{3x^{2}-110x+63}{42x^{3}+105x^{2}-63x}\)
 
Thanks for your reply.

So based on the solutions here, would it be correct to say that one fraction can be decomposed in different ways?
 
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