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I want to convert $$\frac{2x^3+4}{x^2-1}$$ into partial fractions but I've made a mistake somewhere. $$\frac{2x^3+4}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$$ and found that $A=3$ and $B=-1$ so the partial fraction should be $$\frac{2x^3+4}{x^2-1}=\frac{3}{x-1}-\frac{1}{x+1}$$ but the answer is $$\frac{2x^3+4}{x^2-1}=2x+\frac{3}{x-1}-\frac{1}{x+1}$$ and I don't get why.

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    $\begingroup$ You forgot to do long division - first reduce the degree of top $\endgroup$
    – AgentS
    Feb 5, 2018 at 15:57

3 Answers 3

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The degree of numerator must be less than degree of denominator.

So either do the long division, or do the following to do that.

$$\frac{2x^3+4}{x^2-1}=\frac{2x^3-2x+2x+4}{x^2-1}=2x+\frac{2x+4}{x^2-1}$$

Now let $$\frac{2x+4}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$$

to get $$\frac{2x+4}{x^2-1}=\frac{3}{x-1}-\frac{1}{x+1}$$

Now finally;

$$\frac{2x^3+4}{x^2-1}=2x+\frac{2x+4}{x^2-1}=2x+\frac{3}{x-1}-\frac{1}{x+1}$$

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Since the numerator's degree is higher than the denominator's, you must first divide the polynomials:

$$\frac{2x^3+4}{x^2-1}=2x+\frac{2x+4}{x^2-1}$$

and now you do partial fractions with the rightmost fraction (in which the denominator's degree is already higher than the numerator's), and you get the answer.

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hint: at first you must divide $2x^3+4$ by $x^2-1$ the result should be $$2\,x+3\, \left( x-1 \right) ^{-1}- \left( x+1 \right) ^{-1}$$

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