Why is my partial fraction decomp. wrong?

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1. Jan 8, 2016

kostoglotov

1. The problem statement, all variables and given/known data

Decompose $\frac{2(1-2x^2)}{x(1-x^2)}$

I get A = 2, B =-1, C = 1, but this doesn't recompose into the correct equation, and the calculators for partial fraction decomposition online all agree that it should be A = 2, B = 1, C = 1.

Here is one of the online calculator results with steps shown: http://www.emathhelp.net/calculators/algebra-2/partial-fraction-decomposition-calculator/?numer=2(1-2x^2)&denom=x(1-x)(1+x)&steps=on

2. Relevant equations

3. The attempt at a solution

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

$$\frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} = \frac{2(1-2x^2)}{x(1-x)(1+x)}$$

$$A(1-x)(1+x) + Bx(1+x) + Cx(1-x) = 2(1-2x^2)$$

Let (i) x = 0, (ii) x = +1, (iii) x = -1

$$x = 0, A = 2(1-0), A = 2$$
$$x = 1, B(1)(1+(1)) = 2(1-2) = -2, 2B = -2, B = -1$$
$$x = -1, C(-1)(1-(-1)) = 2(1-2) = -2, -2C = -2, C = 1$$

This of course recomposes to

$$\frac{2(1-x^2)-2x^2}{x(1-x)(1+x)}$$

not the original equation.

The only thing the online calculator really did differently from me is change a term in the denominator from $1-x^2$ to $x^2-1$. I used a shortcut of subbing in the roots of the denominator in order to quickly find the constant numerator values, but even if I use the long method of matching the term coefficients as the online calculator used, I still get my wrong values for A,B and C...why does changing $1-x^2$ to $x^2-1$ matter?

Last edited: Jan 8, 2016
2. Jan 8, 2016

Samy_A

Is there a reason you use $x-1$ instead of $1-x$ under the "B"?:
$$\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} = \frac{2(1-2x^2)}{x(1-x)(1+x)}$$

3. Jan 8, 2016

kostoglotov

typo, will fix, thanks.

4. Jan 8, 2016

Samy_A

As $\frac{2(1-x^2)-2x^2}{x(1-x)(1+x)}=\frac{2-2x^2-2x^2}{x(1-x)(1+x)}=\frac{2-4x^2}{x(1-x)(1+x)}=\frac{2(1-2x^2)}{x(1-x²)}$, I don't see why you say that it recomposes to something different from the original expression.

Your result looks correct to me.

5. Jan 8, 2016

kostoglotov

Selective blindness strikes again...that's worse news actually, because that means it wasn't the part frac decomp where I've messed up the solution to this larger diff eq problem.