# 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.