Why is my partial fraction decomp. wrong?

[kostoglotov]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[/B]
$$\frac{2(1-2x^2)}{x(1-x^2)} = \frac{2(1-2x^2)}{x(1-x)(1+x)}$$

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

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

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?

 

[Samy_A]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[/B]
$$\frac{2(1-2x^2)}{x(1-x^2)} = \frac{2(1-2x^2)}{x(1-x)(1+x)}$$

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

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

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?

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

 

[kostoglotov]

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

typo, will fix, thanks.

 

[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.

 

[kostoglotov]

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.

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.