# Partial fractions

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## Homework Statement

Use the method of partial fractions to show that:

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

, may be written as:

$$-2+\frac{1}{1-x}+\frac{1}{1+x}$$

, where $$\lvert x\rvert\neq1$$.

## The Attempt at a Solution

I obviously know how to do it but in the solution am I allowed to do, let x = 1, to cancel some of the terms to figure out the constants in the partial fractions? Or do I have to do simultaneous equations by comparing the coefficients?

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## Homework Statement

Use the method of partial fractions to show that:$$\frac{2x^2}{(1-x(1+x)}$$, may be written as:
$$-2+\frac{1}{1-x}+\frac{1}{1+x}$$, where $$\lvert x\rvert\neq1$$.

## The Attempt at a Solution

I obviously know how to do it but in the solution am I allowed to do, let x = 1, to cancel some of the terms to figure out the constants in the partial fractions? Or do I have to do simultaneous equations by comparing the coefficients?
It depends upon the stage in the solution at which you would set x = 1, or x = -1 .

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MHB
Hi trollcast!

I guess you could let x=1, but I don't see how that would help you...
Neither expression is defined if you do that, and you won't be able to properly compare them.

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Ok I'll type my solution out to check its right:

$$Let \frac{2x^2}{(1-x)(1+x)} = A + \frac{B}{(1-x)}+\frac{C}{(1+x)}$$
$$\frac{A(1+x)(1-x)+B(1+x)+C(1-x)}{(1-x)(1+x)}=2x^2$$
$$\Rightarrow A(1+x)(1-x)+B(1+x)+C(1-x) = 2x^2$$
Compare coefficients of x^2: $$-A=2\\A=-2$$
Let x = 1: $$2B=2\\B=1$$
Let x = -1: $$2c=2\\c=1$$

Is that ok?

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MHB
Nice!

Slightly unconventional, but it seems fine to me.

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Ok I'll type my solution out to check its right:

$$Let \frac{2x^2}{(1-x)(1+x)} = A + \frac{B}{(1-x)}+\frac{C}{(1+x)}$$
$$\frac{A(1+x)(1-x)+B(1+x)+C(1-x)}{(1-x)(1+x)}=2x^2$$
The above line should be:
$\displaystyle \frac{A(1+x)(1-x)+B(1+x)+C(1-x)}{(1-x)(1+x)}=\frac{2x^2}{(1-x)(1+x)}$
$$\Rightarrow A(1+x)(1-x)+B(1+x)+C(1-x) = 2x^2$$
Compare coefficients of x^2: $$-A=2\\A=-2$$
Let x = 1: $$2B=2\\B=1$$
Let x = -1: $$2c=2\\c=1$$

Is that ok?
It's a fairly well known method (trick).

Another variation is to do those substitutions for x immediately after you have

$\displaystyle A(1+x)(1-x)+B(1+x)+C(1-x)=2x^2\$

to get B and C.

Then plug those values in & let x be some convenient number, like x = 0, to find A .

Last edited:
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Thanks, I just wasn't sure whether the |x|=/= 1 , bit meant I couldn't use let x = 1.

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MHB
Thanks, I just wasn't sure whether the |x|=/= 1 , bit meant I couldn't use let x = 1.

Well spotted.
In this case you can get away with it.

When you have your final equation, the solution for A, B, and C is unique and valid for any x, not just for |x|≠1.
After that the same solution applies if you restrict x to |x|≠1.
Formally, you should write something like that down.

Btw, the ultimate check is when you substitute your A, B, and C, and verify they form a proper solution.
When you have that, the steps how you got there become irrelevant (although your teacher may want to see them ;)).

Gold Member
Well spotted.
In this case you can get away with it.

When you have your final equation, the solution for A, B, and C is unique and valid for any x, not just for |x|≠1.
After that the same solution applies if you restrict x to |x|≠1.
Formally, you should write something like that down.

Btw, the ultimate check is when you substitute your A, B, and C, and verify they form a proper solution.
When you have that, the steps how you got there become irrelevant (although your teacher may want to see them ;)).

So the where |x|≠1 is only so it makes mathematical sense otherwise you would imply the denominator could be 0.

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MHB
So the where |x|≠1 is only so it makes mathematical sense otherwise you would imply the denominator could be 0.

Yes.

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Thanks, I just wasn't sure whether the |x|=/= 1 , bit meant I couldn't use let x = 1.
The way I like to think of this is as follows.

If there are values of A, B, and C which make the following true for all x ,
$\displaystyle A(1+x)(1-x)+B(1+x)+C(1-x)=2x^2\,,$​
then certainly the same values of A, B, and C will make the same equation true if we exclude x = ±1 .

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