# Another type of Partial Fractions

• Jun 9th 2010, 03:28 AM
Cthul
Another type of Partial Fractions
The question starts as

$\frac{4x}{(1+x)(1-x)^{2}}$

I've gotten this far and don't know how to solve the rest.

$\frac{4x}{(1+x)(1-x)^{2}}\equiv\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^{2}}$

$4x=A(1-x)^{2}+B(1-x)(1+x)+c(1+x)$

Substitute
$x=1$

$2=c$

Then I don't know how to solve the rest, when I substitute an x, I get a zero for both, it's completely different from this.
• Jun 9th 2010, 05:15 AM
Quote:

Originally Posted by Cthul
The question starts as

$\frac{4x}{(1+x)(1-x)^{2}}$

I've gotten this far and don't know how to solve the rest.

$\frac{4x}{(1+x)(1-x)^{2}}\equiv\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^{2}}$

$4x=A(1-x)^{2}+B(1-x)(1+x)+c(1+x)$

Substitute
$x=1$

$2=c$

Then I don't know how to solve the rest, when I substitute an x, I get a zero for both, it's completely different from this.

Hi Cthul,

now that you have found C, place $x=-1$ to bypass B and discover A.

Then as you have both A and C, you can place x=0 to find B.
• Jun 10th 2010, 12:23 AM
Cthul
It won't work, it doesn't when I sub in zero, I get 2 variables.
• Jun 10th 2010, 12:32 AM
Prove It
Quote:

Originally Posted by Cthul
It won't work, it doesn't when I sub in zero, I get 2 variables.

Try substituting $x = -1$ first to find $A$.

Once you have $A$, then you can let $x = 0$ to find $B$.

My preferred method is expanding all the brackets, equating coefficients of like powers of $x$ and solving the system.
• Jun 10th 2010, 01:56 AM
Quote:

Originally Posted by Cthul
It won't work, it doesn't when I sub in zero, I get 2 variables.

No,

substitute in the values you obtained for A and C before finding B.

x=1, C=2.

x=-1, 4A=-4, A=-1

x=0, 0=A+B+C=-1+B+2=B+1, so B=-1
• Jun 10th 2010, 03:05 AM
Cthul
I thought when x is substituted with zero, you'd get... $(1-(-1))^{2}$ which makes .... oh, I see. Thanks. The signs get confusing.
• Jun 10th 2010, 05:12 AM
Quote:

Originally Posted by Cthul
I thought when x is substituted with zero, you'd get... $(1-(-1))^{2}$ which makes .... oh, I see. Thanks. The signs get confusing.

Thinking of subtracting negative numbers as the difference between temperatures can be helpful...

5-2 is the difference between 5 and 2 degrees, which is 3 degrees.

There is a 6 degree difference between between 5 and -1,
5 degrees down to zero and another degree down to -1.
That's a drop of 6 degrees.

So 5-(-1) is that difference which we know is 6,
subtracting the lower temperature from the bigger one.

Since 5+1=6, we can say 5--1=5+1,
so 1--1=1+1 and so on...

-2-3 is subtract 2, then subtract 3, which is subtract 5 or -(2+3)

-2--3=-2-(-3), the difference between -2 degrees and -3 degrees which is 1 degree
as -2 is greater than -3,

so -2--3=1=3-2 or -2+3 etc
• Jun 10th 2010, 12:09 PM
Circulation
Partial fractions appear to be a very interesting problem.