# Contingent factors method

• Sep 10th 2010, 12:10 PM
Revy
Partial Fractions decomposition
I have an example of Partial Fractions decomposition, but cant understand it.

Here we have $\frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}$
$\frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A}{(x-2)^2}+\frac{B}{x-1}+\frac{C_x+D}{x^2+x+1}$
$\frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2}{(x-1)^2(x^2+x+1)}$

$2x^3+4x^2+x+2=A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2$

And whats goes next, I totally don't get
x^3 $B+C=2$
x^2 $A-2C+D=4$
x^1 $A+C-2D=1$
x^0 $A-B+D=2$

$A=3$; $B=2$; $C=0$; $D=1$

So could anyone explain me that most imprtant part?

Now that I think more, it seems that I posted it on wrong section. It's just that later that fromula will be integrated, so it should be known in calculus.
Seems I used wrong name for this method
• Sep 10th 2010, 01:03 PM
yeKciM
Quote:

Originally Posted by Revy
I have an example of Partial Fractions decomposition, but cant understand it.

Here we have $\frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}$
$\frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A}{(x-2)^2}+\frac{B}{x-1}+\frac{C_x+D}{x^2+x+1}$
$\frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2}{(x-1)^2(x^2+x+1)}$

$2x^3+4x^2+x+2=A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2$

And whats goes next, I totally don't get
x^3 $A+B=2$
x^2 $A-2C+D=4$
x^1 $A+C-2D=1$
x^0 $A-B+D=2$

$A=3$; $B=2$; $C=0$; $D=1$

So could anyone explain me that most imprtant part?

Now that I think more, it seems that I posted it on wrong section. It's just that later that fromula will be integrated, so it should be known in calculus.
Seems I used wrong name for this method

well first of all that isn't true ...
you have the line ( I didn't check all your work up there) that says :

$A+B = 2$

and you have there results $A= 3$ and $B=2$

if you are to put that solutions in that equation you get :

$3+2 = 2$

which isn't true :D

when done correctly you just put those A,B,C,D,E .... coefficient back to the equation and you will get partial fractions :D

note that you can always check your work, if you are just that what did you get (partial fractions) solve :

$\displaystyle \frac {A}{something} \pm \frac {B}{something \; else } \pm ....$

and you must get starting fraction ... if that's not the case than you made some errors while working :D
• Sep 10th 2010, 01:09 PM
Revy
Quote:

Originally Posted by yeKciM
if that's not the case than you made some errors while working :D

Yeah =D you're right, I made mistake (thats why I do hate those undersized laptops)
Sorry, $A+B=2$ should be $B+C=2$
now it's fixed on original post

Seems I made this thread uselessly, because after finding out english name of method I found some easy examples (not text-book ones, where you look and see ****). So, finally, at second day of brainstorming I learned something new =D
• Sep 10th 2010, 01:27 PM
yeKciM
Quote:

Originally Posted by Revy
Yeah =D you're right, I made mistake (thats why I do hate those undersized laptops)
Sorry, $A+B=2$ should be $B+C=2$
now it's fixed on original post

now it's okay :D :D

$\displaystyle \frac {2x^3+4x^2+x+2}{(x-1)^2 (x^2+x+1)} =3 \cdot \frac {1}{(x-1)^2} +2\cdot \frac {1}{x-1} + \frac {1}{x^2+x+1}$

Quote:

Originally Posted by Revy
And whats goes next, I totally don't get
x^3 $B+C=2$
x^2 $A-2C+D=4$
x^1 $A+C-2D=1$
x^0 $A-B+D=2$

$A=3$; $B=2$; $C=0$; $D=1$

So could anyone explain me that most imprtant part?

now what you did there is just group those x-es with same power and look at the starting fraction and see is there any with $x^3$ of there is okay, and which coefficient is with him ? okay it's 2 that means those must be equal to 2 that's way you have B+C=2 and so on ... than just solve system of equations and your off :D
• Sep 10th 2010, 01:43 PM
Revy
Ah.. I wanted to do expanding and then collecting on paper, but now I see that it's easy to do linear system just by looking at equation.

So that's what those x^n, 0<=x<=3 meant :D