# Math Help - Partial Fractions Help

1. ## Partial Fractions Help

Ok. I gotta do the fraction decomposition of

$\frac{6x^2+1}{x^2(x-1)^3}$.

~~~~~~~~~~~

My problem is when I get to the part where I figure out what goes into A-E. I'll show the work I have right now below.

$6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2$
$6x^2+1=A(x^4+3x^3+3x^2-x)+B(x^3 + x^2 + 3x-1)+C(x^4-2x^3+x^2)+D(x^3-x^2)+Ex^2$
$6x^2+1=(A+C)x^4+(3A+B-2C+D)x^3+(3A+3B+C-D+E)x^2+(-A-B)x+(-B)$

Given the highest power of the numerator is $x^2$. I'm really at a lost what to do from here. I'm not even sure if I'm doing it right. So some help would be greatly appreciated

EDIT: I'm gonna go rework this, something tells me I'm really close to the answer.

2. $\frac{7}{(x-1)^{3}}-\frac{2}{(x-1)^{2}}+\frac{3}{x-1}-\frac{3}{x}-\frac{1}{x^{2}}$

3. Originally Posted by mathgeek777
Ok. I gotta do the fraction decomposition of

$\frac{6x^2+1}{x^2(x-1)^3}$.

~~~~~~~~~~~

My problem is when I get to the part where I figure out what goes into A-E. I'll show the work I have right now below.

$6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2$
$6x^2+1=A(x^4+3x^3+3x^2-x)+B(x^3 + x^2 + 3x-1)+C(x^4-2x^3+x^2)+D(x^3-x^2)+Ex^2$
$6x^2+1=(A+C)x^4+(3A+B-2C+D)x^3+(3A+3B+C-D+E)x^2+(-A-B)x+(-B)$

Given the highest power of the numerator is $x^2$. I'm really at a lost what to do from here. I'm not even sure if I'm doing it right. So some help would be greatly appreciated

EDIT: I'm gonna go rework this, something tells me I'm really close to the answer.
$6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2$

Since it's gotta work for all values of x, you can substitute at least one simple value of x to make life slightly easier:

x = 0: 1 = -B therefore B = -1.

4. Originally Posted by mr fantastic
$6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2$

Since it's gotta work for all values of x, you can substitute at least one simple value of x to make life slightly easier:

x = 0: 1 = -B therefore B = -1.
wait, what?

you lost me.....

5. Originally Posted by mathgeek777
wait, what?

you lost me.....

What Mr. Fantastic is saying is that the Equation is true for ALL values of x.

so you can pick some to make your computations easier. Like x=0

$
6(0)^2+1=A[0(0-3)]^3+B(0-1)^3+C[0^2(0-1)^2]+D[0^2(x-1)]+E(0)^2
$

Simplifying

$1=-B$

6. Originally Posted by mr fantastic
Since it's gotta work for all values of x, you can substitute at least one simple value of x to make life slightly easier:

x = 0: 1 = -B therefore B = -1.
$6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2$

is an identity, true for all values of x.

So it has to be true in particular for convenient simple values of x like x = 0. So substitute x = 0:

The left hand side becomes 6(0)^2 + 1 = 1.
The right hand side becomes B(-1)^3 = -B.

This idea can sometimes be a good short cut for getting the values of A, B etc. rather than expanding, equating coefficients, writing down and then solving simultaneous equations etc.

7. Originally Posted by mr fantastic
$6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2$

is an identity, true for all values of x.

So it has to be true in particular for convenient simple values of x like x = 0. So substitute x = 0:

The left hand side becomes 6(0)^2 + 1 = 1.
The right hand side becomes B(-1)^3 = -B.

This idea can sometimes be a good short cut for getting the values of A, B etc. rather than expanding, equating coefficients, writing down and then solving simultaneous equations etc.
Can this be done successfully with the rest of the missing values on partial fractions?

8. My computer froze, please disregard

9. Originally Posted by mathgeek777
Can this be done successfully with the rest of the missing values on partial fractions?
Yes, but you'll get a series of equations to be solved simultaneously rather than any further values dropping out easily.