# Partial Fractions

• Mar 22nd 2010, 01:01 AM
ugkwan
Partial Fractions
The question simply ask to write the partial fraction decomposition of the rational expression.

(x^2) / (x^4 - 2x^2 - 8)

I wrote it out as:

(A / (x-2)) + (B / (x+2)) + ((Cx+D) / (x^2+2))

I got as far as simplifying it as:

Ax^3+2Ax^2+2Ax+4A+Bx^3-2Bx^2+2Bx-4B+Cx^3-4Cx+Dx^2-4D

So I group them up as:

(A+B+C)x^3=0
(2A-2B+D)x^2=1
(2A+2B-4C)x=0
(A-B-D)4=0

Now I'm lost in using the elimination method because I just don't see how to peace together this puzzle, and the use the answer to complete the partial fraction decomposition. OMG I need some assistance please, this problem is driving me mad...
• Mar 22nd 2010, 01:43 AM
Quote:

Originally Posted by ugkwan
The question simply ask to write the partial fraction decomposition of the rational expression.

(x^2) / (x^4 - 2x^2 - 8)

I wrote it out as:

(A / (x-2)) + (B / (x+2)) + ((Cx+D) / (x^2+2))

I got as far as simplifying it as:

Ax^3+2Ax^2+2Ax+4A+Bx^3-2Bx^2+2Bx-4B+Cx^3-4Cx+Dx^2-4D

So I group them up as:

(A+B+C)x^3]=0
(2A-2B+D)x^2=1
(2A+2B-4C)]x=0
(A-B-D)4=0

Now I'm lost in using the elimination method because I just don't see how to peace together this puzzle, and the use the answer to complete the partial fraction decomposition. OMG I need some assistance please, this problem is driving me mad...

hi ugkwan,

continuing on, this means

A+B+C=0 (1)
2A-2B+D=1 (2)
2A+2B-4C=0 (3)
A-B-D=0 (4)

This is a system of 4 simultaneous equations,
for which there is enough information to solve for the 4 values A, B, C, D

Combining (1) and (3)....2A+2B+2C=0 and 2A+2B-4C=0

subtracting these gives 6C=0, so \$\displaystyle C=0\$

Hence 2A+2B=0 gives \$\displaystyle A=-B\$

Hence, from (4) we have 2A-D=0, so \$\displaystyle 2A=D\$

therefore, using (2).....2A-2B+D=1 gives 4A+D=1 so \$\displaystyle 2D+D=1\$ so \$\displaystyle D=1/3\$

Hence \$\displaystyle A=1/6\$ and \$\displaystyle B=-1/6\$