# Math Help - Partial fractions

1. ## Partial fractions

Alright i am working on homework and got stuck on a problem, looked it up online and im unsure of how they did a particular point.

Problem:

A(x-2)(x2+2) + B(x+2)(x2+2) + (Cx+D)(x2-4) = x2

(A+B+C)x3 + (-2A+2B+D)x2 + (2A+2B-4C)x + (-4A+4B-4D) = x2

A+B+C=0 (1) Use 4.(2) + (4)
-2A+2B+D=1 (2) >> -3A+3B=1 (5)
2A+2B-4C=0 (3) Use 4.(1) + (3)
-4A+4B-4D=0 (4) >> 6A+6B = 0 (6)

I know the work is correct, however i am unsure of how you get
(A+B+C)x3 + (-2A+2B+D)x2 + (2A+2B-4C)x + (-4A+4B-4D)

And how you decide the numbers in
A+B+C=0 (1) Use 4.(2) + (4)
-2A+2B+D=1 (2) >> -3A+3B=1 (5)
2A+2B-4C=0 (3) Use 4.(1) + (3)
-4A+4B-4D=0 (4) >> 6A+6B = 0 (6)

For instance, how do you know A+B+C=0? why not A+B+C=1?

2. Originally Posted by Juggalomike
Alright i am working on homework and got stuck on a problem, looked it up online and im unsure of how they did a particular point.

Problem:

A(x-2)(x2+2) + B(x+2)(x2+2) + (Cx+D)(x2-4) = x2

(A+B+C)x3 + (-2A+2B+D)x2 + (2A+2B-4C)x + (-4A+4B-4D) = x2

A+B+C=0 (1) Use 4.(2) + (4)
-2A+2B+D=1 (2) >> -3A+3B=1 (5)
2A+2B-4C=0 (3) Use 4.(1) + (3)
-4A+4B-4D=0 (4) >> 6A+6B = 0 (6)

I know the work is correct, however i am unsure of how you get
(A+B+C)x3 + (-2A+2B+D)x2 + (2A+2B-4C)x + (-4A+4B-4D)

And how you decide the numbers in
A+B+C=0 (1) Use 4.(2) + (4)
-2A+2B+D=1 (2) >> -3A+3B=1 (5)
2A+2B-4C=0 (3) Use 4.(1) + (3)
-4A+4B-4D=0 (4) >> 6A+6B = 0 (6)

For instance, how do you know A+B+C=0? why not A+B+C=1?

$A(x-2)(x^2+2)+B(x+2)(x^2+2)+(Cx+D)(x^2-4) = x^4-2x^2-8$

the factor of x^4 equal in the two sides for and the factor of x^3 equal in the two sides and it is same for x^2 and x and the constant so for example

x^2 factor in right side is -2 so the factor of x^2 in left side should be same I think after you simplify left side x^2 factor will be

-2A+2B+D should equal -2 .......you can continue