Resolve in partial form (x^2+15)/(x+3)^2(x^2+3)
You want to get
A/(x+3) + B/(x+3)^2 + (Cx+D)/(x^2+3)
I get
A = 1/2
B = 2
C = -1/2
D = 1/2
compute your operations
I already did. Don't you have some work of your own to show?
In order to get the values, you want to add up the partial fractions to get the original function, with a common denominator:
A/(x+3) + B/(x+3)^2 + (Cx+D)/(x^2+3)
A(x+3)(x^2+3) + B(x^2+3) + (Cx+D)(x+3)^2 = x^2+15
Now expand the left side, collect like terms, and set the coefficients equal to those on the right side.
You will get four equations in four unknowns. Just solve for A,B,C,D.
If you get stuck, show some of your work. I already know how to do it.
from Steve's line of
A(x+3)(x^2+3) + B(x^2+3) + (Cx+D)(x+3)^2 = x^2+15
since this is an identify, thus valid for all values of x, pick some values that eliminate some of the terms, or makes calculations easy
e.g. let x = -3
0 + 12B + 0 = 24
B = 2 *
let x = 0
9A + 3B + 9D = 15
9A + 9D = 9
A + D = 1 **
let x = 1
16A + 4B + 16C + 16D = 16
16A + 8 + 16C + 16D = 16
16A + 16C + 16D = 8
2A + 2C + 2D = 1 ***
let x = -1
8A + 4B - 4C + 4D = 16
2A + B - C + D = 4
2A - C + D = 2 ****
double ** : 2A + 2D = 2
leave as *** 2A + 2C + 2D = 1
subtract them :
2C = -1
C = -1/2
subtract *** - ****
3C + D = -1
-3/2 + D = -1
D = 1/2
As Steve stated at the top:
A = 1/2
B = 2
C = -1/2
D = 1/2