Start with:
(6x+5)/(x+2)^4 = A/(x+2)^4 + B/(x+2)^3+C/(x+2)^2+D/(x+2)
Multiply both sides by (x+2)^4:
6x+5 = A + B(x+2)+C(x+2)^2+D(x+2)^3
Let x = -2
6(-2)+5 = A
A = -7
Add 7 to both sides of equation:
6x+12 = B(x+2)+C(x+2)^2+D(x+2)^3
Let x = 0:
[
(2,4,8,|,12),
]
Let x = -1
[
(2,4,8,|,12),
(1,1,1,|,6)
]
Let x = 1
[
(2,4,8,|,12),
(1,1,1,|,6),
(3,9,27,|,18)
]
R_1harrR_2
[
(1,1,1,|,6),
(2,4,8,|,12),
(3,9,27,|,18)
]
R_2-2R_1toR_2
[
(1,1,1,|,6),
(0,2,6,|,0),
(3,9,27,|,18)
]
R_3-3R_1toR_3
[
(1,1,1,|,6),
(0,2,6,|,0),
(0,6,24,|,0)
]
R_2/2toR_2
[
(1,1,1,|,6),
(0,1,3,|,0),
(0,6,24,|,0)
]
R_3/6toR_3
[
(1,1,1,|,6),
(0,1,3,|,0),
(0,1,4,|,0)
]
R_3-R_3toR_3
[
(1,1,1,|,6),
(0,1,3,|,0),
(0,0,1,|,0)
]
It is obvious that C and D are 0 so we can clear the matrix:
[
(1,0,0,|,6),
(0,1,0,|,0),
(0,0,1,|,0)
]
B = 6 and C = D = 0
(6x+5)/(x+2)^4 = -7/(x+2)^4 + 6/(x+2)^3