How do you write the partial fraction decomposition of the rational expression #(6x+5) / (x+2) ^4#?

1 Answer
May 27, 2017

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#