The partial fraction decomposition of #(5x+1)/((x-1)(x+1)(x+2))# is given by #A/(x-1)+B/(x+1)+C/(x+2)# . What are the constants of A, B, and C?

2 Answers
Sean
May 6, 2018

#A=1#

#B=2#

#C=-3#

Explanation:

.

#(5x+1)/((x-1)(x+1)(x+2))=A/(x-1)+B/(x+1)+C/(x+2)#

#A/(x-1)+B/(x+1)+C/(x+2)=(A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1))/((x-1)(x+1)(x+2))#

#=(Ax^2+3Ax+2A+Bx^2+Bx-2B+Cx^2-C)/((x-1)(x+1)(x+2))=#

#((A+B+C)x^2+(3A+B)x+(2A-2B-C))/((x-1)(x+1)(x+2))#

This means:

#(A+B+C)x^2+(3A+B)x+(2A-2B-C)=5x+1#

Comparing coefficients, we get:

#A+B+C=0# because there is no #x^2# term on the right side.

#3A+B=5#

#2A-2B-C=1#

We have three equations and three unknowns.

Let's multiply the first equation by #2#:

#2A+2B+2C=0#

Let's add it to the third equation:

#4A+C=1#

#C=1-4A#

From the second equation:

#B=5-3A#

Let's plug these into the first equation:

#A+5-3A+1-4A=0#

#-6A=-6#

#A=1#

#B=2#

#C=-3#

Ananda Dasgupta
May 6, 2018

While the answer in https://socratic.org/s/aQEPEWkD is perfect and the method works in general, there is a shortcut that works rather well in examples like this one - in which all the factors of the denominator are monomials.

Explanation:

Starting with

#(5x+1)/((x-1)(x+1)(x+2)) = A/(x-1)+B/(x+1)+C/(x+2)#

multiply both sides by #(x-1)(x+1)(x+2)# we get

#5x+1 = A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)#

Since this is an identity, it should be valid for all values of #x#.

  • substituting #x=1#, gives

#5times 1+1 = Atimes 2 times 3 implies color(red)(A = 1)#

  • substituting #x=-1#, gives

#5times (-1)1+1 = Btimes (-2) times 1 implies color(red)(B = 2)#

  • substituting #x=-2#, gives

#5times (-2)+1 = Ctimes (-3) times (-1) implies color(red)(C = -3)#

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