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
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

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)