How do you write the partial fraction decomposition of the rational expression?

Question

(x-4) / [x^2(x-1)]

Answer 1

You would write this as A/x+B/x^2+C/(x-1) then solve for the coefficients.

One way to solve for the coefficients is to clear the denominators to get A*x*(x-1)+B*(x-1)+C*x^2=x-4 then you can choose one of two options:

A) equate coefficients of equal degrees----
x^2: A+C=0
x^1: -A+B=1
x^0: -B=-4

so that B=4 then A=3 and C=-3

B) use the "special value" approach
plug x=0 into equation to get B*-1=-4 to get B=4
plug x=1 into equation to get C*1=-3
Unfortunately you need to plug something in that doesnt cancel for the remaining varaible, like when x=2 you get
A*2*1+B*1+C*4=-2 but with B and C you can find A

Otherwise, you might be able to "see" the solution with inspiration.

Answer 2

The partial decomposition would be written as follows:

(x - 4) / [x^2(x - 1)] = A/x + B/x^2 + C/(x - 1)

A rule of thumb is, whenever you have a repeated linear factor in a fraction (in this case, it is x^2), then you will have partial fractions progressively reaching that power.

I'll give you a new example: (3x+5)/[(x - 5)^3 (x - 1)]

Here, we have a linear factor which is cubed. Therefore, the partial fraction decomposition would be

(3x+5)/[(x - 5)^3 (x - 1)] = A/(x - 5) + B/(x - 5)^2 + C/(x - 5)^3 + D/(x - 1)

Check your Calculus book for the various cases of partial fraction decomposition. It covers the "repeated linear factors" case, and it also covers "irreducible quadratics".

Answer 3

Looking at the denominator: x^2(x-1), you can see two factorable terms x^2 and x-1.
Partial fraction decomposition requires you to create fractions with every possible degree of each factor, giving you x-1 (from x-1) and x & x^2 (from x^2), putting all of them into the denominator of separate summed fractions.

Therefore, the proper partial fraction decomposition would be:
A/(x) + B/(x^2) + C/(x+1) = (x-4)/[x^2(x-1)]