What is the partial-fraction decomposition of (x^2+2x+7)/(x(x-1)^2)?

1 Answer
Jim H
Jun 24, 2015

(x^2+2x+7)/(x(x-1)^2) = 7/x-6/(x-1)+10/(x-1)^2

Explanation:

The factors on the denominator are obvious. They are all linear, but one of them is a double factor. So we want A, B, and C so that:

(x^2+2x+7)/(x(x-1)^2) = A/x+B/(x-1)+C/(x-1)^2

Combining the rations on the right, we get a numerator pf:
A(x^2-2x+1) +B(x^2-x) +Cx = Ax^2-2Ax+A+Bx^2-Bx+Cx

= (A+B)x^2 +(-2A-B+C)x+A.

Setting the coefficients equal to those of the original numerator, x^2+2x+7, we get:

A+B=1
-2A-B+C = 2
A=7

It is immediate that A=7 and from that and the first equation (the coefficients of x^2), we get B=-6. Substituting in the middle equation and solving for C, we get C=10.

(x^2+2x+7)/(x(x-1)^2) = 7/x-6/(x-1)+10/(x-1)^2

It is a good idea to check the answer by getting the common denominator. (I did that on paper, but I'm not going to type it up.)

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