How do you express $2/(x^3-x^2)$ in partial fractions?

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Answer 1 · 2016-03-23T00:13:46

$2/(x^3-x^2) = 2/(x-1)-2/x-2/x^2$

The denominator splits into factors as: $x^3-x^2 = x^2(x-1)$

So we solve for a partial fraction decomposition of the form:

$2/(x^3-x^2)$

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

$=(Ax^2+Bx(x-1)+C(x-1))/(x^3-x^2)$

$=((A+B)x^2+(C-B)x-C)/(x^3-x^2)$

Equating coefficients, we get:

${ (A+B = 0), (C-B = 0), (-C=2) :}$

Hence we find:

${ (C = -2), (B = -2), (A = 2) :}$

So:

$2/(x^3-x^2) = 2/(x-1)-2/x-2/x^2$

How do you express 2/(x^3-x^2)2/(x^3-x^2) in partial fractions?