How do you write $2/(x^3-x^2)$ as a partial fraction decomposition?

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Answer 1 · 2016-11-08T08:25:05

The result is $=-2/x^2-2/x+2/(x-1)$

Let's factorise the denominator, $x^3-x^2=x^2(x-1)$
So, $2/(x^3-x^2)=2/(x^2(x-1))=A/x^2+B/x+C/(x-1)$
$=(A(x-1)+Bx(x-1)+Cx^2)/(x^2(x-1))$

$:. 2=A(x-1)+Bx(x-1)+Cx^2$
If $x=0$ $=>$ $2=-A$ $=>$ $A=-2$
Coefficients of $x$ , $0=A-B$ $=>$ $B=-2$
Coefficients of $x^2$ , 0=B+C $=>$ $C=2$

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

How do you write 2/(x^3-x^2) 2/(x^3-x^2) as a partial fraction decomposition?