How do you write the partial fraction decomposition of the rational expression $1/(x^3-6x^2+9x)$ ?

Question

1553 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-14T03:53:53

$1/(9x)-1/(9(x-3))+1/(3(x-3)^2$

Factor the denominator.

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

$1=A(x-3)^2+B(x^2-3x)+C(x)$

$1=Ax^2-6Ax+9A+Bx^2-3Bx+Cx$

$1=x^2(A+B)+x(-6A-3B+C)+1(9A)$

Thus, ${(A+B=0),(-6A-3B+C=0),(9A=1):}$

Solve to see that ${(A=1/9),(B=-1/9),(C=1/3):}$

Therefore,

$1/(x^3-6x^2+9x)=1/(9x)-1/(9(x-3))+1/(3(x-3)^2$

How do you write the partial fraction decomposition of the rational expression 1/(x^3-6x^2+9x) 1/(x^3-6x^2+9x) ?