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

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Answer 1 · 2016-02-20T20:49:57

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

Start off by factoring the denominator as follows

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

We have a repeated linear factor in the denominator, so our decomposition will take the following form:

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

Multiplying both sides by $(x-3)^2$

$3x=A(x-3)+B$

Distributing the $A$

$3x=Ax-3A+B$

Now equate

$Ax=3x$ so $A=3$

$-3A+B=0$

$-3(3)+B=0$

$-9+B=0$ so $B=9$

Now write in $A$ and $B$ into our decomposition

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

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