How do you express $[(3x)/(x^2-6x+9)]$ in partial fractions?

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Answer 1 · 2016-02-27T06:25:10

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

We can write this as follows

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

Now we need to find the values of real constants $A,B$ hence we have that

for $x=0$ we have that $0=-A/3+B/9=>B=3A$

for $x=2$ we have that $6=-A+B=>6=-A+3A=>A=3$

and $B=9$

so finally we have that

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

How do you express [(3x)/(x^2-6x+9)] [(3x)/(x^2-6x+9)] in partial fractions?