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#