How do you express $(3x) / (x-3)^2$ in partial fractions?

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Answer 1 · 2016-10-24T16:29:10

$(3x)/(x-3)^2hArr3/(x-3)+9/(x-3)^2$

Partial fractions of $(3x)/(x-3)^2$ are in the form $A/(x-3)+B/(x-3)^2$

i.e. $(3x)/(x-3)^2hArrA/(x-3)+B/(x-3)^2$

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

Comparing like terms $A=3$ and $-3A+B=0$ i.e. $B=3A=9$

Hence $(3x)/(x-3)^2hArr3/(x-3)+9/(x-3)^2$

How do you express (3x) / (x-3)^2(3x) / (x-3)^2 in partial fractions?