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

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x + 6) (x^{2} + 3)}$ $1/((x+6)(x^2+3))$ ?

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Answer 1 · 2016-02-27T06:25:10

It is $\frac{3 x}{x^{2} - 6 x + 9} = \frac{3}{x - 3} + \frac{9}{{(x - 3)}^{2}}$ $(3x)/(x^2-6x+9)=3/(x-3)+9/(x-3)^2$

We can write this as follows

$\frac{3 x}{x^{2} - 6 x + 9} = \frac{A}{x - 3} + \frac{B}{{(x - 3)}^{2}}$ $(3x)/(x^2-6x+9)=A/(x-3)+B/(x-3)^2$

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

for $x = 0$ $x=0$ we have that $0 = - \frac{A}{3} + \frac{B}{9} \Rightarrow B = 3 A$ $0=-A/3+B/9=>B=3A$

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

and $B = 9$ $B=9$

so finally we have that

$\frac{3 x}{x^{2} - 6 x + 9} = \frac{3}{x - 3} + \frac{9}{{(x - 3)}^{2}}$ $(3x)/(x^2-6x+9)=3/(x-3)+9/(x-3)^2$

How do you express [3xx2−6x+9] [(3x)/(x^2-6x+9)] in partial fractions?