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

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Answer 1 · 2016-02-20T20:49:57

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

Start off by factoring the denominator as follows

$\frac{3 x}{{(x - 3)}^{2}}$ $(3x)/(x-3)^2$

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

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

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

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

Distributing the $A$ $A$

$3 x = A x - 3 A + B$ $3x=Ax-3A+B$

Now equate

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

$- 3 A + B = 0$ $-3A+B=0$

$- 3 (3) + B = 0$ $-3(3)+B=0$

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

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

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

How do you write the partial fraction decomposition of the rational expression [3xx2−6x+9] [(3x)/(x^2-6x+9)]?