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

Feb 20, 2016

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

#### Explanation:

Start off by factoring the denominator as follows

$\frac{3 x}{x - 3} ^ 2$

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

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

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

$3 x = A \left(x - 3\right) + B$

Distributing the $A$

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

Now equate

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

$- 3 A + B = 0$

$- 3 \left(3\right) + B = 0$

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

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

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