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

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

#### Explanation:

We can write this as follows

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

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

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

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

and $B = 9$

so finally we have that

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