# How do you write the partial fraction decomposition of the rational expression (2x-6)/((x-1)(x-2)^2)?

Apr 9, 2018

$\textcolor{b l u e}{\frac{4}{x - 2} - \frac{2}{{\left(x - 2\right)}^{2}} - \frac{4}{x - 1}}$

#### Explanation:

$\frac{2 x - 6}{\left(x - 1\right) {\left(x - 2\right)}^{2}}$

We expect the form of the partial fractions to be:

$\frac{2 x - 6}{\left(x - 1\right) {\left(x - 2\right)}^{2}} \equiv \frac{A}{x - 1} + \frac{B}{{\left(x - 2\right)}^{2}} + \frac{C}{x - 2}$

Adding $R H S$

$2 x - 6 \equiv A {\left(x - 2\right)}^{2} + B \left(x - 1\right) + C \left(x - 1\right) \left(x - 2\right)$

Because this is an identity numerators will be equal:

We are looking to find the values of A B and C.

We now set $x$ to values that will eliminate most of the unknowns we are looking for.

$2 x - 6 \equiv A {\left(x - 2\right)}^{2} + B \left(x - 1\right) + C \left(x - 1\right) \left(x - 2\right)$

Setting $x = 2$

$2 \left(2\right) - 6 \equiv A {\left(\left(2\right) - 2\right)}^{2} + B \left(\left(2\right) - 1\right) + C \left(\left(2\right) - 1\right) \left(\left(2\right) - 2\right)$

$- 2 \equiv B$

$B = - 2$

Setting $x = 1$

$2 \left(1\right) - 6 \equiv A {\left(\left(1\right) - 2\right)}^{2} + B \left(\left(1\right) - 1\right) + C \left(\left(1\right) - 1\right) \left(\left(1\right) - 2\right)$

$- 4 \equiv A$

$A = - 4$

Setting $x = 0$

$2 \left(0\right) - 6 \equiv A {\left(\left(0\right) - 2\right)}^{2} + B \left(\left(0\right) - 1\right) + C \left(\left(0\right) - 1\right) \left(\left(0\right) - 2\right)$

$- 6 \equiv 4 A - B + 2 C$

$4 A - B + 2 C = - 6$

We already know A and B:

$4 \left(- 4\right) - \left(- 2\right) + 2 C = - 6$

$C = 4$

So our partial fractions are:

$\textcolor{b l u e}{\frac{4}{x - 2} - \frac{2}{{\left(x - 2\right)}^{2}} - \frac{4}{x - 1}}$