Partial fraction composition of $f(x) := \frac{2-3x}{(x-1)(x-\frac{1}{2})}$

Question

We want to develop the function

$$f(x) := \frac{2-3x}{(x-1)(x-\frac{1}{2})}$$

into a power series at $0$.

What I'm struggling with is the partial fraction decomposition.

The idea is that we split $f(x)$ into

$$\frac{2-3x}{(x-1)(x-\frac{1}{2})} = \frac{A}{x-1} + \frac{B}{x-\frac{1}{2}}$$

What I don't get is how we can solve this for $A$ and $B$.

I tried multiplying both sides with $(x-1)(x-\frac{1}{2})$ and that would give

$$2-3x = A (x-\frac{1}{2}) + B(x-1)$$

but how do we continue from this?

Answer 1

Since $(x-1), (x - \frac{1}{2})$, are relatively prime, there is a solution and furthermore, $deg(A) < deg(x-1)=1, deg(B) < deg(x-\frac{1}{2})=1$.

So $A, B$ are rational numbers.

Writing

\begin{align} 2−3x=A(x - \frac{1}{2})+B(x−1) \end{align}

and evaluating at $x=\frac{1}{2}, x= 1$ gives $A=-2, B=-1$