Suppose we’re given the following expression

$$ \begin{align*} \frac{- 3x + 4}{x^2 - 2x} \end{align*} $$

and we want to decompose it into multiple terms. We first want to factor the denominator. This is easy. \((x^2 - 2x = x(x - 2))\). Next we’ll write the following:

$$ \begin{align*} \frac{- 3x + 4}{x(x - 2)} = &= \frac{A}{x} + \frac{B}{x-2} \end{align*} $$

The goal is to find \(A\) and \(B\) such that this works. From the earlier post (solving a rational equation), we know that the easiest way to do this is to just to multiply both sides by the LCD (least common multiple) of all denominators so,

$$ \begin{align*} x(x - 2)\frac{- 3x + 4}{x(x - 2)} = &= x(x - 2)\big( \frac{A}{x} + \frac{B}{x-2} \big) \\ - 3x + 4 &= A(x - 2)+ Bx \end{align*} $$

Now we have a nice looking equation. An easy way to find what \(A\) and \(B\) are is to substitute some values for \(x\). Let \(x\) be \(0\),

$$ \begin{align*} - 3(0) + 4 &= A(0 - 2)+ B(0) \\ 4 &= -2A \\ A & = -2 \end{align*} $$

And let’s substitute 2 for \(x\), next

$$ \begin{align*} - 3(2) + 4 &= A(2 - 2)+ B(2) \\ -2 &= 2B \\ B & = -1 \end{align*} $$

Putting everything back together we have,

$$ \begin{align*} \frac{- 3x + 4}{x^2 - 2x} = &= - \frac{2}{x} - \frac{1}{x-2} \end{align*} $$


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