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Here is an example of how to use partial fractions to decompose the rational function \( \dfrac{2x-5}{(x-2)(x+1)} \):
$$\begin{aligned} \frac{2x-5}{(x-2)(x+1)} &= \frac{A}{x-2} + \frac{B}{x+1} \\ &= \frac{A(x+1) + B(x-2)}{(x-2)(x+1)} \end{aligned} $$Multiplying both sides by the common denominator gives:
$$2x-5 = A(x+1) + B(x-2)$$Substituting \(x = 2\) gives:
$$-1 = 3A$$ $$A = -\dfrac{1}{3}$$Substituting \(x = -1\) gives:
$$-7 = -3B$$ $$B = \dfrac{7}{3}$$Therefore, we have:
$$\frac{2x-5}{(x-2)(x+1)} = -\frac{1}{3(x-2)} + \frac{7}{3(x+1)}$$How do you teach this topic? Do you have any tips or suggestions for other teachers? It is always useful to receive feedback and helps make these free resources even more useful for Maths teachers anywhere in the world. Click here to enter your comments.
