\( \DeclareMathOperator{cosec}{cosec} \)
Sign In | Starter Of The Day | Tablesmaster | Fun Maths | Maths Map | Topics | More
Here are some specific activities, investigations or visual aids we have picked out. Click anywhere in the grey area to access the resource.
Here are some exam-style questions on this statement:
Click on a topic below for suggested lesson Starters, resources and activities from Transum.
See the main Partial Fractions section of the syllabus in Number and Algebra.
Here is an example of how to use partial fractions to decompose the rational function \( \dfrac{2x-5}{(x-2)(x+1)^2} \):
$$\begin{aligned} \frac{2x-5}{(x-2)(x+1)^2} &= \frac{A}{x-2} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \\ &= \frac{A(x+1)^2 + B(x-2)(x+1) + C(x-2)}{(x-2)(x+1)^2} \end{aligned} $$Multiplying both sides by the common denominator gives:
$$2x-5 = A(x+1)^2 + B(x-2)(x+1) + C(x-2)$$Substituting \(x = 2\) gives:
$$-1 = 9A$$ $$A = -\dfrac{1}{9}$$Substituting \(x = -1\) gives:
$$-7 = -3C$$ $$C = \dfrac{7}{3}$$Finally, substituting \(x = 0\) gives:
$$-5 = A - 2B -2C$$Substituting A and C gives:
$$-5 = \left(-\frac{1}{9}\right) - 2B + -2\left( \frac{7}{3} \right)$$ $$B = \dfrac{1}{9}$$Therefore, we have:
$$\frac{2x-5}{(x-2)(x+1)^2} = -\frac{1}{9(x-2)} + \frac{1}{9(x+1)} + \frac{7}{3(x+1)^2}$$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.