\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Number and Algebra

Syllabus Content

Partial fractions

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Official Guidance, clarification and syllabus links:

Maximum of two distinct linear terms in the denominator, with degree of numerator less than the degree of the denominator.


$$ \dfrac{2x+1}{x^2+x-2} \equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x+2)} $$

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:

$$\frac{2x-5}{(x-2)(x+1)} = -\frac{1}{3(x-2)} + \frac{7}{3(x+1)}$$

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