Unit 3 Review Exercises

Partial Fractions

Paper 1 means no calculator. Partial fractions are always done by hand.

1 Paper 1 Express \(\dfrac{6}{x^2 - 9}\) in partial fractions.

Solution

\(x^2 - 9 = (x - 3)(x + 3)\). \(\;6 \equiv A(x + 3) + B(x - 3)\): \(x = 3\) gives \(A = 1\); \(x = -3\) gives \(B = -1\).

\[ \frac{1}{x - 3} - \frac{1}{x + 3} \]

2 Paper 1 Express \(\dfrac{x + 11}{x^2 + x - 12}\) in partial fractions.

Solution

\((x + 4)(x - 3)\). \(\;x + 11 \equiv A(x - 3) + B(x + 4)\): \(x = 3\): \(14 = 7B\), \(B = 2\); \(x = -4\): \(7 = -7A\), \(A = -1\).

\[ -\frac{1}{x + 4} + \frac{2}{x - 3} \]

3 Paper 1 Express \(\dfrac{9 - x}{(1 + x)(3 - x)}\) in partial fractions.

Solution

\(9 - x \equiv A(3 - x) + B(1 + x)\). \(x = 3\): \(6 = 4B\), \(B = \frac{3}{2}\). \(x = -1\): \(10 = 4A\), \(A = \frac{5}{2}\).

\[ \frac{5}{2(1 + x)} + \frac{3}{2(3 - x)} \]

4 Paper 1 Express \(\dfrac{10x}{(2x + 1)(3x - 2)}\) in partial fractions.

Solution

\(10x \equiv A(3x - 2) + B(2x + 1)\). \(x = \frac{2}{3}\): \(\frac{20}{3} = \frac{7}{3}B\), \(B = \frac{20}{7}\). \(\;x = -\frac{1}{2}\): \(-5 = -\frac{7}{2}A\), \(A = \frac{10}{7}\).

\[ \frac{10}{7(2x + 1)} + \frac{20}{7(3x - 2)} \]

Fractions in the answer are fine — always check by testing a value, e.g. \(x = 0\): both sides give \(0\) ✓.

5 Paper 1 Given that \(\dfrac{7x - 1}{(x - 1)(x + 2)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x + 2}\), find \(A\) and \(B\), and hence write \(\dfrac{7x - 1}{(x - 1)(x + 2)} - \dfrac{2}{x - 1}\) as a single fraction.

Solution

\(7x - 1 \equiv A(x + 2) + B(x - 1)\): \(x = 1\): \(6 = 3A\), \(A = 2\); \(x = -2\): \(-15 = -3B\), \(B = 5\).

So the expression is \(\dfrac{2}{x - 1} + \dfrac{5}{x + 2} - \dfrac{2}{x - 1} = \dfrac{5}{x + 2}\).

6 Paper 1 Express \(\dfrac{2}{(r + 1)(r + 3)}\) in partial fractions.

Solution

\(2 \equiv A(r + 3) + B(r + 1)\). Put \(r = -1\): \(2 = 2A\), so \(A = 1\). Put \(r = -3\): \(2 = -2B\), so \(B = -1\).

\[ \frac{2}{(r + 1)(r + 3)} \equiv \frac{1}{r + 1} - \frac{1}{r + 3} \]

See 3.1.