Partial fractions let you break a complicated rational expression into a sum of simpler fractions. This technique is essential in FSc and A Level maths, especially for integration and the binomial expansion. In this guide we cover all four standard cases, each with a fully worked example, so you can recognise the pattern and solve any question with confidence.
What Are Partial Fractions?
Adding fractions combines them into one:
$$ \frac{2}{x - 1} + \frac{3}{x + 2} = \frac{5x + 1}{(x - 1)(x + 2)} $$
Partial fractions do the reverse: starting from the right side, we recover the simpler fractions on the left.
Proper vs Improper Fractions
A rational fraction \(\frac{P(x)}{Q(x)}\) is proper if the degree of \(P\) is less than the degree of \(Q\). If not, it is improper, and you must divide first (see Case 4).
The Four Cases at a Glance
| Factor in denominator | Partial fraction form |
|---|---|
| Linear \((ax + b)\) | \(\frac{A}{ax + b}\) |
| Repeated \((ax + b)^2\) | \(\frac{A}{ax + b} + \frac{B}{(ax + b)^2}\) |
| Irreducible quadratic \((ax^2 + bx + c)\) | \(\frac{Ax + B}{ax^2 + bx + c}\) |
Case 1: Distinct Linear Factors
Worked Example 1
Resolve \(\dfrac{5x + 1}{(x - 1)(x + 2)}\) into partial fractions.
Step 1: Write the form:
$$ \frac{5x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} $$
Step 2: Multiply both sides by \((x - 1)(x + 2)\):
$$ 5x + 1 = A(x + 2) + B(x - 1) $$
Step 3: Substitute values that make a bracket zero.
Put \(x = 1\): \(5 + 1 = 3A\), so \(A = 2\).
Put \(x = -2\): \(-10 + 1 = -3B\), so \(-9 = -3B\) and \(B = 3\).
Answer:
$$ \frac{5x + 1}{(x - 1)(x + 2)} = \frac{2}{x - 1} + \frac{3}{x + 2} $$
Case 2: Repeated Linear Factors
A repeated factor \((ax + b)^2\) needs one fraction for each power.
Worked Example 2
Resolve \(\dfrac{3x + 5}{(x + 1)^2}\).
Step 1: Write the form:
$$ \frac{3x + 5}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} $$
Step 2: Multiply by \((x + 1)^2\):
$$ 3x + 5 = A(x + 1) + B $$
Step 3: Put \(x = -1\): \(-3 + 5 = B\), so \(B = 2\).
Step 4: Compare coefficients of \(x\): \(3 = A\).
Answer:
$$ \frac{3x + 5}{(x + 1)^2} = \frac{3}{x + 1} + \frac{2}{(x + 1)^2} $$
Check: \(3(x + 1) + 2 = 3x + 5\). Correct.
Worked Example 3: Repeated and Distinct Factors Together
Resolve \(\dfrac{x^2 + 1}{x^2(x - 1)}\).
Step 1: Form:
$$ \frac{x^2 + 1}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1} $$
Step 2: Multiply through:
$$ x^2 + 1 = Ax(x - 1) + B(x - 1) + Cx^2 $$
Step 3: Put \(x = 0\): \(1 = -B\), so \(B = -1\).
Put \(x = 1\): \(2 = C\), so \(C = 2\).
Step 4: Compare coefficients of \(x^2\): \(1 = A + C\), so \(A = 1 - 2 = -1\).
Answer:
$$ \frac{x^2 + 1}{x^2(x - 1)} = -\frac{1}{x} - \frac{1}{x^2} + \frac{2}{x - 1} $$
Check at \(x = 2\): left side \(\frac{5}{4}\); right side \(-\frac{1}{2} - \frac{1}{4} + 2 = \frac{5}{4}\). Correct.
Case 3: Irreducible Quadratic Factors
If a quadratic factor cannot be factorized (its discriminant is negative), its numerator must be linear: \(Ax + B\).
Worked Example 4
Resolve \(\dfrac{3x^2 + x + 4}{(x + 1)(x^2 + 1)}\).
Step 1: Form:
$$ \frac{3x^2 + x + 4}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} $$
Step 2: Multiply through:
$$ 3x^2 + x + 4 = A(x^2 + 1) + (Bx + C)(x + 1) $$
Step 3: Put \(x = -1\): \(3 - 1 + 4 = 2A\), so \(A = 3\).
Step 4: Compare coefficients.
\(x^2\): \(3 = A + B\), so \(B = 0\).
Constant: \(4 = A + C\), so \(C = 1\).
Answer:
$$ \frac{3x^2 + x + 4}{(x + 1)(x^2 + 1)} = \frac{3}{x + 1} + \frac{1}{x^2 + 1} $$
Check the \(x\) coefficient: right side gives \(B + C = 0 + 1 = 1\), matching the left. Correct.
Case 4: Improper Fractions
If the numerator's degree is equal to or greater than the denominator's, divide first.
Worked Example 5
Resolve \(\dfrac{x^2 + 2x + 3}{(x + 1)(x + 2)}\).
Step 1: The denominator expands to \(x^2 + 3x + 2\). Both have degree 2, so divide:
$$ \frac{x^2 + 2x + 3}{x^2 + 3x + 2} = 1 + \frac{-x + 1}{(x + 1)(x + 2)} $$
Step 2: Resolve the proper part:
$$ \frac{-x + 1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} $$
$$ -x + 1 = A(x + 2) + B(x + 1) $$
Put \(x = -1\): \(2 = A\). Put \(x = -2\): \(3 = -B\), so \(B = -3\).
Answer:
$$ \frac{x^2 + 2x + 3}{(x + 1)(x + 2)} = 1 + \frac{2}{x + 1} - \frac{3}{x + 2} $$
Common Mistakes
- Skipping division for improper fractions.
- Using a constant numerator over a quadratic. An irreducible quadratic always needs \(Ax + B\).
- Missing a term for repeated factors. \((x + 1)^2\) needs both \(\frac{A}{x + 1}\) and \(\frac{B}{(x + 1)^2}\).
- Not checking. Substitute a convenient value of \(x\) into both sides.
Key Takeaways
- Make sure the fraction is proper; divide first if it is not.
- Each linear factor gets a constant numerator; each power of a repeated factor gets its own fraction.
- Irreducible quadratic factors take a linear numerator \(Ax + B\).
- Use substitution of convenient values plus comparison of coefficients to find the constants.
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