Decompose distinct linear factors and equal-degree rational functions
Choose one constant numerator per distinct linear factor, find the coefficients and rebuild the original fraction.
Section 3.1 · Algebra
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.1. Paper 1 knowledge is assumed. Shared modulus, polynomial division, factor and remainder lessons are linked from both sections 2.1 and 3.1. Paper 3 adds partial fractions and rational-power binomial expansions. Partial-fraction numerators may have the same degree as the denominator, but higher-degree numerators are excluded. General-term questions for the rational binomial expansion are not included. Use the stated strict validity condition. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Factor polynomials and use a quotient–remainder identity.
- Solve simultaneous linear equations and evaluate polynomials at signed or fractional inputs.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At which input is 2x − 1 zero?
What you will learn
- Decompose proper fractions with up to three distinct linear factors, including non-monic factors.
- Extract a constant quotient when the numerator and denominator have equal degree; retain all original domain exclusions.
Choose the form before finding its coefficients
For a denominator containing three distinct linear factors L₁, L₂ and L₃, use A/L₁ + B/L₂ + C/L₃, with constants A, B and C. A two-factor case uses two terms. A proper fraction has a numerator of lower degree than its denominator. Multiply by the full denominator to obtain a polynomial identity; this identity holds at all real inputs even though the original fractions exclude the denominator's zeros.
For (5x + 1)/[(x − 1)(x + 2)], the identity is 5x + 1 = A(x + 2) + B(x − 1). At x = 1, 6 = 3A, so A = 2. At x = −2, −9 = −3B, so B = 3. Hence the fraction is 2/(x − 1) + 3/(x + 2), for x ≠ 1, −2. Recombining the numerators gives 2(x + 2) + 3(x − 1) = 5x + 1.
A non-monic factor changes its root, not the partial-fraction rule
For (5x − 1)/[(2x − 1)(x + 1)], start with A/(2x − 1) + B/(x + 1). The identity is 5x − 1 = A(x + 1) + B(2x − 1). Use x = 1/2 to get A = 1, and x = −1 to get B = 2. Substituting 1 instead of 1/2 into the first root calculation would not eliminate the other coefficient. Always solve the linear factor's actual equation.
If the numerator and denominator have equal degree, first divide to remove a constant quotient, then decompose the proper remainder fraction. For (x² + 4x − 1)/[(x − 1)(x + 2)], the denominator is x² + x − 2, so division gives 1 + (3x + 1)/[(x − 1)(x + 2)]. The remainder decomposes as (4/3)/(x − 1) + (5/3)/(x + 2). A sum of proper partial fractions alone cannot reproduce the leading constant 1. Higher-degree numerators are outside this section's specified partial-fraction cases.
Pause and explain
What must precede proper partial fractions for (x² + 4x − 1)/(x² + x − 2)?
Worked example
Decompose (5x + 1)/[(x − 1)(x + 2)] and state the exclusions.
Show the worked solution
- Write A/(x − 1) + B/(x + 2) and clear the denominator.
- The identity 5x + 1 = A(x + 2) + B(x − 1) gives A = 2 at 1 and B = 3 at −2.
- Recombine to confirm 5x + 1 and retain x ≠ 1, −2.
Answer 2/(x − 1) + 3/(x + 2), with x ≠ 1, −2.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Decompose (5x + 1)/[(x − 1)(x + 2)] and verify its numerator.
Give me a hint
Write both constant-numerator terms and substitute each factor's zero into the cleared identity.
Compare my reasoning
- Write 5x + 1 = A(x + 2) + B(x − 1).
- At 1, A = 2; at −2, B = 3.
- The rebuilt numerator is 2(x + 2) + 3(x − 1) = 5x + 1; exclude both denominator roots.
2/(x − 1) + 3/(x + 2), x ≠ 1, −2.
Look for these in your work
- Find both constants from the identity.
- Recombine the numerator and retain the original domain.
Decompose (2x² + 6x − 2)/[x(x − 1)(x + 2)].
Give me a hint
Use a term for each of the three distinct linear factors and evaluate at their three roots.
Compare my reasoning
- Set the numerator equal to A(x − 1)(x + 2) + Bx(x + 2) + Cx(x − 1).
- At the three factor roots 0, 1 and −2, obtain A = 1, B = 2 and C = −1.
- Expanding (x − 1)(x + 2) + 2x(x + 2) − x(x − 1) gives 2x² + 6x − 2; exclude 0, 1 and −2.
1/x + 2/(x − 1) − 1/(x + 2), x ≠ 0, 1, −2.
Look for these in your work
- Include all three denominator factors.
- Verify the complete polynomial numerator, including its constant term.
Express (x² + 4x − 1)/[(x − 1)(x + 2)] as a constant plus partial fractions, and evaluate it at x = 0.
Give me a hint
Divide the equal-degree polynomials first, then decompose only the remainder fraction.
Compare my reasoning
- Division gives constant 1 and remainder numerator 3x + 1.
- From 3x + 1 = A(x + 2) + B(x − 1), the constants are A = 4/3 and B = 5/3.
- At 0, 1 − 4/3 + 5/6 = 1/2, matching (−1)/(−2). The excluded inputs remain 1 and −2.
1 + 4/[3(x − 1)] + 5/[3(x + 2)]; value 1/2 at 0; x ≠ 1, −2.
Look for these in your work
- Include the constant quotient before decomposing.
- Check the original fraction at an allowed input.
Common mistakes
- Use one term for each distinct factor, even if one coefficient later turns out to be zero.
- Find the actual root of a non-monic factor.
- An equal-degree fraction needs a constant quotient before its proper part is decomposed.
What can you explain now?
Decompose (5x − 1)/[(2x − 1)(x + 1)].
Compare with the explanation
1/(2x − 1) + 2/(x + 1), x ≠ 1/2, −1.
Clearing the denominator gives (x + 1) + 2(2x − 1) = 5x − 1; factor roots are 1/2 and −1.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why each distinct factor needs a constant numerator, how a cleared identity can be evaluated at excluded inputs, and why equal-degree division comes first.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.