Include every power of a repeated linear factor in partial fractions
Use both simple and squared denominator terms, then obtain the remaining coefficient by comparing the identity.
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
- Decompose fractions with distinct linear factors.
- Expand quadratic products and compare corresponding coefficients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many terms are needed for the general denominator (x − 1)(x + 1)²?
What you will learn
- Use A/L₁ + B/L₂ + C/L₂² when one linear factor is repeated.
- Find all coefficients, including those not isolated by root substitution, and verify the resulting identity.
Repeated factors need a term for each power
For a denominator (x − 1)(x + 1)², the complete form is A/(x − 1) + B/(x + 1) + C/(x + 1)². Omitting the simple (x + 1) term usually makes the identity impossible. After clearing the denominator, the numerator is A(x + 1)² + B(x − 1)(x + 1) + C(x − 1). The squared-denominator term contributes only one remaining linear factor to this identity.
For numerator 2x² − x + 3, substitute x = 1 to get 4 = 4A, so A = 1. Substitute x = −1 to get 6 = −2C, so C = −3. Neither substitution finds B: both make its product zero. Comparing x² coefficients gives 2 = A + B, hence B = 1. The decomposition is 1/(x − 1) + 1/(x + 1) − 3/(x + 1)², excluding 1 and −1.
Finish and verify the whole polynomial identity
Expand the result: (x + 1)² + (x − 1)(x + 1) − 3(x − 1) = 2x² − x + 3. This checks all coefficients, not only the two substituted values. A different convenient input, such as 0, also supplies the final equation if coefficient comparison is inconvenient. Keep signs attached to the full products when expanding.
For a denominator (x − 1)² alone, use A/(x − 1) + B/(x − 1)². Thus (3x − 1)/(x − 1)² has identity 3x − 1 = A(x − 1) + B, giving A = 3 and B = 2. A non-monic repeated factor follows the same template; its zero is found by solving its linear equation. Repeated denominator factors do not restore any excluded input, even if some coefficient is zero.
Pause and explain
In the cleared identity, root substitution gives A = 1 and C = −3. How can B be found?
Worked example
Decompose (2x² − x + 3)/[(x − 1)(x + 1)²].
Show the worked solution
- Use A/(x − 1) + B/(x + 1) + C/(x + 1)² and clear the denominator.
- At x = 1 obtain A = 1; at x = −1 obtain C = −3.
- Compare x² coefficients to get B = 1, then expand to verify the numerator.
Answer 1/(x − 1) + 1/(x + 1) − 3/(x + 1)², x ≠ 1, −1.
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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 (3x − 1)/(x − 1)².
Give me a hint
Include both denominator powers and compare the linear coefficient and constant term.
Compare my reasoning
- Write A/(x − 1) + B/(x − 1)².
- Clear the denominator: 3x − 1 = A(x − 1) + B; the x coefficient gives A = 3.
- The constant equation −1 = −3 + B gives B = 2. Recombination recovers 3x − 1.
3/(x − 1) + 2/(x − 1)², x ≠ 1.
Look for these in your work
- Include both first and second powers.
- Match both coefficients in the cleared identity.
Decompose (2x² − x + 3)/[(x − 1)(x + 1)²], explaining how the middle coefficient is determined.
Give me a hint
After root substitution, use the leading coefficient or another allowed input for the remaining constant.
Compare my reasoning
- The complete form has coefficients A, B and C over x − 1, x + 1 and (x + 1)².
- At 1 obtain A = 1; at −1 obtain C = −3. Compare x² coefficients: 2 = A + B, so B = 1.
- Rebuild the numerator as (x + 1)² + (x − 1)(x + 1) − 3(x − 1) = 2x² − x + 3.
1/(x − 1) + 1/(x + 1) − 3/(x + 1)², x ≠ 1, −1.
Look for these in your work
- Find the coefficient not isolated by root evaluation.
- Verify all polynomial coefficients and domain exclusions.
Decompose (5x² − 2x + 2)/[(2x − 1)(x + 1)²].
Give me a hint
Use the actual non-monic root 1/2, then compare leading coefficients to finish.
Compare my reasoning
- Clear the denominator using A(x + 1)² + B(2x − 1)(x + 1) + C(2x − 1).
- At 1/2 obtain A = 1; at −1 obtain C = −3. The x² equation is 5 = A + 2B, so B = 2.
- Expand (x + 1)² + 2(2x − 1)(x + 1) − 3(2x − 1) to recover 5x² − 2x + 2.
1/(2x − 1) + 2/(x + 1) − 3/(x + 1)², x ≠ 1/2, −1.
Look for these in your work
- Handle the non-monic linear factor without a scale error.
- Use the leading coefficient to complete and verify the form.
Common mistakes
- Do not skip a lower power of a repeated factor.
- Root substitution may leave a coefficient undetermined; add an independent equation.
- Check the entire polynomial identity after finding all coefficients.
What can you explain now?
What is the full form for a proper fraction with denominator (2x − 1)(x + 1)²?
Compare with the explanation
A/(2x − 1) + B/(x + 1) + C/(x + 1)².
The distinct linear factor needs one term and the repeated linear factor needs both its first and second powers.
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Make it stick
Write the complete repeated-factor template from memory, then explain why two factor-root substitutions can leave the middle coefficient unknown.
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.