Use a linear numerator over an irreducible quadratic factor
Combine a constant numerator over a linear factor with Bx + C over a quadratic, and check its real domain.
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
- Expand and compare polynomial coefficients.
- Determine whether a quadratic has real roots and retain denominator exclusions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the general numerator over the irreducible factor x² + 1?
What you will learn
- Decompose a proper fraction with denominator (ax + b)(cx² + d), using a linear numerator over a real irreducible quadratic.
- Solve the coefficient equations and distinguish a positive quadratic from one that factors over the reals.
The numerator degree must be lower than its factor degree
For a denominator (x − 1)(x² + 1), use A/(x − 1) + (Bx + C)/(x² + 1). A constant alone over the quadratic is not the general form: the linear term Bx is needed even though it can occasionally turn out to be zero. The factor x² + 1 is positive for every real x, so it is irreducible over the reals and introduces no real excluded input; x = 1 remains excluded.
For numerator 3x² + x − 2, clearing the denominator gives 3x² + x − 2 = A(x² + 1) + (Bx + C)(x − 1). At x = 1, 2 = 2A, so A = 1. The x² coefficient gives 3 = A + B, so B = 2; the x coefficient gives 1 = C − B, so C = 3. The constant equation −2 = A − C checks the result. Thus the fraction is 1/(x − 1) + (2x + 3)/(x² + 1).
Check the factor type before choosing its template
For (2x + 1)(x² + 4), the quadratic is again positive for every real input, so use A/(2x + 1) + (Bx + C)/(x² + 4) and exclude only x = −1/2. By contrast x² − 4 factors into (x − 2)(x + 2), so a complete distinct-real-linear form uses separate constant terms over those factors. A quadratic written compactly is not necessarily irreducible.
Comparing coefficients avoids pretending an irreducible quadratic has a real root at which to substitute. One linear-root evaluation plus the remaining coefficient equations determines A, B and C. Keep exact fractions if necessary, then multiply through to verify every coefficient. If the whole rational function has equal numerator and denominator degrees, extract its constant quotient before applying this proper template.
Pause and explain
For denominator (2x + 1)(x² + 4), which real input is excluded?
Worked example
Decompose (3x² + x − 2)/[(x − 1)(x² + 1)].
Show the worked solution
- Use A/(x − 1) + (Bx + C)/(x² + 1). At 1, A = 1.
- Comparing x² and x coefficients gives B = 2 and C = 3.
- Verify A − C = −2 and the full numerator; exclude x = 1 only over the reals.
Answer 1/(x − 1) + (2x + 3)/(x² + 1), x ≠ 1.
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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 (x² + x + 2)/[(x + 1)(x² + 1)].
Give me a hint
Include Bx + C and allow a coefficient to become zero after solving the equations.
Compare my reasoning
- Write x² + x + 2 = A(x² + 1) + (Bx + C)(x + 1).
- At −1 obtain A = 1. The leading coefficient gives B = 0, and the x coefficient gives C = 1.
- The rebuilt numerator is (x² + 1) + (x + 1) = x² + x + 2, with x ≠ −1.
1/(x + 1) + 1/(x² + 1), x ≠ −1; B happens to be 0.
Look for these in your work
- Start with the full linear-numerator form.
- Accept a zero coefficient only after solving and verifying it.
Decompose (2x² + 3x + 4)/[(x − 1)(x² + 1)].
Give me a hint
Use the real linear-factor root, then match the leading and linear coefficients exactly.
Compare my reasoning
- At x = 1 the cleared identity gives 9 = 2A, so A = 9/2.
- The x² coefficient gives B = 2 − 9/2 = −5/2; the x coefficient gives C = 3 + B = 1/2.
- Check the constant A − C = 4 and rebuild the numerator. Only x = 1 is excluded over the reals.
9/[2(x − 1)] + (−5x + 1)/[2(x² + 1)], x ≠ 1.
Look for these in your work
- Keep exact fractional coefficients in both numerators.
- Verify the constant equation and real domain.
Decompose (3x² + 5x + 6)/[(2x + 1)(x² + 4)] and justify its real domain.
Give me a hint
The quadratic has no real root; the linear root −1/2 isolates A.
Compare my reasoning
- At −1/2 the numerator and quadratic are both 17/4, giving A = 1.
- The leading coefficient equation 3 = A + 2B gives B = 1; the x equation 5 = B + 2C gives C = 2.
- The rebuilt numerator is (x² + 4) + (x + 2)(2x + 1) = 3x² + 5x + 6. Exclude −1/2 only, since x² + 4 > 0.
1/(2x + 1) + (x + 2)/(x² + 4), x ≠ −1/2.
Look for these in your work
- Use both coefficients over the positive quadratic.
- Verify the full identity and justify every real exclusion.
Common mistakes
- Keep Bx + C over the quadratic instead of guessing a constant numerator.
- Do not substitute a nonexistent real zero of x² + 1.
- Factor a reducible quadratic into its real linear factors before selecting the general form.
What can you explain now?
Which denominator factor has no real zero: x² + 4 or x² − 4?
Compare with the explanation
x² + 4 has no real zero; x² − 4 has zeros −2 and 2.
A real square is non-negative, so adding 4 is strictly positive. Subtracting 4 permits x² = 4.
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Make it stick
Explain why an irreducible quadratic needs a linear numerator, and why a quadratic written as cx² + d must still be checked for real factorisation.
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.