Use conjugate pairs to solve real coefficient polynomials
Pair non-real roots of a real-coefficient polynomial, form a real quadratic factor and solve the remaining cubic or quartic factors.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.9. Paper 1 algebra and radian trigonometry are assumed. Use Cartesian working for products, quotients and the two square roots of a non-zero complex number. Arguments usually use −π < θ ≤ π; an alternative interval 0 ≤ θ < 2π is acceptable unless specified. The argument of zero is undefined. General nth-root methods are beyond this mapped sequence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Apply the factor theorem and divide a polynomial.
- Solve quadratic equations and expand Cartesian products.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What real quadratic factor comes from roots 2 ± i?
What you will learn
- Explain why non-real roots occur in conjugate pairs for real coefficients.
- Use a supplied non-real root to solve cubic and quartic equations.
Real coefficients are essential
Conjugating a sum or product conjugates each complex term; a real coefficient is unchanged. Thus if a polynomial p with real coefficients has p(a + bi) = 0, conjugating the equality gives p(a − bi) = 0. For b ≠ 0 these are distinct roots. Their product factor is [z − (a + bi)][z − (a − bi)] = (z − a)² + b², a quadratic with real coefficients.
This conclusion depends on real coefficients. The polynomial z − i has the root i but not −i, since its constant coefficient is complex. Repeated non-real roots also pair with the same multiplicity for real coefficients. Knowing a conjugate pair does not determine every other root without using the remaining coefficients or polynomial factor.
Reduce a cubic or quartic before solving
For z³ − 5z² + 9z − 5 = 0, a supplied root 2 + i gives the second root 2 − i and factor z² − 4z + 5. Polynomial division leaves z − 1, so the three roots are 1 and 2 ± i. Expand (z² − 4z + 5)(z − 1) to check every coefficient. This is stronger than merely guessing a remaining root from one term.
For a quartic a conjugate pair similarly removes a quadratic factor. If the remaining factor is another real quadratic, solve it with completing the square or the quadratic formula. It may give real roots or another conjugate pair. When unknown coefficients are real, express the polynomial as the conjugate quadratic times the remaining real factor and equate all relevant coefficients.
Pause and explain
A real-coefficient cubic has root 1 + 2i. What other root is guaranteed?
Worked example
Given that 2 + i is a root of z³ − 5z² + 9z − 5 = 0, find all roots.
Show the worked solution
- Real coefficients imply the conjugate root 2 − i.
- The pair gives (z − 2)² + 1 = z² − 4z + 5.
- Division gives the remaining factor z − 1; expansion verifies the complete factorisation.
Answer z = 1, 2 + i, 2 − i.
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.
Solve z² + 4z + 13 = 0 over the complex numbers and check the pair.
Give me a hint
Complete the square and use the two square roots of a negative real number.
Compare my reasoning
- Write (z + 2)² + 9 = 0, so (z + 2)² = −9.
- Hence z + 2 = ±3i and z = −2 ± 3i.
- The roots are conjugates; their factors multiply back to z² + 4z + 13.
z = −2 + 3i and −2 − 3i.
Look for these in your work
- Keep both roots of the squared expression.
- Check the real quadratic factor.
Given the root 1 + 2i of z⁴ + 3z² + 6z + 10 = 0, find all four roots.
Give me a hint
Remove its real conjugate quadratic before solving the remaining quadratic.
Compare my reasoning
- The partner is 1 − 2i, giving factor z² − 2z + 5.
- Division leaves z² + 2z + 2 = (z + 1)² + 1.
- Its roots are −1 ± i; the two quadratic factors multiply to the stated quartic.
z = 1 ± 2i and −1 ± i.
Look for these in your work
- Use the real-coefficient conjugate partner.
- Solve and verify both quadratic factors.
The real-coefficient cubic z³ + az² + bz + 15 = 0 has root 1 + 2i. Find real a and b and the third root.
Give me a hint
The pair gives z² − 2z + 5; use the constant term of the remaining linear factor.
Compare my reasoning
- Write the cubic as (z² − 2z + 5)(z − c), where c is the third root.
- The constant −5c = 15 gives c = −3.
- Expansion with z + 3 gives z³ + z² − z + 15, so a = 1 and b = −1.
a = 1, b = −1; third root −3.
Look for these in your work
- Form a real quadratic from the non-real pair.
- Use and check all coefficients, including the constant sign.
Common mistakes
- Check the coefficients are real before asserting a conjugate partner.
- Form (z − a)² + b² with the right signs.
- Solve and check the remaining factor; one conjugate pair may not finish the equation.
What can you explain now?
Does the root i of z − i force −i to be a root?
Compare with the explanation
No: the coefficient −i is not real.
The conjugate-pair theorem requires every polynomial coefficient to be real.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State the real-coefficient condition, form a conjugate quadratic and use it to solve or reconstruct a polynomial.
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.