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Cubic equations with complex roots

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

A cubic with real coefficients has three roots. Either all three are real, or one is real and the other two are a complex conjugate pair. Given one complex root, you can find the rest.

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The conjugate root rule

If a polynomial has real coefficients and z = p + qi is a root, then z* = p - qi is also a root. So the non-real roots of a real cubic come in a pair, and there is always at least one real root.

Method from a complex root

1. Write down the conjugate root.

2. Multiply (z - α)(z - α*) = z2 - 2Re(α)z + |α|2 to get a real quadratic factor.

3. Find the linear factor by dividing, or by comparing coefficients (the z3 terms and constant terms are quickest).

4. State all three roots.

Method from a real root

If no complex root is given, use the factor theorem to find a real root z = a (try factors of the constant term), divide by (z - a), then solve the quadratic, which may have complex roots.

Worked example

Given that 2 + 3i is a root of z3 - 6z2 + 21z - 26 = 0, find the other roots.

  1. 2 - 3i is also a root, as the coefficients are real.
  2. (z - 2 - 3i)(z - 2 + 3i) = z2 - 4z + 13.
  3. z3 - 6z2 + 21z - 26 = (z2 - 4z + 13)(z - 2): compare z3 and the constants: 13 × (-2) = -26.

Answer: The roots are 2 + 3i, 2 - 3i and 2

Key idea: Real coefficients: complex roots come in conjugate pairs. Pair them into a real quadratic factor, then find the remaining factor.

Check yourself

  1. -1 + 2i is a root of z3 + z2 + 3z - 5 = 0. Find the real root.
    Show answer

    1. -1 - 2i is also a root, giving the factor z2 + 2z + 5. Comparing constants: 5 × (real factor constant) = -5, so the factor is z - 1 and the real root is 1.

  2. 2 - 3i is a root of z3 - 3z2 + 9z + 13 = 0. Find the real root.
    Show answer

    -1. The quadratic factor is z2 - 4z + 13. 13 × c = 13 gives the factor z + 1, so the real root is -1.

  3. 3 + i is a root of a cubic with real coefficients. Find the real quadratic factor of the cubic, in the form z2 + bz + c.
    Show answer

    z2 - 6z + 10. The roots 3 ± i add to 6 and multiply to 9 + 1 = 10.

  4. z3 + pz2 + qz + 10 = 0, with p and q real, has a root 1 + 2i. Find q.
    Show answer

    1. The quadratic factor is z2 - 2z + 5. Since 5 × 2 = 10, the cubic is (z2 - 2z + 5)(z + 2) = z3 + z + 10, so p = 0 and q = 1.

  5. A cubic with real coefficients has roots 3 and 1 - i. What is the third root?
    • -1 + i
    • 1 + i
    • -3
    • -1 - i
    Show answer

    1 + i. Non-real roots of a real polynomial come in conjugate pairs, so 1 + i must be a root.

  6. Solve z3 - 7z + 6 = 0.
    Show answer

    1, 2, -3. z = 1 is a root (1 - 7 + 6 = 0). Dividing: (z - 1)(z2 + z - 6) = (z - 1)(z - 2)(z + 3).

  7. What is the smallest possible number of real roots of a cubic equation with real coefficients?
    Show answer

    1. There are 3 roots in total and the non-real ones come in pairs, so 1 or 3 of them are real.

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Last updated 8 November 2026. Spotted a mistake? Tell us.