One complex root → its conjugate is free: If a polynomial has real coefficients and a + bi is a root, then a − bi is also a root.
Complex roots always come in conjugate pairs — so you get the second one for free.
A conjugate root pair sits symmetrically above and below the real axis.
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IB-style question — the partner root
A cubic with real coefficients has 2 + i as one root.
Write down another root, and the real quadratic factor it gives.
Step by step
- Real coefficients → the conjugate is also a root.
- Their quadratic factor: sum = 4, product = 2² + 1² = 5.
Final answer
Another root is 2 − i; the real quadratic factor is z² − 4z + 5.
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Pair → real quadratic → divide out: A conjugate pair a ± bi multiplies to the real quadratic z² − 2az + (a² + b²). Divide the polynomial by it to find the remaining factor (and roots).
IB-style question — find all the roots
Given that 1 + 2i is a root of z³ − 3z² + 7z − 5 = 0, find all three roots.
Step by step
- Conjugate is also a root.
- Their quadratic factor: sum = 2, product = 1² + 2² = 5.
- Divide z³ − 3z² + 7z − 5 by z² − 2z + 5.
- The last factor gives the real root.
Final answer
Roots: 1 + 2i, 1 − 2i and 1.