Learning goal and prerequisites
Use conjugate pairing and multiplicity to recover a real polynomial. Before starting, review function change.
Method and conditions
For real coefficients, each nonreal zero a+bi brings its conjugate a−bi with the same multiplicity. Multiply the pair before adding real factors; the imaginary cross terms cancel.
Worked example: show the decisions
Zeros 2+3i and 2−3i give (x−2)²+9=x²−4x+13. Adding zero −1 gives (x²−4x+13)(x+1)=x³−3x²+9x+13. Check the x² coefficient against the negative sum of all three roots: −(2+3i+2−3i−1)=−3.
Contrast: why a plausible shortcut fails
A listed zero 2+3i does not count as one isolated nonreal root of a real polynomial. A repeated real zero adds its multiplicity, whereas a listed conjugate pair should not be counted twice again.
Transfer: try before revealing the answer
A real monic cubic has zeros 1+2i, 1−2i and 4. Find its x² coefficient and constant term.
Transfer solution and check
The sum is 6, so the x² coefficient is −6. The pair contributes x²−2x+5; multiplying by x−4 gives constant −20.
Let z=3+5i, where i²=−1. What is the real number z times its complex conjugate?
- Multiply the conjugates; i²=−1 turns the product of imaginary parts positive.
- Using these quantities: 3^2+5^2 = 34.
- Check the original domain, requested quantity and units; do not substitute a related parameter.
Compare the other answer choices
Add the conjugates instead of multiplying them.
Add component magnitudes instead of multiplying the complex numbers.
Use i²=+1 in the conjugate product.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Precalculus Course at a Glance