Overview
The symbol i extends real arithmetic by satisfying i² = −1. Complex-number questions often become simpler when you keep conjugate pairs together: their imaginary parts cancel in sums and their product is real. We will connect completing the square, conjugate multiplication and the roots of a real-coefficient polynomial, then use an extra evaluation to determine the remaining factor.
Complete the square before introducing i
To solve x² − 6x + 13 = 0, isolate a square: x² − 6x + 9 = −4, so (x − 3)² = −4. Taking the two complex square roots gives x − 3 = ±2i, hence x = 3 ± 2i. The center of the pair is 3, and the imaginary displacement has magnitude 2.
Each part of the expression carries information. The coefficient −6 determines the shift 3 when completing the square; the remaining −4 determines the imaginary part. The option 3 ± 4i forgets to take the square root of 4, while −3 ± 2i shifts in the wrong direction. Substituting one root into the completed-square form is an efficient check because (±2i)² = −4 immediately. Keeping the square visible explains the two roots and their symmetry without requiring a long expansion.
Multiply conjugates and use the sign of i squared
The product (2 + 3i)(2 − 3i) is a difference of squares: 2² − (3i)². Since (3i)² = 9i² = −9, the result is 4 − (−9) = 13. The cross terms cancel because the two imaginary parts have opposite signs.
A common wrong answer is −5, obtained by treating i² as positive 1. Write the substitution i² = −1 explicitly until that sign is reliable. The broader pattern is (a + bi)(a − bi) = a² + b² for real a and b. This sum of squares is real and nonnegative, which provides a useful check on a calculated product. Conjugate multiplication is not an unrelated trick: it is the same algebraic cancellation that lets pairs of nonreal roots contribute real coefficients to a polynomial.
Reconstruct the real factor from a conjugate pair
Suppose a monic cubic with real coefficients has root 1 + 2i. Its conjugate 1 − 2i must also be a root. The corresponding factors multiply to (x − 1 − 2i)(x − 1 + 2i) = (x − 1)² + 4 = x² − 2x + 5. Because the cubic is monic, the remaining factor can be written x − r.
We also know P(0) = −20. At zero, the quadratic factor is 5 and the linear factor is −r, so −5r = −20 and r = 4. The polynomial is therefore (x² − 2x + 5)(x − 4). At x = 2, these factors have values 5 and −2, giving P(2) = −10. This solution uses the real-coefficient condition to supply a second root and the evaluation condition to identify the third. Neither piece alone would determine the requested value.
The conjugate-root condition comes specifically from real coefficients. Conjugating an evaluation of a real-coefficient polynomial leaves every coefficient unchanged, so a zero at one nonreal value implies a zero at its conjugate. This gives a reason for the paired factor rather than an unexplained rule. The remaining root is real here because dividing the real cubic by the real quadratic leaves a real linear factor.
Check which facts make the polynomial unique
The distractor −20 repeats the given value P(0), but a polynomial does not have the same value at every input. The question provides that value to determine the unknown root, not to replace the final evaluation. The condition “monic” matters too: without it, an additional leading multiplier could change the relationship between the constant term and the remaining root.
For transfer, sort the information into roots, coefficient restrictions and known evaluations. Pair nonreal conjugates when coefficients are real, preserve any leading multiplier until it is determined, and use the remaining data to solve for what is missing. In practice, check the reconstructed polynomial against the supplied root structure and the given value before calculating the target. This approach turns a complex-number problem into ordinary real polynomial arithmetic while retaining the reason that the conjugate root had to be present.
Apply the method to a new case
A real-coefficient monic quadratic has one root 2 + 3i. Find its constant term. Its other root must be 2 − 3i. Their product is (2 + 3i)(2 − 3i) = 4 − 9i² = 13. Therefore the quadratic is x² − 4x + 13 and its constant term is 13. Adding the roots gives 4, which is related to the linear coefficient, not the constant. Multiplying only the visible imaginary coefficients and reporting 9 omits the real contribution. Check by evaluating at 2 + 3i: its square is −5 + 12i, subtracting four times the root gives −13, and adding 13 gives zero. The assumption of real coefficients is essential; without it, one nonreal root would not determine the other root or the constant term.