A-Level Complex numbers in cartesian form and the Argand diagram: worked solution

Question

The cubic f(x) = a x^3 + b x^2 + c x + d has real coefficients a, b, c, d. Given that both 4 + i and -1 satisfy f(x) = 0, express b, c and d in terms of a. [4]

Step 1. Use the conjugate root theorem. The coefficients a, b, c, d are real, so any non-real root must come paired with its complex conjugate. 4 + i is a root, so 4 − i is a root as well. Step 2. Count the roots. A cubic has exactly three roots, and all three are now known: 4 + i, 4 − i and −1. So f(x) factorises completely as f(x) = a(x − (4 + i))(x − (4 − i))(x − (−1)) f(x) = a(x − 4 − i)(x − 4 + i)(x + 1) The leading coefficient a must be there so that the x³ term is a x³. Step 3. Multiply the conjugate pair first, because the i terms cancel and leave real coefficients. Write A = x − 4, so the pair is (A − i)(A + i) = A² − i² = A² + 1, since i² = −1. (x − 4)² + 1 = x² − 8x + 16 + 1 = x² − 8x + 17 Step 4. Multiply by the remaining factor (x + 1). (x² − 8x + 17)(x + 1) = x³ + x² − 8x² − 8x + 17x + 17 = x³ + (1 − 8)x² + (−8 + 17)x + 17 = x³ − 7x² + 9x + 17 Step 5. Multiply through by a and compare coefficients with a x³ + b x² + c x + d. f(x) = a x³ − 7a x² + 9a x + 17a Comparing the x², x and constant terms: b = −7a, c = 9a, d = 17a