Complex Numbers: Real polynomial, complex root — RVHS 2025 H2 Math Prelim Paper 1
What this question tests
Question
The function \(\mathrm{f}\) is defined by \(\mathrm{f}(x) = ax^3 + bx^2 + cx + d\) where \(a\), \(b\), \(c\) and \(d\) are real numbers. Given that \(4+\mathrm{i}\) and \(-1\) are roots of \(\mathrm{f}(x) = 0\), find \(b\), \(c\) and \(d\) in terms of \(a\).
Show full worked solution▾
Method 1 — Using factor theorem and system of equations
Since \(-1\) is a root, \(\mathrm{f}(-1) = 0\): \[ a(-1)^3 + b(-1)^2 + c(-1) + d = 0 \implies -a + b - c + d = 0 \implies b - c + d - a = 0 \quad\cdots(1) \]
Since \(4+\mathrm{i}\) is a root, \(\mathrm{f}(4+\mathrm{i}) = 0\): \[ a(4+\mathrm{i})^3 + b(4+\mathrm{i})^2 + c(4+\mathrm{i}) + d = 0 \] \[ a(52+47\mathrm{i}) + b(15+8\mathrm{i}) + c(4+\mathrm{i}) + d = 0 \]
Comparing real parts: \(52a + 15b + 4c + d = 0 \quad\cdots(2)\)
Comparing imaginary parts: \(47a + 8b + c = 0 \quad\cdots(3)\)
Solving \((1)\), \((2)\) and \((3)\) using the GC:
Method 2 — Conjugate root theorem
Since all coefficients of \(\mathrm{f}(x)\) are real, \(4+\mathrm{i}\) and \(4-\mathrm{i}\) are conjugate roots of \(\mathrm{f}(x)=0\). \[\begin{aligned} \mathrm{f}(x) &= a\bigl[x-(4+\mathrm{i})\bigr]\bigl[x-(4-\mathrm{i})\bigr]\bigl[x+1\bigr]\\ &= a\bigl[x^2 - 8x + 17\bigr]\bigl[x+1\bigr]\\ &= a\bigl[x^3 - 7x^2 + 9x + 17\bigr]\\ &= ax^3 - 7ax^2 + 9ax + 17a \end{aligned}\]