Complex Numbers: conjugate root theorem & finding real polynomial coefficients — RVHS 2025 H2 Math Prelim P1
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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 \(a\).
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Identify the third root via the conjugate root theorem
Because all coefficients are real, any non-real root comes paired with its complex conjugate.
Since all coefficients of \(\mathrm{f}(x)\) are real, \(4+\mathrm{i}\) and \(4-\mathrm{i}\) are conjugate roots of \(\mathrm{f}(x)=0\). Together with the given root \(-1\), these are the three roots.
Build the factorised form and expand
Multiply the conjugate pair first to clear the imaginary unit, then expand and compare with \(ax^3+bx^2+cx+d\).
\[ \begin{aligned} \mathrm{f}(x) &= a\bigl[x-(4+\mathrm{i})\bigr]\bigl[x-(4-\mathrm{i})\bigr]\bigl[x+1\bigr]\\ &= a\bigl[x^2 - 8x + 17\bigr]\bigl[x+1\bigr]\\ &= a\bigl[x^3 - 7x^2 + 9x + 17\bigr]\\ &= ax^3 - 7ax^2 + 9ax + 17a \end{aligned} \]Comparing coefficients gives \(b = -7a\), \(c = 9a\) and \(d = 17a\).