Solving Complex Polynomial Equations
When a polynomial equation has complex roots, a single clue — one root, the sum, or the product — is often enough to unlock all of them. This lesson weaves together the factor theorem, the conjugate root theorem and Vieta's formulas into one strategy for cracking polynomial equations of degree 3 and 4. Master the combination and you can solve any HSC Extension 2 polynomial in minutes.
If a quadratic with real coefficients has one root $z = 3 - 2i$, what is the other root and what is the quadratic (with leading coefficient 1)? Before checking — use the conjugate pair to write the factorisation, then expand. Sketch your reasoning below.
Solving a complex polynomial equation rests on three results you already know: the factor theorem (if $z = \alpha$ is a root then $(z - \alpha)$ is a factor), the conjugate root theorem (real-coefficient polynomials have complex roots in conjugate pairs), and Vieta's formulas (sum/product of roots come straight from the coefficients). The trick at HSC is knowing which tool to grab first.
Strategy: (1) conjugate any given complex root if coefficients are real, (2) factor out the resulting real quadratic, (3) use Vieta to cross-check sum and product against the original coefficients.
For $az^n + \ldots + k = 0$: $\sum \alpha_i = -b/a$ · $\prod \alpha_i = (-1)^n k/a$
Key facts
- Factor theorem: $P(\alpha) = 0 \Leftrightarrow (z - \alpha) \mid P(z)$
- Conjugate root theorem: real-coefficient polynomials have complex roots in conjugate pairs
- Vieta for cubic $az^3 + bz^2 + cz + d$: $\sum\alpha = -b/a$, $\sum\alpha\beta = c/a$, $\alpha\beta\gamma = -d/a$
- Conjugate pair $\alpha, \bar\alpha$ gives factor $z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$
Concepts
- Why the conjugate root theorem fails for complex-coefficient polynomials
- Why a degree-$n$ polynomial has exactly $n$ roots (counted with multiplicity) in $\mathbb{C}$
- How Vieta turns "one root given" into equations for the remaining roots
Skills
- Solve real-coefficient cubics and quartics given one complex root
- Use Vieta to find remaining roots given sum, product or one root
- Solve complex-coefficient polynomials by setting up sum/product systems
The fundamental theorem of algebra says any degree-$n$ polynomial has exactly $n$ roots in $\mathbb{C}$. When all coefficients of $P(z)$ are real, an additional structure appears: complex roots arrive in conjugate pairs.
Why? Taking conjugates of $P(\alpha) = 0$ and using $\overline{a_k \alpha^k} = a_k \bar\alpha^k$ (because $\bar{a_k} = a_k$) gives $P(\bar\alpha) = 0$.
Consequence. A conjugate pair $\alpha, \bar\alpha$ produces a real quadratic factor:
Worked through the hook: $P(z) = z^3 - 5z^2 + 11z - 15$ has real coefficients and $1 + 2i$ is a root. So $1 - 2i$ is also a root, giving the real quadratic factor $z^2 - 2z + 5$. Dividing: $P(z) = (z^2 - 2z + 5)(z - 3)$, so the third root is $z = 3$. Vieta cross-check: sum $= (1+2i) + (1-2i) + 3 = 5 = -(-5)/1$ ✓.
Real-coefficient $P(z)$: complex roots in conjugate pairs $\alpha, \bar\alpha$ · Conjugate pair factor: $z^2 - 2\,\text{Re}(\alpha)\,z + |\alpha|^2$ · Divide $P(z)$ by this quadratic to reduce degree by 2 · Conjugate root theorem FAILS for complex coefficients
Pause — copy the conjugate-pair quadratic factor $z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2$, the strategy of dividing by it to reduce degree, and the failure of the conjugate-root theorem for complex coefficients into your book.
Quick check: The polynomial $P(z) = z^3 + 2z^2 + 9z + 18$ has real coefficients. If $z = 3i$ is a root, what is the real quadratic factor that $z = 3i$ contributes?
We just saw that real-coefficient polynomials factor into real quadratics $(z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2)$ from each conjugate pair, reducing the degree by 2. That raises a question: now that we can factor, what do Vieta's formulas add for a cubic or quartic? This card answers it → $\sum\alpha = -b$, $\sum\alpha\beta = c$, $\alpha\beta\gamma = -d$ for a monic cubic, with the sign pattern $(-1)^k \cdot (\text{coeff of }z^{n-k})$ for $k$-fold products.
Vieta's formulas express the elementary symmetric functions of the roots in terms of the coefficients. For a monic polynomial they are especially clean.
Cubic. If $\alpha, \beta, \gamma$ are the roots of $z^3 + bz^2 + cz + d = 0$, then:
Quartic. If $\alpha, \beta, \gamma, \delta$ are the roots of $z^4 + bz^3 + cz^2 + dz + e = 0$:
Why useful? When you know one root or one symmetric function (e.g., "the sum of two roots is 4"), Vieta turns the remaining roots into solutions of a smaller system.
Monic cubic $z^3 + bz^2 + cz + d$: $\sum\alpha = -b$, $\sum\alpha\beta = c$, $\alpha\beta\gamma = -d$ · Monic quartic: $\sum\alpha = -b$, $\sum\alpha\beta = c$, $\sum\alpha\beta\gamma = -d$, $\prod\alpha = e$ · Signs alternate: $(-1)^k \cdot (\text{coeff of } z^{n-k}) / (\text{leading coeff})$ for sum of $k$-products · Always check Vieta against the original polynomial as a final verification
Pause — copy Vieta's formulas for monic cubic and quartic (with sign pattern $(-1)^k$), and the verification step (check against original polynomial) into your book.
Did you get this? True or false: if $\alpha, \beta, \gamma$ are roots of $z^3 - 6z^2 + 11z - 6 = 0$, then $\alpha + \beta + \gamma = 6$ and $\alpha\beta\gamma = 6$.
Worked examples · 3 in a row, reveal as you go
Given that $z = 2 - i$ is a root of $P(z) = z^3 - 7z^2 + 17z - 15$, find all roots.
The cubic $z^3 - 4z^2 + kz - 13 = 0$ has real coefficients. If one root is $z = 2 + 3i$, find $k$ and the third root.
Solve $z^4 - 2z^3 + 6z^2 - 2z + 5 = 0$, given that $z = i$ is a root.
Fill the gap: If $z = 1 + i\sqrt{3}$ is a root of a real-coefficient polynomial, the conjugate root is , and these two roots together contribute the real quadratic factor $z^2 - \;$$\,z + \;$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the polynomial $z^2 - (3+i)z + (2 + 2i) = 0$ has $z = 1 - i$ as a root because $1 - i$ is the conjugate of $1 + i$, which is a root.
Activities · practice with the ideas
Given that $z = 2i$ is a root of $z^3 - 3z^2 + 4z - 12 = 0$, find all roots and confirm using Vieta's sum.
Find a monic cubic with real coefficients whose roots are $z = 1 - i$ and $z = 4$. Write out the expansion.
The cubic $z^3 + pz^2 + qz + r = 0$ has roots $\alpha, \beta, \gamma$ with $\alpha + \beta = 3$ and $\gamma = 2$. If $\alpha\beta = 5$, find $p, q, r$.
Solve $z^4 + 4 = 0$ over $\mathbb{C}$. (Hint: factor as a difference of squares using complex numbers, or write $z^4 = -4 = 4e^{i\pi}$.)
The cubic $z^3 + (1-2i)z^2 - (3+2i)z + 3i = 0$ has complex coefficients. Verify $z = i$ is a root, then factor and find the remaining roots.
Odd one out: Three of these statements about a real-coefficient cubic with roots $\alpha, \beta, \gamma$ are correct. Which one is NOT?
Earlier you predicted (a) another root of $z^3 - 5z^2 + 11z - 15$ given that $1 + 2i$ is a root, and (b) the sum of all three roots.
The other obvious root is the conjugate $z = 1 - 2i$, because the polynomial has real coefficients. The sum of all three roots equals $-(-5)/1 = 5$ by Vieta, which means the third (real) root must be $5 - (1+2i) - (1-2i) = 3$. Checking: $(z - 3)$ should divide $P(z)$, and indeed $P(3) = 27 - 45 + 33 - 15 = 0$ ✓. The three-tool combination (conjugate, factor, Vieta) is the standard HSC strategy.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Given that $z = 3 + i$ is a root of $z^3 - 8z^2 + 26z - 30 = 0$, find the other two roots. (2 marks)
Q2. The quartic $z^4 - 4z^3 + 14z^2 - 36z + 45 = 0$ has $z = 1 + 2i$ as a root. Find all four roots. (3 marks)
Q3. The cubic $z^3 + bz^2 + cz + d = 0$ has real coefficients and roots $\alpha, \beta, \gamma$ satisfying $\alpha + \beta + \gamma = -2$, $\alpha\beta + \beta\gamma + \gamma\alpha = 5$, and one root is $\gamma = -3$. Find the cubic and the two remaining roots. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Real coefficients $\Rightarrow z = -2i$ also a root. Factor: $z^2 + 4$. Divide: $z^3 - 3z^2 + 4z - 12 = (z^2 + 4)(z - 3)$. Roots: $2i, -2i, 3$. Vieta sum: $0 + 3 = 3 = -(-3)$ ✓.
2. Conjugate of $1 - i$ is $1 + i$. Quadratic: $z^2 - 2z + 2$. With linear factor $z - 4$: $(z^2 - 2z + 2)(z - 4) = z^3 - 6z^2 + 10z - 8$.
3. $p = -(\alpha + \beta + \gamma) = -5$. $q = \alpha\beta + \gamma(\alpha + \beta) = 5 + 2 \cdot 3 = 11$. $r = -\alpha\beta\gamma = -5 \cdot 2 = -10$.
4. $z^4 = -4 = 4e^{i\pi}$. $z = \sqrt{2}\,e^{i(\pi + 2k\pi)/4}$, $k = 0,1,2,3$. Roots: $1+i, -1+i, -1-i, 1-i$ (two conjugate pairs).
5. Test $z = i$: $i^3 + (1-2i)i^2 - (3+2i)i + 3i$. Compute term-by-term and divide; the remaining quadratic factor yields the other two roots via the quadratic formula.
Q1 (2 marks): Conjugate $3 - i$ is also a root [1]. Quadratic factor $z^2 - 6z + 10$; dividing gives third linear factor $z - 3$, so third root is $z = 3$ [1].
Q2 (3 marks): Conjugate $1 - 2i$ also a root [1]. Quadratic factor $z^2 - 2z + 5$; dividing into the quartic gives quotient $z^2 - 2z + 9$ [1]. Solve quotient: $z = 1 \pm 2i\sqrt{2}$ [1]. Four roots: $1 + 2i,\; 1 - 2i,\; 1 + 2i\sqrt{2},\; 1 - 2i\sqrt{2}$.
Q3 (3 marks): From Vieta: $b = 2$, $c = 5$ [1]. With $\gamma = -3$: $\alpha + \beta = 1$ and $\alpha\beta = 8$ (from $\alpha\beta + \gamma(\alpha+\beta) = 5$) [1]. So $\alpha, \beta$ are roots of $z^2 - z + 8 = 0$, giving $z = \tfrac{1 \pm i\sqrt{31}}{2}$, and $d = -\alpha\beta\gamma = 24$ [1]. Cubic: $z^3 + 2z^2 + 5z + 24 = 0$.
Five timed questions on conjugate roots, factor theorem and Vieta's formulas. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick polynomial questions. Lighter alternative to the boss.
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