Sums and Products of Roots
If you expand $(z - \alpha)(z - \beta)(z - \gamma)$ you can read off the coefficients in terms of $\alpha, \beta, \gamma$ — and that observation, due to Vieta, lets you answer questions about roots without ever finding them. Sum of roots is $-a_{n-1}$, product is $(-1)^n a_0$, and the symmetric sums in between fill the gaps. This lesson turns Vieta into a fast tool for missing coefficients and identity proofs.
Expand $(z - \alpha)(z - \beta)$. Before checking — read off the coefficient of $z$ and the constant term in terms of $\alpha$ and $\beta$. What does that tell you about the quadratic $z^2 + bz + c$ whose roots are $\alpha, \beta$? Sketch your reasoning below.
Every Vieta question rewards two habits: make the polynomial monic first (divide through by the leading coefficient), then read off the elementary symmetric sums $e_1, e_2, \ldots, e_n$ from the coefficients with alternating signs. Once you have $e_1, e_2, \ldots$ you can compute any symmetric expression in the roots without solving the equation.
The monic-read-combine routine: (1) divide so the leading coefficient is 1, (2) read off $\sum \alpha_i = -a_{n-1}$, $\sum_{i $z^n + a_{n-1}z^{n-1} + \cdots + a_0 = 0 \;\Rightarrow\; \sum \alpha_i = -a_{n-1},\; \prod \alpha_i = (-1)^n a_0$
Key facts
- For monic $z^n + a_{n-1}z^{n-1} + \cdots + a_0 = 0$: $\sum \alpha_i = -a_{n-1}$, $\prod \alpha_i = (-1)^n a_0$
- Cubic $z^3 + bz^2 + cz + d$: $\sum \alpha = -b$, $\sum_{i
- Quartic $z^4 + bz^3 + cz^2 + dz + e$: $\sum \alpha = -b$, $\sum_{i
- $\sum \alpha_i^2 = e_1^2 - 2e_2$ and $\sum 1/\alpha_i = e_{n-1}/e_n$ (for monic)
- Quartic $z^4 + bz^3 + cz^2 + dz + e$: $\sum \alpha = -b$, $\sum_{i
Concepts
- Why Vieta's formulas follow from expanding $(z - \alpha_1)\cdots(z - \alpha_n)$
- Why symmetric expressions in roots are always polynomial functions of the coefficients
- Why you must make the polynomial monic before applying the formulas
Skills
- Find missing coefficients given partial information about the roots
- Compute $\sum \alpha^2$, $\sum 1/\alpha$ and related identities without solving
- Construct a polynomial whose roots are a transformation of given roots
If the monic polynomial $P(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ has roots $\alpha_1, \alpha_2, \ldots, \alpha_n$ (with multiplicity), then $P(z) = (z - \alpha_1)(z - \alpha_2) \cdots (z - \alpha_n)$. Expanding the right side and matching coefficients gives Vieta's formulas:
Concrete cases.
- Quadratic $z^2 + bz + c$: $\alpha + \beta = -b$ and $\alpha\beta = c$.
- Cubic $z^3 + bz^2 + cz + d$: $\alpha + \beta + \gamma = -b$, $\;\alpha\beta + \beta\gamma + \gamma\alpha = c$, $\;\alpha\beta\gamma = -d$.
- Quartic $z^4 + bz^3 + cz^2 + dz + e$: $\sum \alpha = -b$, $\sum_{i
Worked through the hook: $z^3 - 4z^2 + pz - 6 = 0$ is monic with $b = -4$, $d = -6$. Vieta: $\alpha + \beta + \gamma = -(-4) = 4$ ✓, $\alpha\beta\gamma = -(-6) = 6$ ✓. The missing piece is $\alpha\beta + \beta\gamma + \gamma\alpha = c = p$ (the coefficient of $z$). With only the information given, $p$ isn't determined uniquely — but if we additionally knew, say, $\sum \alpha^2 = 6$, then $p = \tfrac{1}{2}(e_1^2 - \sum \alpha^2) = \tfrac{1}{2}(16 - 6) = 5$.
Monic $z^n + a_{n-1}z^{n-1} + \cdots + a_0 = 0$ $\Rightarrow$ $\sum \alpha = -a_{n-1}$, $\prod \alpha = (-1)^n a_0$ · Cubic $z^3 + bz^2 + cz + d$: $\sum = -b$, $\sum_{i
Pause — copy Vieta's formulas for monic degree $n$ ($\sum\alpha = -a_{n-1}$, $\prod\alpha = (-1)^n a_0$), the alternating-sign pattern, and the cubic/quartic sign table into your book.
Quick check: The cubic $z^3 - 6z^2 + 11z - 6 = 0$ has roots $\alpha, \beta, \gamma$. What is the value of $\alpha\beta + \beta\gamma + \gamma\alpha$?
We just saw Vieta's formulas for a monic polynomial: $\sum \alpha = -a_{n-1}$, $\prod \alpha = (-1)^n a_0$, with signs alternating by degree. That raises a question: how do we compute symmetric combinations like $\sum \alpha_i^2$ or $\sum 1/\alpha_i$ directly from the coefficients? This card answers it → use $\sum \alpha_i^2 = e_1^2 - 2e_2$ and $\sum 1/\alpha_i = e_{n-1}/e_n$.
Many exam targets are not $e_1, e_2, \ldots$ themselves but related symmetric expressions. The two most common identities — and the ones you should commit to memory:
- Sum of squares. $\displaystyle \sum_i \alpha_i^2 = \left(\sum_i \alpha_i\right)^2 - 2\sum_{i
- Sum of reciprocals. $\displaystyle \sum_i \frac{1}{\alpha_i} = \frac{\sum_{i
- Sum of reciprocals. $\displaystyle \sum_i \frac{1}{\alpha_i} = \frac{\sum_{i
These follow from the algebra of expanding $(\sum \alpha_i)^2$ and combining $1/\alpha_1 + \cdots + 1/\alpha_n$ over the common denominator $\prod \alpha_i$.
$\sum \alpha_i^2 = e_1^2 - 2e_2$ (sum of squares identity) · $\sum 1/\alpha_i = e_{n-1}/e_n$ when all $\alpha_i \neq 0$ · For non-monic $az^n + bz^{n-1} + \cdots$: $\sum \alpha = -b/a$, $\prod \alpha = (-1)^n a_0/a$ · Vieta works for any symmetric expression in roots
Pause — copy the sum-of-squares identity $\sum\alpha_i^2 = e_1^2 - 2e_2$, the reciprocal-sum formula $\sum 1/\alpha_i = e_{n-1}/e_n$, and the non-monic adjustment ($\sum\alpha = -b/a$) into your book.
Did you get this? True or false: if $\alpha, \beta, \gamma$ are roots of $z^3 - 2z^2 + 3z - 4 = 0$, then $\alpha^2 + \beta^2 + \gamma^2 = -2$.
Worked examples · 3 in a row, reveal as you go
The cubic $z^3 + pz^2 + qz - 12 = 0$ has roots $\alpha, \beta, \gamma$ with $\alpha + \beta + \gamma = 2$ and $\alpha\beta + \beta\gamma + \gamma\alpha = -5$. Find $p$ and $q$.
If $\alpha, \beta, \gamma$ are the roots of $z^3 - 3z^2 + 4z - 1 = 0$, find $\alpha^2 + \beta^2 + \gamma^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ without solving the equation.
The quartic $2z^4 - 8z^3 + cz^2 + 6z - 4 = 0$ has roots whose sum is $4$ and whose product is $-2$. Find $c$ given additionally that $\sum \alpha^2 = 10$.
Fill the gap: For the monic cubic $z^3 + bz^2 + cz + d$ with roots $\alpha, \beta, \gamma$: $\alpha + \beta + \gamma = $, $\;\alpha\beta + \beta\gamma + \gamma\alpha = $, $\;\alpha\beta\gamma = $.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for the non-monic cubic $2z^3 - 6z^2 + 4z - 8 = 0$, the sum of the roots is $-6$.
Activities · practice with the ideas
For $z^3 - 5z^2 + 8z - 4 = 0$ with roots $\alpha, \beta, \gamma$, write down $\alpha + \beta + \gamma$, $\alpha\beta + \beta\gamma + \gamma\alpha$, $\alpha\beta\gamma$. Hence find $\alpha^2 + \beta^2 + \gamma^2$.
The quartic $z^4 + pz^3 + qz^2 + 4z + r = 0$ has roots whose sum is $-3$, whose sum of pairwise products is $5$, and whose product is $-2$. Find $p$, $q$ and $r$.
For $z^3 - 4z^2 + 5z - 2 = 0$, find $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ without solving.
The roots of $z^3 + 2z^2 - z + 5 = 0$ are $\alpha, \beta, \gamma$. Find the monic cubic whose roots are $\alpha + 2, \beta + 2, \gamma + 2$. (Hint: substitute $w = z + 2$, i.e., $z = w - 2$.)
If $\alpha, \beta, \gamma$ are roots of $z^3 - 6z^2 + 11z - 6 = 0$, show that $\alpha^3 + \beta^3 + \gamma^3 = 36$. (Use Newton's identity $\sum \alpha^3 = e_1^3 - 3 e_1 e_2 + 3 e_3$.)
Odd one out: Three of these are valid Vieta identities for the monic cubic $z^3 + bz^2 + cz + d$ with roots $\alpha, \beta, \gamma$. Which one is NOT?
Earlier you considered $z^3 - 4z^2 + pz - 6 = 0$ with $\sum \alpha = 4$ and $\alpha\beta\gamma = 6$, and were asked for $\alpha\beta + \beta\gamma + \gamma\alpha$ and $p$.
Vieta gives $\alpha\beta + \beta\gamma + \gamma\alpha = c = p$ directly — but the value depends on extra information (such as $\sum \alpha^2$). With $\sum \alpha^2 = 6$, $p = \tfrac{1}{2}(16 - 6) = 5$. The big idea: Vieta turns the symmetric structure of the roots into linear access to the coefficients, so you can move between roots and coefficients in either direction without ever solving the polynomial.
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. The cubic $z^3 - 7z^2 + kz - 5 = 0$ has roots $\alpha, \beta, \gamma$ with $\alpha\beta + \beta\gamma + \gamma\alpha = 12$. Find $k$ and state $\alpha + \beta + \gamma$ and $\alpha\beta\gamma$. (2 marks)
Q2. If $\alpha, \beta, \gamma$ are roots of $z^3 - 4z^2 + 7z - 3 = 0$, compute $\alpha^2 + \beta^2 + \gamma^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ without solving the equation. (3 marks)
Q3. The quartic $z^4 + 2z^3 - 3z^2 + pz + q = 0$ has roots whose sum is $-2$, sum of pairwise products $-3$, and sum of triple products $1$. Find the product of the roots, $p$ and $q$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $e_1 = 5$, $e_2 = 8$, $e_3 = 4$. $\sum \alpha^2 = e_1^2 - 2 e_2 = 25 - 16 = 9$.
2. From the monic quartic: $-p = -3 \Rightarrow p = 3$. $q = 5$. Coefficient of $z$ is $-(\sum_{i 3. $\sum 1/\alpha_i = e_2/e_3 = 5/2$. 4. Substituting $z = w - 2$ into $z^3 + 2z^2 - z + 5$: $(w-2)^3 + 2(w-2)^2 - (w-2) + 5 = w^3 - 6w^2 + 12w - 8 + 2(w^2 - 4w + 4) - w + 2 + 5 = w^3 - 4w^2 + 3w + 7$. Roots are $\alpha + 2$, etc. 5. $e_1 = 6$, $e_2 = 11$, $e_3 = 6$. By Newton: $\sum \alpha^3 = e_1^3 - 3 e_1 e_2 + 3 e_3 = 216 - 198 + 18 = 36$. (Sanity check: $1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$ since the roots are $1, 2, 3$.) Q1 (2 marks): Monic cubic, Vieta: $k = \alpha\beta + \beta\gamma + \gamma\alpha = 12$ [1]. $\sum \alpha = 7$ and $\alpha\beta\gamma = 5$ [1]. Q2 (3 marks): $e_1 = 4$, $e_2 = 7$, $e_3 = 3$ from Vieta [1]. $\sum \alpha^2 = e_1^2 - 2 e_2 = 16 - 14 = 2$ [1]. $\sum 1/\alpha = e_2/e_3 = 7/3$ [1]. Q3 (3 marks): Monic quartic. $e_1 = -b = -2$ ✓, $e_2 = c = -3$ ✓ [1]. $e_3 = -d = 1 \Rightarrow d = -1$, so $p = -1$ [1]. The product $e_4 = q$ is not pinned by the given data — additional information (e.g., $\sum \alpha^2$ or a specific root) is needed; if the problem intended $\sum \alpha^2 = 10$, then $q$ remains free as no constraint involves $e_4$ via the sum-of-squares identity. Stating "$q$ is undetermined by the data given" earns the final mark [1].
Five timed questions on Vieta's formulas, symmetric sums and missing coefficients. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick Vieta questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.