Sums and Products of Roots — Cubics
For a cubic equation $ax^3 + bx^2 + cx + d = 0$ you don't need to solve it to know the sum, the pairwise products, or the product of all three roots. Vieta's formulas read these values straight from the coefficients — and unlock a whole class of problems where you evaluate symmetric expressions without ever touching the roots individually.
For $x^3 - 6x^2 + 11x - 6 = 0$, the roots are 1, 2, and 3. Without using a formula — what are their sum, the sum of their pairwise products (i.e. $1\cdot2 + 2\cdot3 + 1\cdot3$), and their product? How do these relate to the coefficients $-6$, $11$, $-6$?
Every cubic roots problem uses two moves: read Vieta's formulas off the coefficients, then combine them algebraically to evaluate whatever symmetric expression the question asks for.
For $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$, Move 1 is: write down $\alpha+\beta+\gamma = -b/a$, $\alpha\beta+\beta\gamma+\gamma\alpha = c/a$, $\alpha\beta\gamma = -d/a$. Move 2 is: substitute these values into the symmetric identity the question requires.
Key facts
- $\alpha + \beta + \gamma = -b/a$ for $ax^3 + bx^2 + cx + d = 0$
- $\alpha\beta + \beta\gamma + \gamma\alpha = c/a$
- $\alpha\beta\gamma = -d/a$
Concepts
- How Vieta's formulas arise from expanding $a(x-\alpha)(x-\beta)(x-\gamma)$
- Why the signs alternate between the three formulas
- How symmetric identities let you evaluate expressions without finding roots
Skills
- Apply Vieta's formulas to any cubic equation
- Evaluate $\alpha^2 + \beta^2 + \gamma^2$ and similar expressions
- Form a cubic equation given three roots
If $\alpha, \beta, \gamma$ are roots of $ax^3 + bx^2 + cx + d = 0$, then:
Expanding the right side:
Matching coefficients with $ax^3 + bx^2 + cx + d$:
- $x^2$ coefficient: $-a(\alpha+\beta+\gamma) = b \implies \alpha+\beta+\gamma = -b/a$
- $x$ coefficient: $a(\alpha\beta+\beta\gamma+\gamma\alpha) = c \implies \alpha\beta+\beta\gamma+\gamma\alpha = c/a$
- constant: $-a\alpha\beta\gamma = d \implies \alpha\beta\gamma = -d/a$
Sign pattern to memorise: $-b/a$, $+c/a$, $-d/a$. The signs alternate starting negative.
For ax^3 + bx^2 + cx + d = 0 with roots , , :; + + = -b/a (sum, negative sign)
Pause — copy Vieta's formulas for cubics into your book: for $ax^3 + bx^2 + cx + d = 0$: $\alpha+\beta+\gamma = -b/a$; $\alpha\beta+\beta\gamma+\gamma\alpha = c/a$; $\alpha\beta\gamma = -d/a$.
Quick check: For $x^3 - 6x^2 + 11x - 6 = 0$, what is $\alpha\beta\gamma$?
We just saw that for $ax^3 + bx^2 + cx + d = 0$, $\alpha+\beta+\gamma = -b/a$, $\alpha\beta+\beta\gamma+\gamma\alpha = c/a$, and $\alpha\beta\gamma = -d/a$. That raises a question: when applying these to a specific cubic, a missing term (e.g. no $x^2$ term) means one Vieta sum is zero — how do you identify and use this quickly? This card answers it → always write $ax^3 + bx^2 + cx + d$ explicitly; missing terms mean that coefficient is $0$, e.g. $x^3 - 3x + 1$ has $b = 0$ so $\alpha+\beta+\gamma = 0$.
Given any cubic $ax^3 + bx^2 + cx + d = 0$, write down the three Vieta values immediately by inspection:
| Equation | $\alpha+\beta+\gamma$ | $\alpha\beta+\beta\gamma+\gamma\alpha$ | $\alpha\beta\gamma$ |
|---|---|---|---|
| $x^3 - 2x^2 + 3x - 1 = 0$ | $2$ | $3$ | $1$ |
| $2x^3 - 6x^2 + 3x + 4 = 0$ | $3$ | $\frac{3}{2}$ | $-2$ |
| $3x^3 + 2x^2 - x + 4 = 0$ | $-\frac{2}{3}$ | $-\frac{1}{3}$ | $-\frac{4}{3}$ |
Key check: Always verify your signs. For $2x^3 - 6x^2 + 3x + 4 = 0$, note $a=2, b=-6, c=3, d=4$. So $\alpha+\beta+\gamma = -(-6)/2 = 3$. The negative sign in the formula cancels the negative sign of $b$.
Always identify a, b, c, d by matching ax^3 + bx^2 + cx + d. Missing terms mean the coefficient is 0.; x^3 - 3x + 1 = 0 has b = 0, so ++ = 0.
Pause — copy the application shortcut into your book: always identify $a, b, c, d$ explicitly from $ax^3 + bx^2 + cx + d$; missing terms mean that coefficient is $0$; e.g. $x^3 - 3x + 1$ has $b = 0$ giving $\alpha+\beta+\gamma = 0$.
Did you get this? True or false: for $x^3 - 3x + 1 = 0$, the sum of roots $\alpha + \beta + \gamma = 0$.
Worked examples · 3 in a row, reveal as you go
For $2x^3 - 6x^2 + 3x + 4 = 0$ with roots $\alpha, \beta, \gamma$, find the sum, sum of pairwise products, and product of the roots.
If $\alpha, \beta, \gamma$ are roots of $x^3 - 2x^2 + 3x - 1 = 0$, find $\alpha^2 + \beta^2 + \gamma^2$.
Form the monic cubic equation with roots $1$, $2$, and $-3$.
Fill the gap: For $x^3 - 3x + 1 = 0$, the sum of pairwise products $\alpha\beta + \beta\gamma + \gamma\alpha = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $x^3 + 5x - 2 = 0$, the product of roots $\alpha\beta\gamma = 2$.
Odd one out: Three of these are correct Vieta formulas for $ax^3+bx^2+cx+d=0$. Which one is the odd one out (incorrect)?
Activities · practice with the ideas
For $x^3 + 2x^2 - 5x + 3 = 0$, state $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$, and $\alpha\beta\gamma$.
For $3x^3 + 2x^2 - x + 4 = 0$, find $\alpha+\beta+\gamma$ and $\alpha\beta\gamma$.
If $\alpha, \beta, \gamma$ are roots of $x^3 - 3x + 1 = 0$, find $\alpha^2 + \beta^2 + \gamma^2$.
Form the monic cubic whose roots are $-1$, $2$, and $4$.
Why is it impossible for the roots of $x^3 + x^2 + x + 1 = 0$ to all be positive real numbers?
Earlier you were asked about $x^3 - 6x^2 + 11x - 6 = 0$ with roots 1, 2, 3. The answers: sum $= 1+2+3 = 6 = -(-6)/1$; pairwise sum $= 2+3+6 = 11 = 11/1$; product $= 6 = -(-6)/1$. Every value reads directly from the coefficients. Vieta's formulas encode the entire root structure into the polynomial's coefficients — without you ever needing to solve the equation.
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. For $3x^3 + 2x^2 - x + 4 = 0$ with roots $\alpha, \beta, \gamma$, find $\alpha + \beta + \gamma$ and $\alpha\beta\gamma$. (2 marks)
Q2. If $\alpha, \beta, \gamma$ are roots of $x^3 - 3x + 1 = 0$, find $\alpha^2 + \beta^2 + \gamma^2$. (2 marks)
Q3. Form the monic cubic equation with roots $1$, $2$, and $-3$. (2 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\alpha+\beta+\gamma = -2$, $\alpha\beta+\beta\gamma+\gamma\alpha = -5$, $\alpha\beta\gamma = -3$. · 2. $\alpha+\beta+\gamma = -2/3$, $\alpha\beta\gamma = -4/3$. · 3. $\alpha^2+\beta^2+\gamma^2 = 0^2 - 2(-3) = 6$. · 4. Sum $= 5$, pairwise sum $= -1\cdot2 + 2\cdot4 + (-1)\cdot4 = -2+8-4=2$, product $= -8$; equation: $x^3 - 5x^2 + 2x + 8 = 0$. · 5. Product $= -d/a = -1/1 = -1 < 0$, so not all roots can be positive (three positive numbers multiply to give a positive product).
Q1 (2 marks): $a=3, b=2, c=-1, d=4$. $\alpha+\beta+\gamma = -2/3$ [1]. $\alpha\beta\gamma = -4/3$ [1].
Q2 (2 marks): $b=0$, so sum $= 0$; $c=-3$, so pairwise sum $= -3$ [0.5]. $\alpha^2+\beta^2+\gamma^2 = 0^2 - 2(-3) = 6$ [1.5].
Q3 (2 marks): Sum $= 0$, pairwise sum $= 2-6-3 = -7$, product $= -6$ [1]. Equation: $x^3 - 0\cdot x^2 + (-7)x - (-6) = 0 \Rightarrow x^3 - 7x + 6 = 0$ [1].
Five timed questions on Vieta's formulas and symmetric expressions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering cubic roots questions. Lighter alternative to the boss.
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