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.
If $\alpha, \beta, \gamma$ are roots of $ax^3 + bx^2 + cx + d = 0$, then:
Expanding the right side:
Matching coefficients with $ax^3 + bx^2 + cx + d$:
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)
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].
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