Roots and Coefficients, General Polynomials
When Newton and Vieta uncovered the hidden language connecting a polynomial's roots to its coefficients, they cracked open a powerful shortcut still used in every branch of modern mathematics. In this lesson you'll generalise Vieta's formulas to polynomials of any degree, unlocking the alternating sign pattern and the elementary symmetric sums that tie roots to coefficients.
For a quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ with roots $\alpha, \beta, \gamma, \delta$, without looking up a formulawhat patterns do you expect for the sums and products of roots? Think about what you already know from quadratics and cubics.
Everything in this lesson flows from two core ideas. First: any degree-$n$ polynomial with roots $\alpha_1, \dots, \alpha_n$ can be written as $a_n(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$. Second: expanding that product and comparing coefficients gives Vieta's formulas, the alternating-sign relations between roots and coefficients.
Every Vieta question uses one of two moves: read the symmetric sum directly from $-a_{n-k}/a_n$ (with the correct sign), or build a required expression from the elementary symmetric sums using algebraic identities.
Key facts
- Vieta's formulas for a degree-$n$ polynomial
- The alternating sign pattern: $-, +, -, +, \dots$
- $\alpha_1\alpha_2\cdots\alpha_n = (-1)^n \dfrac{a_0}{a_n}$
Concepts
- Why expanding $(x-\alpha_1)\cdots(x-\alpha_n)$ produces the symmetric sums
- The connection between each coefficient and an elementary symmetric sum
- Why the sign depends on whether the degree is even or odd
Skills
- Read off any symmetric sum directly from the polynomial's coefficients
- Find the sum, product, and sum of products taken two at a time for any degree
- Apply general Vieta formulas in HSC-style problems
For $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$ with roots $\alpha_1, \alpha_2, \dots, \alpha_n$:
Expanding and comparing coefficients with the standard form gives the elementary symmetric sums:
- $\displaystyle e_1 = \sum_i \alpha_i = -\dfrac{a_{n-1}}{a_n}$
- $\displaystyle e_2 = \sum_{i<j} \alpha_i\alpha_j = +\dfrac{a_{n-2}}{a_n}$
- $\displaystyle e_3 = \sum_{i<j<k} \alpha_i\alpha_j\alpha_k = -\dfrac{a_{n-3}}{a_n}$
- $\vdots$
- $\displaystyle e_n = \alpha_1\alpha_2\cdots\alpha_n = (-1)^n\dfrac{a_0}{a_n}$
Sign pattern: The signs strictly alternate, starting with negative for $e_1$. In compact form: $e_k = (-1)^k \dfrac{a_{n-k}}{a_n}$.
e_1 = _i = -a_{n-1}{a_n} (sum of roots); e_2 = _{i<j}_i_j = +a_{n-2}{a_n} (sum of products taken 2 at a time)
Pause, copy the general Vieta formulas into your book: $e_1 = \sum \alpha_i = -a_{n-1}/a_n$; $e_2 = \sum_{i<j} \alpha_i\alpha_j = +a_{n-2}/a_n$; signs alternate; $e_k$ has $\binom{n}{k}$ terms.
Quick check: For $P(x) = x^3 - 6x^2 + 11x - 6$, what is the sum of the roots?
We just saw the general Vieta formulas: $e_1 = -a_{n-1}/a_n$ (sum), $e_2 = a_{n-2}/a_n$ (sum of pairwise products), and so on, alternating signs. That raises a question: for a quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ with four roots, what are the four Vieta expressions and how many terms does each one have? This card answers it → $e_1 = -b/a$ (4 terms), $e_2 = c/a$ (6 terms), $e_3 = -d/a$ (4 terms), $e_4 = e/a$ (1 term); $e_k$ has $\binom{4}{k}$ terms.
For a quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ with roots $\alpha, \beta, \gamma, \delta$:
Note that $(-1)^4 = +1$, so the product of all four roots is $+e/a$ (positive).
Each $e_k$ is a sum of $\binom{n}{k}$ terms, for $n=4$: $e_1$ has 4 terms, $e_2$ has 6, $e_3$ has 4, $e_4$ has 1.
Quartic ax^4+bx^3+cx^2+dx+e: e_1=-b/a, e_2=c/a, e_3=-d/a, e_4=e/a; e_k has n{k} terms, for degree 4: 4, 6, 4, 1
Pause, copy the quartic Vieta formulas into your book: $e_1 = -b/a$, $e_2 = c/a$, $e_3 = -d/a$, $e_4 = e/a$; number of terms: 4, 6, 4, 1 respectively (binomial coefficients $\binom{4}{k}$).
Did you get this? True or false: for a degree-4 polynomial, the product of all roots equals $+a_0/a_n$.
Worked examples · 3 in a row, reveal as you go
For $2x^4 - 3x^3 + x^2 - 5x + 7 = 0$, find (a) the sum of all roots and (b) the product of all roots.
For $x^5 + 3x^4 - 2x^3 + x - 7 = 0$, find $e_1$, $e_2$, and $e_5$ (the product of all roots).
For $3x^4 + 2x^3 - x^2 + 4x - 5 = 0$, find the sum of products of roots taken three at a time.
Fill the gap: For $x^4 - 2x^3 + 3x^2 - x + 4 = 0$, the product $\alpha\beta\gamma\delta = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for a degree-5 polynomial, the product of all roots equals $-a_0/a_5$.
Activities · practice with the ideas
For $x^4 - 5x^3 + 6x^2 - 3x + 2 = 0$, write down $e_1$, $e_2$, $e_3$, $e_4$ directly from the coefficients.
For $2x^5 - x^4 + 3x^3 - 7x + 4 = 0$, find the sum of all roots and the product of all roots.
A monic quartic has roots whose sum is $-4$, sum of products in pairs is $5$, sum of products in triples is $-2$, and product is $3$. Write down the quartic.
If $\alpha, \beta, \gamma, \delta$ are roots of $x^4 - 2x^3 + 3x^2 - x + 4 = 0$, find $\alpha\beta\gamma\delta$ and $\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta$.
Explain in your own words why the sign pattern for Vieta's formulas alternates, starting with a negative for the sum of roots.
Earlier you were asked about the patterns for a quartic. Now you know: $\alpha+\beta+\gamma+\delta = -b/a$ and $\alpha\beta\gamma\delta = +e/a$. The sign of the product is positive for even degree, which may surprise students who expect a negative. The sign alternates from $e_1$ to $e_4$ as $-, +, -, +$.
Why does the sign of the product depend on whether the degree is even or odd? The factor $(-1)^n$ comes directly from expanding $(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)$: each $-\alpha_i$ contributes one factor of $-1$, giving $(-1)^n$ in the constant term.
Odd one out: Which of the following is NOT a valid Vieta relation for a quartic $ax^4+bx^3+cx^2+dx+e=0$?
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. For $3x^4 + 2x^3 - x^2 + 4x - 5 = 0$, find the sum of roots and the product of roots. (2 marks)
Q2. If $\alpha, \beta, \gamma, \delta$ are roots of $x^4 - 2x^3 + 3x^2 - x + 4 = 0$, find $\alpha\beta\gamma\delta$. (1 mark)
Q3. For $x^5 + 3x^4 - 2x^3 + x - 7 = 0$, write down the sum of the roots and the sum of products of roots taken two at a time. (2 marks)
Comprehensive answers (click to reveal)
Activity drills: 1. $e_1=5,\;e_2=6,\;e_3=-3,\;e_4=2$ · 2. Sum $=\tfrac{1}{2}$, Product $=-\tfrac{4}{2}=-2$ · 3. $x^4+4x^3+5x^2+2x+3$ · 4. Product $=4$, Sum of products in pairs $=3$ · 5. Expanding $(x-\alpha_1)\cdots$ gives $(-1)^k e_k$ for the coefficient of $x^{n-k}$, so $e_k = (-1)^k a_{n-k}/a_n$.
Q1 (2 marks): $a_4=3$. Sum $= -\tfrac{2}{3}$ [1]. Product $= (-1)^4\cdot\tfrac{-5}{3} = -\tfrac{5}{3}$ [1].
Q2 (1 mark): $\alpha\beta\gamma\delta = (-1)^4\cdot\tfrac{4}{1} = 4$ [1].
Q3 (2 marks): $a_5=1$, $a_4=3$, $a_3=-2$, $a_2=0$. Sum $= -3$ [1]. $e_2 = +\tfrac{a_3}{a_5} = -2$ [1].
Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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