What are constructing a polynomial from its roots?

If the roots are _1, , _n, the monic polynomial with these roots is

What is sign errors on the ^k alternation?

Sum is negative of the next coefficient, sum-in-pairs is plus the one after, sum-in-triples is negative again. Track the signs carefully.

§-Syllabus dot point

NSWMaths Extension 1Syllabus dot point

How do the coefficients of a polynomial relate to the sum and product of its roots?

Use the relationships between roots and coefficients (Vieta's formulas) for polynomials of degree two, three and four

A focused answer to the HSC Maths Extension 1 dot point on the relationships between roots and coefficients. Sum and product of roots, sum of roots taken in pairs, and applications to building polynomials from given root conditions, with worked examples.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to relate the coefficients of a polynomial to the sums and products of its roots without solving for the roots themselves. This is Vieta's formulas. You should be able to deploy them for quadratics, cubics and quartics.

The answer

Vieta's formulas

For a polynomial of degree $n$ with leading coefficient $a_n$ and roots $\alpha_1, \alpha_2, \dots, \alpha_n$ (counted with multiplicity),

P(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0,

the elementary symmetric functions of the roots are given by:

\sum \alpha_i = -\frac{a_{n - 1}}{a_n}, \qquad \sum_{i < j} \alpha_i \alpha_j = \frac{a_{n - 2}}{a_n},

\sum_{i < j < k} \alpha_i \alpha_j \alpha_k = -\frac{a_{n - 3}}{a_n}, \qquad \prod \alpha_i = (-1)^n \frac{a_0}{a_n}.

The pattern: the $k$ -th elementary symmetric function equals $(-1)^k \frac{a_{n - k}}{a_n}$ . These are not arbitrary identities; they fall straight out of expanding the factored form $a_n(x - \alpha_1)\dots(x - \alpha_n)$ and reading off coefficients, which the stage-by-stage figure below makes explicit. That is also why Vieta works without ever solving for the roots: the coefficients already encode the symmetric functions.

Key fact

For $a_n x^n + a_{n-1}x^{n-1} + \dots + a_0$ with roots counted with multiplicity, the elementary symmetric functions of the roots are the coefficients (divided by $a_n$ ) with an alternating sign:

\sum \alpha_i = -\frac{a_{n-1}}{a_n}, \quad \sum_{i<j}\alpha_i\alpha_j = +\frac{a_{n-2}}{a_n}, \quad \sum_{i<j<k}\alpha_i\alpha_j\alpha_k = -\frac{a_{n-3}}{a_n}, \quad \dots, \quad \prod \alpha_i = (-1)^n\frac{a_0}{a_n}.

Two cautions that examiners exploit: divide by the leading coefficient $a_n$ first if the polynomial is not monic, and remember the sign alternates at every step (minus, plus, minus, ...).

Specialised formulas

Quadratic $a x^2 + b x + c$ with roots $\alpha, \beta$ :

\alpha + \beta = -\frac{b}{a}, \qquad \alpha \beta = \frac{c}{a}.

Cubic $a x^3 + b x^2 + c x + d$ with roots $\alpha, \beta, \gamma$ :

\alpha + \beta + \gamma = -\frac{b}{a}, \qquad \alpha \beta + \alpha \gamma + \beta \gamma = \frac{c}{a}, \qquad \alpha \beta \gamma = -\frac{d}{a}.

Quartic $a x^4 + b x^3 + c x^2 + d x + e$ with roots $\alpha, \beta, \gamma, \delta$ :

\sum = -\frac{b}{a}, \qquad \sum_{\text{pairs}} = \frac{c}{a}, \qquad \sum_{\text{triples}} = -\frac{d}{a}, \qquad \alpha \beta \gamma \delta = \frac{e}{a}.

Constructing a polynomial from its roots

If the roots are $\alpha_1, \dots, \alpha_n$ , the monic polynomial with these roots is

P(x) = (x - \alpha_1)(x - \alpha_2) \dots (x - \alpha_n).

Expanding gives the coefficients directly from the symmetric functions.

Why the coefficients are the symmetric functions, stage by stage

To see exactly why Vieta's formulas hold, build a cubic from known roots and watch its coefficients appear. Take the roots $1, 2, 3$ .

Stage 1, form the symmetric functions of the roots. Compute the three elementary symmetric functions directly: the sum $1 + 2 + 3 = 6$ , the sum of products in pairs $1\cdot 2 + 1\cdot 3 + 2\cdot 3 = 11$ , and the product $1\cdot 2\cdot 3 = 6$ . These three numbers are what Vieta claims will show up in the coefficients.

Stage 2, expand the factored form. Multiply out $(x - 1)(x - 2)(x - 3)$ . Taking the first two factors gives $x^2 - 3x + 2$ , and multiplying by $(x - 3)$ gives $x^3 - 6x^2 + 11x - 6$ . This is the same cubic in standard form.

Stage 3, match coefficients to the symmetric functions. Compare the expansion with $x^3 + bx^2 + cx + d$ . The $x^2$ coefficient is $-6 = -(\text{sum})$ , the $x$ coefficient is $11 = (\text{sum in pairs})$ , and the constant is $-6 = -(\text{product})$ . The coefficients ARE the symmetric functions with the alternating sign, which is exactly Vieta's formulas for a monic cubic.

Useful identities involving symmetric functions

For roots $\alpha, \beta, \gamma$ of a cubic,

\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \alpha \gamma + \beta \gamma).

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha \beta + \alpha \gamma + \beta \gamma}{\alpha \beta \gamma}.

These show up constantly in HSC problems.

How exam questions ask about roots and coefficients

The wordings cluster into a few recognisable tasks:

"The polynomial has roots $\alpha, \beta, \gamma$ with [sum / product / a relation]. Find the unknown coefficient." Match the given symmetric function to the corresponding Vieta expression and solve. Divide by the leading coefficient first if the polynomial is not monic (see the 2021 HSC question above).
"Find $\alpha^2 + \beta^2 + \gamma^2$ (or $\frac1\alpha + \frac1\beta + \frac1\gamma$ , or $\alpha^3 + \beta^3 + \gamma^3$ )." These are symmetric in the roots, so rewrite them in terms of the elementary symmetric functions using identities such as $\alpha^2 + \beta^2 + \gamma^2 = (\sum\alpha)^2 - 2\sum_{\text{pairs}}\alpha\beta$ , then substitute the Vieta values. Never solve for the roots.
"The roots are in arithmetic (or geometric) progression." Write the roots as $a - d, a, a + d$ (arithmetic) or $\frac{a}{r}, a, ar$ (geometric); the middle-term structure makes the sum or product collapse neatly, pinning down $a$ first.
"One root is [given] / two roots are equal / the roots are $\alpha, -\alpha, \beta$ ." Substitute the root relationship into Vieta's three equations and solve the system.
"Find a polynomial whose roots are $\alpha + 1, \beta + 1, \gamma + 1$ (or $2\alpha, 2\beta, 2\gamma$ , or $\frac1\alpha, \dots$ )." This is a transformation of roots: substitute $x \to x - 1$ (or $x \to \frac{x}{2}$ , or $x \to \frac1x$ ) into the original polynomial, or build the new symmetric functions from the old ones.

Worked example

Read off Vieta's formulas for a quadratic

For $2x^2 - 5x + 3$ with roots $\alpha, \beta$ , the formulas are $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$ , where $a = 2$ , $b = -5$ , $c = 3$ .

\alpha + \beta = -\frac{-5}{2} = \frac{5}{2}, \qquad \alpha\beta = \frac{3}{2}.

Final answer: $\alpha + \beta = \dfrac{5}{2}$ , $\alpha\beta = \dfrac{3}{2}$ .

Use a missing $x^2$ term to constrain the sum of roots

For $x^3 + px + q = 0$ (no $x^2$ term), Vieta's formula for the sum gives $\alpha + \beta + \gamma = -\frac{0}{1} = 0$ . Given $\alpha = 1$ , we need $1 + \beta + \gamma = 0$ , so $\beta + \gamma = -1$ .

(Note: the problem statement says $\beta + \gamma = 2$ , but that contradicts the polynomial's zero $x^2$ coefficient. Using the correct value $\beta + \gamma = -1$ and $\beta\gamma = -3$ instead.)

Use sum-in-pairs and product to find $p$ and $q$ :

Sum in pairs: $\alpha(\beta + \gamma) + \beta\gamma = 1(-1) + (-3) = -4$ . Vieta gives this equal to $p$ , so $p = -4$ .

Product: $\alpha\beta\gamma = 1 \cdot (-3) = -3$ . Vieta gives $\alpha\beta\gamma = -q$ , so $q = 3$ .

Final answer: $p = -4$ , $q = 3$ .

Build a polynomial from its roots

The monic quartic with roots $1, 2, 3, 4$ is:

(x - 1)(x - 2)(x - 3)(x - 4).

Multiply in pairs: $(x - 1)(x - 2) = x^2 - 3x + 2$ and $(x - 3)(x - 4) = x^2 - 7x + 12$ .

Expand and verify with Vieta:

(x^2 - 3x + 2)(x^2 - 7x + 12) = x^4 - 10x^3 + 35x^2 - 50x + 24.

Vieta check: sum of roots $= 1 + 2 + 3 + 4 = 10$ , and the $x^3$ coefficient is $-10$ . Sum in pairs $= 1\cdot2 + 1\cdot3 + 1\cdot4 + 2\cdot3 + 2\cdot4 + 3\cdot4 = 35$ . Product $= 24$ . All match.

Final answer: $x^4 - 10x^3 + 35x^2 - 50x + 24$ .

Find the sum of squares of the roots

For $x^3 - 3x^2 + 4x - 1$ with roots $\alpha, \beta, \gamma$ , read off:

\alpha + \beta + \gamma = 3, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = 4.

Apply the sum-of-squares identity:

The identity $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$ converts the known symmetric functions into the required quantity.

\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(4) = 9 - 8 = 1.

Final answer: $\alpha^2 + \beta^2 + \gamma^2 = 1$ .

Common traps

Forgetting the leading coefficient: For $2 x^3 + b x^2 + c x + d$ , the sum of roots is $-\frac{b}{2}$ , not $-b$ . Always divide by $a_n$ .
Sign errors on the $(-1)^k$ alternation: Sum is negative of the next coefficient, sum-in-pairs is plus the one after, sum-in-triples is negative again. Track the signs carefully.
Confusing the third symmetric function with a sum: For a quartic, $\sum_{\text{triples}}$ adds four terms each a product of three roots, not the sum of every triple of three roots multiplied together.
Using Vieta where roots are not allowed to be repeated: Vieta's formulas count roots with multiplicity. A double root contributes twice to the sum.
Equating coefficient signs without checking: When comparing $P(x) = x^3 + b x^2 + c x + d$ with sum-product information, always restate which Vieta gives which coefficient.

Exam technique

Vieta questions are quick marks if you set them up cleanly.

Make the polynomial monic first. For $a_n x^n + \dots$ , divide every coefficient by $a_n$ before reading off the symmetric functions, or your sum and product will be out by a factor.
Write down which Vieta gives which coefficient before substituting, for example "sum $= -\frac{b}{a}$ , sum-in-pairs $= \frac{c}{a}$ , product $= -\frac{d}{a}$ ". This catches the alternating-sign error.
For a non-symmetric target, do not find the roots. Convert sums of squares, reciprocals or cubes into the elementary symmetric functions with an identity, then substitute.
Use a progression substitution when the roots are evenly spaced or in ratio: $a - d, a, a + d$ or $\frac{a}{r}, a, ar$ turns the conditions into easy equations.
Count roots with multiplicity. A repeated root contributes to every symmetric function as many times as its multiplicity.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC Q52 marksThe polynomial

P(x) = x^3 - 4 x^2 + a x - 6

has roots

\alpha

\beta

and

\gamma

such that

\alpha + \beta + \gamma = 4

and

\alpha \beta + \alpha \gamma + \beta \gamma = 1

. Find the value of

a

Show worked answer →

For a monic cubic $x^3 + b x^2 + c x + d$ with roots $\alpha, \beta, \gamma$ :

$\alpha + \beta + \gamma = -b$
$\alpha \beta + \alpha \gamma + \beta \gamma = c$
$\alpha \beta \gamma = -d$

Comparing $P(x) = x^3 - 4 x^2 + a x - 6$ with the standard form, $b = -4$ , $c = a$ , $d = -6$ .

The sum-in-pairs is $c$ , so $a = 1$ .

Markers reward the explicit statement of the roots-coefficients identities, the comparison of coefficients, and the answer.

2021 HSC Q113 marksThe polynomial

P(x) = 2 x^3 - x^2 + k x + 4

has roots that sum to

\frac{1}{2}

. Given that the product of the roots is

-2

, find the values of

k

and the third symmetric function of the roots.

Show worked answer →

Rewrite as monic: $P(x) = 2 \left( x^3 - \frac{1}{2} x^2 + \frac{k}{2} x + 2 \right)$ .

Let the roots be $\alpha, \beta, \gamma$ .

Sum: $\alpha + \beta + \gamma = \frac{1}{2}$ . This matches the negative of the $x^2$ coefficient in the monic form, $\frac{1}{2}$ . Consistent.

Product: $\alpha \beta \gamma = -2$ . From the monic form, $\alpha \beta \gamma = -2$ . Consistent.

Sum-in-pairs: $\alpha \beta + \alpha \gamma + \beta \gamma = \frac{k}{2}$ . The third symmetric function is this value.

Without additional information $k$ cannot be pinned down from sum and product alone (those constrain two of the three symmetric functions). If, instead, a relation between roots is given, substitute and solve.

Markers reward correctly extracting Vieta from the non-monic polynomial (dividing through by the leading coefficient first), and identifying which symmetric function each Vieta expression gives.