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 OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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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 $(-1)^k \frac{a_{n - k}}{a_n}$ .

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.

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.

Worked example

Quadratic sum and product

The quadratic $2 x^2 - 5 x + 3$ has roots $\alpha, \beta$ . Find $\alpha + \beta$ and $\alpha \beta$ .

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

Cubic with one given root

The cubic $x^3 + p x + q = 0$ has roots $\alpha, \beta, \gamma$ . If $\alpha = 1$ , $\beta + \gamma = 2$ and $\beta \gamma = -3$ , find $p$ and $q$ .

Sum: $\alpha + \beta + \gamma = 1 + 2 = 3$ . But coefficient of $x^2$ is $0$ , so sum should be $0$ . Contradiction unless $\alpha = 1$ is consistent with $\beta + \gamma = -1$ .

Recompute with $\beta + \gamma = -1$ and $\beta \gamma = -3$ : sum is $1 + (-1) = 0$ . Good.

Sum in pairs: $\alpha \beta + \alpha \gamma + \beta \gamma = \alpha(\beta + \gamma) + \beta \gamma = 1(-1) + (-3) = -4$ . So $p = -4$ .

Product: $\alpha \beta \gamma = 1 \cdot (-3) = -3$ . So $q = 3$ (since $\alpha \beta \gamma = -q$ ).

Building a quartic

Find a monic quartic with roots $1, 2, 3, 4$ .

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

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

Check via Vieta: sum is $1 + 2 + 3 + 4 = 10$ , coefficient of $x^3$ is $-10$ . Sum in pairs is $\sum_{i < j} i j = 35$ . Product is $24$ . All match.

Sum of squares of roots

The cubic $x^3 - 3 x^2 + 4 x - 1$ has roots $\alpha, \beta, \gamma$ . Find $\alpha^2 + \beta^2 + \gamma^2$ .

$\sum = 3$ , $\sum_{\text{pairs}} = 4$ .

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

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.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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 $\alpha, \beta, \gamma$ :

IMATH_2
IMATH_3
IMATH_4

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.