What is forming a new equation?

To find the equation whose roots are a transformation of the originals (for example α + 1, or 1α), substitute. If y = α + 1 then α = y - 1, so substituting x = y - 1 into the original polynomial gives the new equation in y. Alternatively, recompute the symmetric functions of the new roots.

What is errors in the transformation substitution?

If y = α + 1 then α = y - 1; substitute x = y - 1, not x = y + 1.

For x^2 - 5x + 6 = 0 with roots α, β, state α + β and αβ. [1 mark]

If α + β = 4 and αβ = 3, find α^2 + β^2. [2 marks]

SEAB Original-style practice: The roots of x^3 - 6x^2 + 11x - 6 = 0 are α, β, γ. Without solving the equation, find α + β + γ, αβ + βγ + γα and αβγ, and hence evaluate α^2 + β^2 + γ^2.

For x^3 + px^2 + qx + r = 0 the symmetric functions are sumα = -p, sumαβ = q, αβγ = -r. Here p = -6, q = 11, r = -6, so $α + β + γ = 6, αβ + βγ + γα = 11, αβγ = 6.$ For the sum of squares use the identity $α^2 + β^2 + γ^2 = (α + β + γ)^2 - 2(αβ + βγ + γα) = 6^2 - 2(11) = 36 - 22 = 14.$ Markers reward the three symmetric functions with correct signs, the identity for the sum of squares, and the value 14.

SingaporeFurther MathsSyllabus dot point

How do the roots of a polynomial relate to its coefficients, and how do complex roots occur in conjugate pairs?

Use the relationships between the roots and coefficients of a polynomial and apply the conjugate root theorem for real polynomials

A focused answer to the H2 Further Mathematics outcome on polynomials. The sum and product of roots, symmetric functions of roots, the conjugate root theorem for real polynomials, and forming new equations whose roots are transformed.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking

SEAB wants you to relate the roots of a polynomial to its coefficients through the symmetric functions (sum of roots, sum of products in pairs, product of roots), to use the conjugate root theorem for polynomials with real coefficients, and to use these relationships to evaluate symmetric expressions in the roots or to build new equations without solving the original.

The answer

Roots and coefficients: the symmetric functions

For a monic quadratic $x^2 + bx + c = 0$ with roots $\alpha, \beta$ :

\alpha + \beta = -b, \qquad \alpha\beta = c.

For a monic cubic $x^3 + px^2 + qx + r = 0$ with roots $\alpha, \beta, \gamma$ :

\alpha + \beta + \gamma = -p, \quad \alpha\beta + \beta\gamma + \gamma\alpha = q, \quad \alpha\beta\gamma = -r.

The signs alternate: each successive symmetric function is $(-1)^k$ times the corresponding coefficient (for a monic polynomial). If the polynomial is not monic, divide through by the leading coefficient first.

Evaluating symmetric expressions

Many quantities reduce to the symmetric functions through standard identities. The most used:

\alpha^2 + \beta^2 + \gamma^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta, \qquad \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}.

You evaluate them from the coefficients without ever finding the individual roots.

The conjugate root theorem

If a polynomial has real coefficients and $a + bi$ (with $b \neq 0$ ) is a root, then its complex conjugate $a - bi$ is also a root. Complex roots of a real polynomial therefore come in conjugate pairs, so a real polynomial of odd degree has at least one real root.

Forming a new equation

To find the equation whose roots are a transformation of the originals (for example $\alpha + 1$ , or $\dfrac{1}{\alpha}$ ), substitute. If $y = \alpha + 1$ then $\alpha = y - 1$ , so substituting $x = y - 1$ into the original polynomial gives the new equation in $y$ . Alternatively, recompute the symmetric functions of the new roots.

Worked example

The roots of $2x^3 - 3x^2 + 4x - 1 = 0$ are $\alpha, \beta, \gamma$ . Find the value of $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ .

Step 1: Make the polynomial monic

Divide by the leading coefficient $2$ : $x^3 - \tfrac{3}{2}x^2 + 2x - \tfrac{1}{2} = 0$ .

Step 2: Read off the symmetric functions

Comparing with $x^3 + px^2 + qx + r$ : $p = -\tfrac{3}{2}$ , $q = 2$ , $r = -\tfrac{1}{2}$ . Hence

\sum\alpha = \tfrac{3}{2}, \quad \sum\alpha\beta = 2, \quad \alpha\beta\gamma = \tfrac{1}{2}.

Step 3: Use the reciprocal-sum identity

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

Step 4: Substitute

= \frac{2}{1/2} = 4.

Step 5: State the answer

$\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = 4$ , found entirely from the coefficients.

Common traps

Wrong signs in the symmetric functions: They alternate: $\sum\alpha = -p$ but $\sum\alpha\beta = +q$ . Misremembering the pattern is the commonest error.
Forgetting to make the polynomial monic: The relations assume a leading coefficient of $1$ ; divide through first if it is not.
Applying the conjugate root theorem to complex coefficients: It holds only for real coefficients; a polynomial with complex coefficients need not have conjugate roots.
Trying to find the roots: The point of these methods is to evaluate symmetric expressions without solving the polynomial; computing roots wastes time and may be impossible exactly.
Errors in the transformation substitution: If $y = \alpha + 1$ then $\alpha = y - 1$ ; substitute $x = y - 1$ , not $x = y + 1$ .

Examples in context

Example 1. Designing a polynomial with set properties. To construct a cubic with roots summing to $5$ and product $6$ , you write the coefficients directly from the symmetric functions, the reverse of root-finding and the basis of designing filters and characteristic polynomials.

Example 2. Real quadratic factors. Because complex roots pair up, a real quartic factorises into real quadratic factors $x^2 - 2ax + (a^2 + b^2)$ , which is why real signals and systems are analysed with real second-order building blocks rather than individual complex roots.

Try this

Q1. For $x^2 - 5x + 6 = 0$ with roots $\alpha, \beta$ , state $\alpha + \beta$ and $\alpha\beta$ . [1 mark]

Cue. $\alpha + \beta = 5$ , $\alpha\beta = 6$ .

Q2. A real cubic has $1 - 2i$ as a root. State another root. [1 mark]

Cue. Its conjugate $1 + 2i$ , by the conjugate root theorem.

Q3. If $\alpha + \beta = 4$ and $\alpha\beta = 3$ , find $\alpha^2 + \beta^2$ . [2 marks]

Cue. $(\alpha + \beta)^2 - 2\alpha\beta = 16 - 6 = 10$ .

Exam-style practice questions

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

Original6 marksThe roots of

x^3 - 6x^2 + 11x - 6 = 0

are

\alpha

\beta

\gamma

. Without solving the equation, find

\alpha + \beta + \gamma

\alpha\beta + \beta\gamma + \gamma\alpha

and

\alpha\beta\gamma

, and hence evaluate

\alpha^2 + \beta^2 + \gamma^2

Show worked answer →

For $x^3 + px^2 + qx + r = 0$ the symmetric functions are $\sum\alpha = -p$ , $\sum\alpha\beta = q$ , $\alpha\beta\gamma = -r$ . Here $p = -6$ , $q = 11$ , $r = -6$ , so

\alpha + \beta + \gamma = 6, \quad \alpha\beta + \beta\gamma + \gamma\alpha = 11, \quad \alpha\beta\gamma = 6.

For the sum of squares use the identity

\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 6^2 - 2(11) = 36 - 22 = 14.

Markers reward the three symmetric functions with correct signs, the identity for the sum of squares, and the value $14$ .

Original6 marksGiven that

2 + 3i

is a root of

x^3 + ax^2 + bx + 13 = 0

where

a

and

b

are real, find the third root and the values of

a

and

b

Show worked answer →

The coefficients are real, so by the conjugate root theorem $2 - 3i$ is also a root. Let the third (real) root be $\rho$ .

Product of roots: $(2 + 3i)(2 - 3i)\rho = -\dfrac{13}{1} = -13$ . Now $(2+3i)(2-3i) = 4 + 9 = 13$ , so $13\rho = -13$ , giving $\rho = -1$ .

Sum of roots: $(2 + 3i) + (2 - 3i) + (-1) = 3 = -a$ , so $a = -3$ .

Sum of products in pairs: $(2+3i)(2-3i) + (2+3i)(-1) + (2-3i)(-1) = 13 + (-2 - 3i) + (-2 + 3i) = 13 - 4 = 9 = b$ .

So the third root is $-1$ , with $a = -3$ and $b = 9$ .

Markers reward invoking the conjugate root theorem, using the product of roots to find $\rho = -1$ , and the sum and pairwise-product relations to find $a$ and $b$ .