Why is "Writing $\sum\alpha^2 = (\sum\alpha)^2$, dropping the $-2\sum\alpha\beta$ term" a common mistake in Roots and Coefficients of polynomial equations?

Why it happens: Pattern-matching to (a+b)^2 without expanding; rushing straight to substitution. How to avoid it: Always write the identity ^2 = ()^2 - 2 in full BEFORE substituting any numbers. Source: 9231 Paper 1 Examiner Reports — recurring comment

Why is "Sign errors in the Vieta relations (e.g. $\alpha\beta\gamma = +\frac{d}{a}$)" a common mistake in Roots and Coefficients of polynomial equations?

Why it happens: Forgetting the alternating -,+,-,+ pattern; not accounting for a negative coefficient. How to avoid it: Memorise: sum = -b/a, product = (-1)^nconsta. If unsure, expand a(x-)(x-)·s to check. Source: 9231 Paper 1 mark schemes

Why is "Reading coefficients before rearranging into standard form" a common mistake in Roots and Coefficients of polynomial equations?

Why it happens: The equation is presented as e.g. 3x^2 = 5x - 2 and the student reads b = 5. How to avoid it: Move everything to one side as ax^n + ·s = 0 with descending powers first, THEN identify a, b, c, Source: 9231 Paper 1 Examiner Reports

Why is "Forgetting to divide by the leading coefficient $a$ when $a \neq 1$" a common mistake in Roots and Coefficients of polynomial equations?

Why it happens: Memorising the relations only for monic cubics ( = -b instead of -b/a). How to avoid it: Every relation has an 1/a in it. For 2x^3 - x^2 + , = 1/2, not 1. Source: 9231 Paper 1 Examiner Reports

Why is "Using $S_0 = 0$ instead of $S_0 = n$ in a power-sum recurrence" a common mistake in Roots and Coefficients of polynomial equations?

Why it happens: Assuming S_0 = ^0 = 0 = 0, forgetting that ^0 = 1. How to avoid it: Each ^0 = 1, so S_0 equals the number of roots (3 for a cubic, 4 for a quartic). Source: 9231 Paper 1 — power-sum questions

Why is "On a 'Show that' power-sum result, writing only the final relation" a common mistake in Roots and Coefficients of polynomial equations?

Why it happens: Treating a 'Show that' like a 'State'; the answer is given so students skip the derivation. How to avoid it: For 'Show that', every step is a mark: write ^3 = , multiply by ^n, then sum. The given answer earns nothing on its own. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Roots and Coefficients of polynomial equations

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Detailed Study Notes

Detailed notes on Roots of Polynomial Equations for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Roots and Coefficients of Polynomial Equations — Cambridge International AS & A Level Further Mathematics 9231 Paper 1 (2026-2027 syllabus)

Vieta's formulae link the roots of a polynomial to its coefficients — without ever solving the equation. Master the sum/product patterns for quadratics, cubics and quartics, then build symmetric functions like $\sum\alpha^2$ and $\sum\frac{1}{\alpha}$ .

At a glance

Vieta's formulae connect roots ( $\alpha, \beta, \gamma\ldots$ ) to coefficients ( $a, b, c\ldots$ ) — no need to find the roots.
Sign rule: signs alternate $-,+,-,+\ldots$ starting from $-\frac{b}{a}$ .
Symmetric functions ( $\sum\alpha^2$ , $\sum\frac{1}{\alpha}$ , $\sum\alpha^3$ ) are rebuilt from the basic sums/products using identities.
Not in MF19 — you MUST memorise every Vieta relation. This is the topic's biggest exam trap.
Core identity: $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.1a — Recall and use the relations between the roots and coefficients of polynomial equations of degree 2, 3 and 4.
1.1b — Evaluate symmetric functions of the roots (e.g. $\sum\alpha^2$ , $\sum\alpha^2\beta$ , $\sum\frac{1}{\alpha}$ ) using these relations.
1.1c — Recognise when a question requires roots-and-coefficients reasoning rather than solving the polynomial.

Why this works — the factor-form idea

A polynomial equals its leading coefficient times the product of (x − each root).

The key idea. Any polynomial can be written in two ways. Comparing them gives Vieta's formulae.

Take a quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$ . It must factorise as:

$a(x - \alpha)(x - \beta) = 0$

Expand the factor form: $a\left[x^2 - (\alpha + \beta)x + \alpha\beta\right] = ax^2 - a(\alpha+\beta)x + a\alpha\beta$

Now compare with $ax^2 + bx + c$ , coefficient by coefficient:

$x$ term: $-a(\alpha + \beta) = b \;\Rightarrow\; \alpha + \beta = -\dfrac{b}{a}$
constant: $a\alpha\beta = c \;\Rightarrow\; \alpha\beta = \dfrac{c}{a}$

That's it — no solving required. The same comparison for cubics and quartics gives every formula below. You don't need to re-derive it in the exam, but understanding why it works stops you mixing up the signs.

Cambridge tip. If you ever forget a sign, mentally expand $a(x-\alpha)(x-\beta)\cdots$ for two seconds — it is faster than guessing.

Polynomial = leading coefficient × product of $(x - \text{root})$ .
Expand and compare coefficients to read off the relations.
Re-deriving the quadratic case fixes any forgotten sign.

See the full worked example for roots and coefficients of polynomial equations →

The Vieta formulae — degree 2, 3 and 4

Alternating signs, starting from −b/a. Memorise the pattern, not 12 separate facts.

Write the polynomial in standard form with the highest power first, then read off using the alternating-sign pattern. ( $\sum$ means "sum of all distinct terms of that type".)

Quadratic $ax^2 + bx + c = 0$ , roots $\alpha, \beta$ : $\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$

Cubic $ax^3 + bx^2 + cx + d = 0$ , roots $\alpha, \beta, \gamma$ : $\sum\alpha = \alpha+\beta+\gamma = -\frac{b}{a}$ $\sum\alpha\beta = \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}$ $\alpha\beta\gamma = -\frac{d}{a}$

Quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ , roots $\alpha, \beta, \gamma, \delta$ : $\sum\alpha = -\frac{b}{a}, \quad \sum\alpha\beta = \frac{c}{a}, \quad \sum\alpha\beta\gamma = -\frac{d}{a}, \quad \alpha\beta\gamma\delta = \frac{e}{a}$

Spot the pattern (this is all you really memorise):

The signs alternate: $-,\,+,\,-,\,+$ as you go from $-\frac{b}{a}$ to the product of all roots.
"Sum of roots one at a time" $= -\frac{b}{a}$ (second coefficient over first, negated).
"Sum of products two at a time" $= +\frac{c}{a}$ .
"Product of all roots" $= (-1)^n \frac{\text{constant}}{a}$ .

The roots α, β, γ are where the curve cuts the x-axis. Vieta's formulae let you combine them (sum, product, etc.) without finding their values.

Cambridge tip. Always rewrite the equation so the leading coefficient sits on the $x^n$ term first. A sloppy " $3x^2 = 5x - 2$ " must become " $3x^2 - 5x + 2 = 0$ " before you read off $\alpha+\beta = \frac{5}{3}$ .

Signs alternate $-,+,-,+$ from $-\frac{b}{a}$ .
Sum of roots $= -\frac{b}{a}$ ; product $= (-1)^n\frac{\text{constant}}{a}$ .
Rearrange to standard form BEFORE reading coefficients.

See the full worked example for roots and coefficients of polynomial equations →

Building symmetric functions of the roots

Rewrite Σα², Σ1/α, Σα³ in terms of the basic Vieta sums and products.

Exam questions rarely ask for $\sum\alpha$ directly — they ask for a symmetric function such as $\sum\alpha^2$ or $\sum\frac{1}{\alpha}$ . The trick is always the same: express it using only the Vieta quantities you already know.

The four you must have at your fingertips (shown for a cubic with $\sum\alpha$ , $\sum\alpha\beta$ , $\alpha\beta\gamma$ ):

1. Sum of squares. Square the sum of roots and subtract the doubled pair-sum: $\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta$

2. Sum of reciprocals. Common denominator is the product of all roots: $\sum\frac{1}{\alpha} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}$

3. Sum of reciprocal pairs. $\sum\frac{1}{\alpha\beta} = \frac{\sum\alpha}{\alpha\beta\gamma}$

4. Sum of cubes (use the identity — it is not given in MF19): $\sum\alpha^3 = \left(\sum\alpha\right)^3 - 3\left(\sum\alpha\right)\left(\sum\alpha\beta\right) + 3\alpha\beta\gamma$

The universal strategy:

Write the target as a polynomial in the symmetric sums.
Substitute the numerical Vieta values.
Simplify carefully (watch signs).

Newton's method (sum of cubes, faster route). If $\alpha$ is a root of $x^3 + px^2 + qx + r = 0$ then $\alpha^3 = -p\alpha^2 - q\alpha - r$ . Summing over all roots: $\sum\alpha^3 = -p\sum\alpha^2 - q\sum\alpha - 3r$ This "substitute the root back into its own equation, then sum" idea generalises to $\sum\alpha^4$ , $\sum\alpha^5$ , … and is often the cleanest route for higher powers.

Cambridge tip. Examiner reports repeatedly note candidates who quote $\sum\alpha^2 = (\sum\alpha)^2$ — forgetting the $-2\sum\alpha\beta$ . Write the identity in full before substituting numbers.

$\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ .
$\sum\frac{1}{\alpha} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}$ (reciprocals over the product).
Higher powers: substitute the root into its own equation, then sum.

See the full worked example for roots and coefficients of polynomial equations →

Power sums and Newton's identities

A recurrence that grinds out Σαᵏ for any power k.

Let $S_k = \sum\alpha^k$ be the $k$ -th power sum of the roots. For a polynomial, the power sums satisfy a recurrence built from the Vieta coefficients.

For a cubic $x^3 + px^2 + qx + r = 0$ (note: leading coefficient $1$ , so $p = -\sum\alpha$ , $q = \sum\alpha\beta$ , $r = -\alpha\beta\gamma$ ):

Every root satisfies $\alpha^3 + p\alpha^2 + q\alpha + r = 0$ . Multiply by $\alpha^{k-3}$ and sum over the three roots: $S_k + pS_{k-1} + qS_{k-2} + rS_{k-3} = 0 \quad (k \geq 3)$

with starting values $S_0 = 3$ (three roots), $S_1 = -p$ , $S_2 = p^2 - 2q$ .

Why it matters. A question asking for $\sum\alpha^4$ or $\sum\alpha^5$ is designed to be done this way — expanding $(\alpha^2)^2$ etc. by brute force wastes time and invites errors. Set up the recurrence and march upward one power at a time.

Worked sketch. If $x^3 - 2x + 4 = 0$ then $p = 0$ , $q = -2$ , $r = 4$ :

$S_1 = 0$
$S_2 = 0^2 - 2(-2) = 4$
$S_3 = -p S_2 - q S_1 - rS_0 = 0 - 0 - 4(3) = -12$
$S_4 = -pS_3 - qS_2 - rS_1 = 0 -(-2)(4) - 4(0) = 8$

Cambridge tip. Make sure the polynomial is monic (leading coefficient $1$ ) before quoting $p, q, r$ — divide through first if not.

$S_k = \sum\alpha^k$ is the $k$ -th power sum.
Cubic recurrence: $S_k + pS_{k-1} + qS_{k-2} + rS_{k-3} = 0$ .
Start from $S_0 = n$ , $S_1 = -p$ , $S_2 = p^2 - 2q$ and climb.

See the full worked example for roots and coefficients of polynomial equations →

Quick recap

Vieta links roots to coefficients with alternating signs from $-\frac{b}{a}$ .
Cubic: $\sum\alpha = -\frac{b}{a}$ , $\sum\alpha\beta = \frac{c}{a}$ , $\alpha\beta\gamma = -\frac{d}{a}$ .
$\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ — never forget the $-2$ term.
$\sum\frac{1}{\alpha} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}$ .
Higher powers: substitute the root into its own equation, or use Newton's recurrence.
Vieta is NOT in MF19 — memorise it.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Quadratic, cubic, quartic Vieta formulae (alternating-sign pattern).
$\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ .
$\sum\frac{1}{\alpha} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}$ and $\sum\frac{1}{\alpha\beta} = \frac{\sum\alpha}{\alpha\beta\gamma}$ .
$\sum\alpha^3 = (\sum\alpha)^3 - 3(\sum\alpha)(\sum\alpha\beta) + 3\alpha\beta\gamma$ .
Newton's recurrence for power sums $S_k$ (substitute root into its own equation).

How it’s examined

Topic 1.1 appears in virtually every Paper 1 (6-10 marks). Part (a) typically asks for a symmetric function such as $\sum\alpha^2$ or $\sum\frac{1}{\alpha}$ ; part (b) flows into 1.2 (finding a related-roots equation). Marks are lost almost entirely through (i) sign errors in the Vieta relations and (ii) quoting $\sum\alpha^2=(\sum\alpha)^2$ without the $-2\sum\alpha\beta$ . Because Vieta is not in MF19, secure recall is the single highest-value preparation.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11 & 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Roots and Coefficients of polynomial equations

Step-by-step solutions to past-paper-style questions on roots and coefficients of polynomial equations, written exactly the way a tutor would explain them at the board.

Question type:

1Sum and product of roots of a quadratic
Getting startedDirect calculation• quadratic, vieta
Question
The equation $2x^2 - 7x + 3 = 0$ has roots $\alpha$ and $\beta$ . Write down $\alpha + \beta$ and $\alpha\beta$ .
Step-by-step solution
1. Step 1
  Identify $a, b, c$ from $ax^2 + bx + c = 0$ . Here $a = 2$ , $b = -7$ , $c = 3$ .
2. Step 2
  Sum of roots $= -\dfrac{b}{a}$ .
  $\alpha + \beta = -\frac{-7}{2} = \frac{7}{2}$
3. Step 3
  Product of roots $= \dfrac{c}{a}$ .
  $\alpha\beta = \frac{3}{2}$
Answer
$\alpha + \beta = \frac{7}{2}$ , $\quad\alpha\beta = \frac{3}{2}$ .
Examiner tip
Note $b = -7$ , so $-\frac{b}{a} = +\frac{7}{2}$ . The double-negative is where beginners slip.
2Reading off the three relations for a cubic
Getting startedDirect calculation• cubic, vieta
Question
The cubic $x^3 - 4x^2 + x + 6 = 0$ has roots $\alpha, \beta, \gamma$ . State $\sum\alpha$ , $\sum\alpha\beta$ and $\alpha\beta\gamma$ .
Step-by-step solution
1. Step 1
  Coefficients of $ax^3 + bx^2 + cx + d$ : $a = 1$ , $b = -4$ , $c = 1$ , $d = 6$ .
2. Step 2
  Apply the alternating pattern $-\frac{b}{a}, +\frac{c}{a}, -\frac{d}{a}$ .
  $\sum\alpha = -\frac{-4}{1} = 4,\quad \sum\alpha\beta = \frac{1}{1} = 1,\quad \alpha\beta\gamma = -\frac{6}{1} = -6$
Answer
$\sum\alpha = 4$ , $\quad\sum\alpha\beta = 1$ , $\quad\alpha\beta\gamma = -6$ .
Examiner tip
The product is $-\frac{d}{a}$ for a cubic — the odd degree forces the minus sign.
3Sum of squares of the roots
Building confidenceMulti-step problem• symmetric-function, cubic
Question
The roots of $x^3 + 2x^2 - 5x + 1 = 0$ are $\alpha, \beta, \gamma$ . Find $\alpha^2 + \beta^2 + \gamma^2$ .
Step-by-step solution
1. Step 1
  Vieta values. $\sum\alpha = -2$ , $\sum\alpha\beta = -5$ .
2. Step 2
  Quote the identity in full before substituting.
  $\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta$
3. Step 3
  Substitute the values.
  $\sum\alpha^2 = (-2)^2 - 2(-5) = 4 + 10 = 14$
Answer
$\alpha^2 + \beta^2 + \gamma^2 = 14$ .
Examiner tip
Writing the identity out first guarantees the $-2\sum\alpha\beta$ term is not dropped — the single most common error in this topic.
4Sum of reciprocals of the roots
Building confidenceMulti-step problem• reciprocal, symmetric-function
Question
The roots of $2x^3 - x^2 + 4x - 3 = 0$ are $\alpha, \beta, \gamma$ . Find $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ .
Step-by-step solution
1. Step 1
  Vieta values (divide by $a = 2$ as you read off): $\sum\alpha = \frac{1}{2}$ , $\sum\alpha\beta = \frac{4}{2} = 2$ , $\alpha\beta\gamma = \frac{3}{2}$ .
2. Step 2
  Combine over a common denominator — the product of all roots.
  $\sum\frac{1}{\alpha} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}$
3. Step 3
  Substitute.
  $\sum\frac{1}{\alpha} = \frac{2}{\,3/2\,} = \frac{4}{3}$
Answer
$\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{4}{3}$ .
Examiner tip
Watch the product: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{-3}{2} = \frac{3}{2}$ . A sign slip here flips the whole answer.
5Sum of cubes via the substitution method
StretchMulti-step problem• sum-of-cubes, newton
Question
The roots of $x^3 - 3x + 1 = 0$ are $\alpha, \beta, \gamma$ . Find $\alpha^3 + \beta^3 + \gamma^3$ .
Step-by-step solution
1. Step 1
  Each root satisfies its own equation: $\alpha^3 - 3\alpha + 1 = 0$ , so $\alpha^3 = 3\alpha - 1$ (same for $\beta, \gamma$ ).
2. Step 2
  Sum over all three roots.
  $\sum\alpha^3 = 3\sum\alpha - 3$
3. Step 3
  Vieta: here $b = 0$ so $\sum\alpha = 0$ .
  $\sum\alpha^3 = 3(0) - 3 = -3$
Answer
$\alpha^3 + \beta^3 + \gamma^3 = -3$ .
Examiner tip
The 'substitute the root into its own equation then sum' technique is far cleaner than the cubic identity here, and generalises to $\sum\alpha^4$ , $\sum\alpha^5$ , …
6A symmetric function of a quartic's roots
StretchMulti-step problem• quartic, symmetric-function
Question
The roots of $x^4 - 2x^3 + 3x^2 - x + 5 = 0$ are $\alpha, \beta, \gamma, \delta$ . Find $\sum\alpha^2$ (the sum of the squares of the roots).
Step-by-step solution
1. Step 1
  Quartic Vieta: $\sum\alpha = -\frac{b}{a} = 2$ and $\sum\alpha\beta = \frac{c}{a} = 3$ .
2. Step 2
  Same identity works for any degree: $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ , where $\sum\alpha\beta$ is the sum of ALL pair-products.
  $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$
3. Step 3
  Substitute.
  $\sum\alpha^2 = (2)^2 - 2(3) = 4 - 6 = -2$
Answer
$\sum\alpha^2 = -2$ .
Examiner tip
A negative $\sum\alpha^2$ is fine — the roots are complex (the quartic has no real roots). The identity is degree-independent; only the number of pair-products changes.

Model Answers — Roots and Coefficients of polynomial equations

High-scoring sample answers for roots and coefficients of polynomial equations on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
The roots of $3x^2 + 5x - 2 = 0$ are $\alpha$ and $\beta$ . Find the value of $\alpha^2 + \beta^2$ .
Model answer
Vieta's formulae for $3x^2 + 5x - 2 = 0$ : $\alpha + \beta = -\frac{5}{3}, \qquad \alpha\beta = -\frac{2}{3}$

Identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Substitute: $\alpha^2 + \beta^2 = \left(-\frac{5}{3}\right)^2 - 2\left(-\frac{2}{3}\right) = \frac{25}{9} + \frac{4}{3} = \frac{25}{9} + \frac{12}{9} = \frac{37}{9}$
Why this scores
M1 for both Vieta values, M1 for the correct identity, M1 for substitution, A1 for $\frac{37}{9}$ . Showing the identity explicitly secures the method marks even if the arithmetic slips.
Question 2
5 marks
The roots of $x^3 + 4x^2 - 7x + 2 = 0$ are $\alpha, \beta, \gamma$ . (a) Write down $\sum\alpha$ , $\sum\alpha\beta$ and $\alpha\beta\gamma$ . (b) Hence find $\sum\alpha^2$ .
Model answer
(a) Comparing with $ax^3 + bx^2 + cx + d = 0$ ( $a=1, b=4, c=-7, d=2$ ): $\sum\alpha = -\frac{b}{a} = -4, \qquad \sum\alpha\beta = \frac{c}{a} = -7, \qquad \alpha\beta\gamma = -\frac{d}{a} = -2$

(b) Using $\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta$ : $\sum\alpha^2 = (-4)^2 - 2(-7) = 16 + 14 = 30$
Why this scores
Part (a): B1 B1 B1 for the three correct relations (signs matter). Part (b): M1 for the identity quoted with the $-2\sum\alpha\beta$ term, A1 for $30$ . 'Hence' signals you must use part (a), not re-solve.
Question 3
6 marks
The roots of $2x^3 - 3x^2 + x - 4 = 0$ are $\alpha, \beta, \gamma$ . Find (a) $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}$ and (b) $\dfrac{1}{\alpha\beta}+\dfrac{1}{\beta\gamma}+\dfrac{1}{\gamma\alpha}$ .
Model answer
Vieta's formulae (with $a = 2$ ): $\sum\alpha = \frac{3}{2}, \qquad \sum\alpha\beta = \frac{1}{2}, \qquad \alpha\beta\gamma = -\frac{-4}{2} = 2$

(a) Sum of reciprocals over the product of all roots: $\sum\frac{1}{\alpha} = \frac{\sum\alpha\beta}{\alpha\beta\gamma} = \frac{1/2}{2} = \frac{1}{4}$

(b) Sum of reciprocal pairs: $\sum\frac{1}{\alpha\beta} = \frac{\sum\alpha}{\alpha\beta\gamma} = \frac{3/2}{2} = \frac{3}{4}$
Why this scores
M1 A1 for each part, plus B1 for a correct product $\alpha\beta\gamma = 2$ . The structure $\frac{\text{pair-sum}}{\text{product}}$ and $\frac{\text{root-sum}}{\text{product}}$ should be instant recall.
Question 4
6 marks
The roots of $x^3 - 2x^2 + 4 = 0$ are $\alpha, \beta, \gamma$ . Find $\alpha^3 + \beta^3 + \gamma^3$ .
Model answer
Vieta's formulae ( $a=1, b=-2, c=0, d=4$ ): $\sum\alpha = 2, \qquad \sum\alpha\beta = 0, \qquad \alpha\beta\gamma = -4$

Method — substitute each root into its own equation. Since $\alpha^3 - 2\alpha^2 + 4 = 0$ : $\alpha^3 = 2\alpha^2 - 4 \quad\Rightarrow\quad \sum\alpha^3 = 2\sum\alpha^2 - 12$

Find $\sum\alpha^2$ : $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = 2^2 - 0 = 4$

Therefore: $\sum\alpha^3 = 2(4) - 12 = -4$
Why this scores
M1 for the substitution $\alpha^3 = 2\alpha^2 - 4$ summed over roots, M1 A1 for $\sum\alpha^2 = 4$ , A1 for $-4$ . The note ' $-12$ ' comes from $3 \times 4$ constant terms — a frequent slip is writing $-4$ instead of $-12$ .
Question 5
7 marks
The roots of $x^4 + 2x^3 - x^2 + 3x - 1 = 0$ are $\alpha, \beta, \gamma, \delta$ . Find (a) $\sum\alpha^2$ and (b) $\sum\dfrac{1}{\alpha}$ .
Model answer
Quartic Vieta ( $a=1, b=2, c=-1, d=3, e=-1$ ): $\sum\alpha = -2, \quad \sum\alpha\beta = -1, \quad \sum\alpha\beta\gamma = -3, \quad \alpha\beta\gamma\delta = -1$

(a) $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ : $\sum\alpha^2 = (-2)^2 - 2(-1) = 4 + 2 = 6$

(b) For a quartic, the sum of reciprocals is the sum of the triple-products over the product of all four roots: $\sum\frac{1}{\alpha} = \frac{\beta\gamma\delta + \alpha\gamma\delta + \alpha\beta\delta + \alpha\beta\gamma}{\alpha\beta\gamma\delta} = \frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta} = \frac{-3}{-1} = 3$
Why this scores
B1 for all four Vieta relations; (a) M1 A1; (b) M1 for recognising $\sum\frac1\alpha = \frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}$ , A1 for $3$ . The reciprocal rule generalises: numerator is the 'one-degree-below-product' symmetric sum.
Question 6
8 marks
The roots of $x^3 + x - 3 = 0$ are $\alpha, \beta, \gamma$ . Let $S_n = \alpha^n + \beta^n + \gamma^n$ . (a) Show that $S_{n+3} = -S_{n+1} + 3S_n$ . (b) Given $S_1 = 0$ and $S_2 = -2$ , find $S_4$ .
Model answer
(a) Each root satisfies its own equation: $\alpha^3 + \alpha - 3 = 0$ , i.e. $\alpha^3 = -\alpha + 3$ .

Multiply by $\alpha^n$ : $\alpha^{n+3} = -\alpha^{n+1} + 3\alpha^n$

The identical relation holds for $\beta$ and $\gamma$ . Summing over all three roots: $S_{n+3} = -S_{n+1} + 3S_n \qquad \text{(as required)}$

(b) We need $S_4$ . First find $S_3$ using $n = 0$ (with $S_0 = 3$ , since there are three roots): $S_3 = -S_1 + 3S_0 = -0 + 3(3) = 9$

Now $n = 1$ : $S_4 = -S_2 + 3S_1 = -(-2) + 3(0) = 2$
Why this scores
Part (a): M1 for $\alpha^3 = -\alpha+3$ , M1 for multiplying by $\alpha^n$ , A1 for summing to the stated result — this is a 'Show that', so every step must appear. Part (b): M1 for $S_3$ using $S_0=3$ , A1; M1 A1 for $S_4 = 2$ . Forgetting $S_0 = 3$ (number of roots) is the classic trap.

Key Formulae — Roots and Coefficients of polynomial equations

The formulae you need to memorise for roots and coefficients of polynomial equations on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Quadratic — roots & coefficients
$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$
$a, b, c$
coefficients of $ax^2 + bx + c = 0$
$\alpha, \beta$
the two roots
When to use
Any quadratic where you need the sum or product of roots without solving.
Cubic — roots & coefficients
$\sum\alpha = -\frac{b}{a}, \quad \sum\alpha\beta = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}$
$a, b, c, d$
coefficients of $ax^3+bx^2+cx+d=0$
When to use
Cubic equations — the workhorse case in Paper 1.
Quartic — roots & coefficients
$\sum\alpha = -\frac{b}{a},\; \sum\alpha\beta = \frac{c}{a},\; \sum\alpha\beta\gamma = -\frac{d}{a},\; \alpha\beta\gamma\delta = \frac{e}{a}$
$a..e$
coefficients of $ax^4+bx^3+cx^2+dx+e=0$
When to use
Quartic equations. Signs alternate $-,+,-,+$ .
Sum of squares of roots
$\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta$
When to use
The most-used symmetric-function identity. Works for any degree — $\sum\alpha\beta$ is the sum of all pair-products.
Sum of reciprocals
$\sum\frac{1}{\alpha} = \frac{\sum\alpha\beta}{\alpha\beta\gamma} \;\;(\text{cubic})$
When to use
Put reciprocals over the product of all roots; numerator is the next symmetric sum down.
Sum of cubes (cubic)
$\sum\alpha^3 = \left(\sum\alpha\right)^3 - 3\left(\sum\alpha\right)\left(\sum\alpha\beta\right) + 3\alpha\beta\gamma$
When to use
Sum of cubes of three roots. Often quicker: substitute each root into its own equation, then sum.

Key Definitions and Keywords — Roots and Coefficients of polynomial equations

Definitions to memorise and the exact keywords mark schemes credit for roots and coefficients of polynomial equations answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Vieta's formulae
Examiner keyword
The relations between the roots of a polynomial and its coefficients, obtained by comparing the expanded factor form with the standard form. NOT given in MF19.
Symmetric function of the roots
Examiner keyword
An expression in the roots whose value is unchanged when the roots are permuted, e.g. $\sum\alpha^2$ , $\sum\frac{1}{\alpha}$ , $\alpha\beta\gamma$ . Always expressible via Vieta's formulae.
Power sum $S_k$
Examiner keyword
$S_k = \sum\alpha^k = \alpha^k + \beta^k + \cdots$ , the sum of the $k$ -th powers of all roots. Note $S_0 = n$ (the number of roots).
$\sum$ notation for roots
Examiner keyword
$\sum\alpha$ means the sum of all roots; $\sum\alpha\beta$ means the sum of all distinct products of roots taken two at a time, and so on.
Newton's identities
A recurrence linking power sums $S_k$ to the polynomial's coefficients, obtained by substituting a root into its own equation and summing.
Monic polynomial
A polynomial whose leading coefficient is $1$ . Newton's recurrence is stated for monic polynomials, so divide through by $a$ first if needed.

Common Mistakes and Misconceptions — Roots and Coefficients of polynomial equations

The traps other students keep falling into on roots and coefficients of polynomial equations questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Writing $\sum\alpha^2 = (\sum\alpha)^2$ , dropping the $-2\sum\alpha\beta$ term
9231 Paper 1 Examiner Reports — recurring comment
Why it happens
Pattern-matching to $(a+b)^2$ without expanding; rushing straight to substitution.
How to avoid it
Always write the identity $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ in full BEFORE substituting any numbers.
✕Sign errors in the Vieta relations (e.g. $\alpha\beta\gamma = +\frac{d}{a}$ )
9231 Paper 1 mark schemes
Why it happens
Forgetting the alternating $-,+,-,+$ pattern; not accounting for a negative coefficient.
How to avoid it
Memorise: sum $= -\frac{b}{a}$ , product $= (-1)^n\frac{\text{const}}{a}$ . If unsure, expand $a(x-\alpha)(x-\beta)\cdots$ to check.
✕Reading coefficients before rearranging into standard form
9231 Paper 1 Examiner Reports
Why it happens
The equation is presented as e.g. $3x^2 = 5x - 2$ and the student reads $b = 5$ .
How to avoid it
Move everything to one side as $ax^n + \cdots = 0$ with descending powers first, THEN identify $a, b, c, \ldots$
✕Forgetting to divide by the leading coefficient $a$ when $a \neq 1$
9231 Paper 1 Examiner Reports
Why it happens
Memorising the relations only for monic cubics ( $\sum\alpha = -b$ instead of $-\frac{b}{a}$ ).
How to avoid it
Every relation has an $\frac{1}{a}$ in it. For $2x^3 - x^2 + \ldots$ , $\sum\alpha = \frac{1}{2}$ , not $1$ .
✕Using $S_0 = 0$ instead of $S_0 = n$ in a power-sum recurrence
9231 Paper 1 — power-sum questions
Why it happens
Assuming $S_0 = \sum\alpha^0 = \sum 0 = 0$ , forgetting that $\alpha^0 = 1$ .
How to avoid it
Each $\alpha^0 = 1$ , so $S_0$ equals the number of roots ( $3$ for a cubic, $4$ for a quartic).
✕On a 'Show that' power-sum result, writing only the final relation
9231 Paper 1 Examiner Reports
Why it happens
Treating a 'Show that' like a 'State'; the answer is given so students skip the derivation.
How to avoid it
For 'Show that', every step is a mark: write $\alpha^3 = \ldots$ , multiply by $\alpha^n$ , then sum. The given answer earns nothing on its own.

Roots and Coefficients of polynomial equations

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Sum and product of roots of a quadratic

2Reading off the three relations for a cubic

3Sum of squares of the roots

4Sum of reciprocals of the roots

5Sum of cubes via the substitution method

6A symmetric function of a quartic's roots

Quadratic — roots & coefficients

Cubic — roots & coefficients

Quartic — roots & coefficients

Sum of squares of roots

Sum of reciprocals

Sum of cubes (cubic)

Vieta's formulae

Symmetric function of the roots

Power sum SkS_kSk​

∑\sum∑ notation for roots

Newton's identities

Monic polynomial

✕Writing ∑α2=(∑α)2\sum\alpha^2 = (\sum\alpha)^2∑α2=(∑α)2, dropping the −2∑αβ-2\sum\alpha\beta−2∑αβ term

✕Sign errors in the Vieta relations (e.g. αβγ=+da\alpha\beta\gamma = +\frac{d}{a}αβγ=+ad​)

✕Reading coefficients before rearranging into standard form

✕Forgetting to divide by the leading coefficient aaa when a≠1a \neq 1a=1

✕Using S0=0S_0 = 0S0​=0 instead of S0=nS_0 = nS0​=n in a power-sum recurrence

✕On a 'Show that' power-sum result, writing only the final relation

Power sum $S_k$

$\sum$ notation for roots

✕Writing $\sum\alpha^2 = (\sum\alpha)^2$ , dropping the $-2\sum\alpha\beta$ term

✕Sign errors in the Vieta relations (e.g. $\alpha\beta\gamma = +\frac{d}{a}$ )

✕Forgetting to divide by the leading coefficient $a$ when $a \neq 1$

✕Using $S_0 = 0$ instead of $S_0 = n$ in a power-sum recurrence