What are the factors of polynomials in the context of Cambridge IGCSE Additional Mathematics?

In Cambridge IGCSE Additional Mathematics, factors of polynomials refer to expressions that can be multiplied together to obtain the original polynomial. For example, if a polynomial can be expressed as the product of two or more polynomials, those polynomials are considered its factors. Understanding how to factor polynomials is crucial for solving polynomial equations and simplifying expressions.

How do I determine the factors of a polynomial using the factor theorem?

The factor theorem is a key concept in determining the factors of a polynomial. It states that if a polynomial f(x) has a factor (x - c), then f(c) = 0. To use this theorem, substitute potential roots into the polynomial to find values that make the polynomial equal zero. These values correspond to factors of the form (x - c). This method is particularly useful for finding linear factors.

What is the role of synthetic division in finding polynomial factors?

Synthetic division is a simplified form of polynomial division that is used to divide a polynomial by a linear factor of the form (x - c). It is a quick and efficient method to determine whether (x - c) is a factor of the polynomial. If the remainder is zero after performing synthetic division, then (x - c) is a factor. This technique is often used in conjunction with the factor theorem to factorize polynomials completely.

Can you explain how to factor quadratic polynomials?

Factoring quadratic polynomials involves expressing them in the form (ax + b)(cx + d) = 0. For a quadratic polynomial ax^2 + bx + c, you can use methods such as splitting the middle term, using the quadratic formula, or applying the difference of squares if applicable. The goal is to rewrite the quadratic as a product of two binomials, which can then be solved to find the roots of the equation.

What are some common techniques for factoring higher-degree polynomials?

For higher-degree polynomials, several techniques can be employed. These include grouping terms to factor by grouping, using the factor theorem and synthetic division to find linear factors, and applying special formulas such as the sum or difference of cubes. Additionally, recognizing patterns like perfect square trinomials or using substitution can simplify the process. Mastery of these techniques is essential for tackling complex polynomial equations in the Cambridge IGCSE Additional Mathematics curriculum.

Why is "For factor $(x + 2)$, testing $f(2)$ instead of $f(-2)$" a common mistake in Factors of Polynomials?

Why it happens: Sign confusion. How to avoid it: (x - a) test f(a). (x + 2) = (x - (-2)) test f(-2). Always match the sign with the bracket. Source: 0606 Examiner Reports 2022-2024

Why is "Stopping after finding one root" a common mistake in Factors of Polynomials?

Why it happens: Question seems answered. How to avoid it: Cubics have up to 3 roots, quartics up to 4. Find one, divide, solve the resulting quadratic.

Why is "Arithmetic errors in polynomial long division" a common mistake in Factors of Polynomials?

Why it happens: Many small steps. How to avoid it: Check by multiplying back: divisor × quotient + remainder = original.

Factors of PolynominalsCambridge IGCSE Additional Mathematics (0606)

Factors of Polynomials

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Factors of Polynomials Study Notes — Cambridge IGCSE Additional Mathematics 0606 (2025-2027 syllabus)

Factor theorem and remainder theorem. Polynomial long division. Solving cubic and quartic equations by spotting a rational root and dividing.

What you’ll learn

Mapped to the Cambridge IGCSE 0606 syllabus (2025-2027).

1.5.1 — State and apply the factor theorem.
1.5.2 — State and apply the remainder theorem.
1.5.3 — Use polynomial long division.
1.5.4 — Solve cubic and higher polynomial equations.

Factor and remainder theorems

Two related theorems for polynomials.

Factor theorem. $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ .

Remainder theorem. When $f(x)$ is divided by $(x - a)$ , the REMAINDER equals $f(a)$ .

The factor theorem is a special case of the remainder theorem (where the remainder is $0$ ).

Why useful. Lets you find roots without actually dividing — just SUBSTITUTE.

How to use to find $k$ . If the question says ' $(x - 3)$ is a factor of $f(x)$ ', set $f(3) = 0$ and solve for the unknown coefficient.

Cambridge tip. Match the sign carefully: $(x - 3) \Rightarrow f(3)$ , $(x + 2) \Rightarrow f(-2)$ .

Factor: $(x - a)$ factor iff $f(a) = 0$ .
Remainder: dividing by $(x - a)$ gives remainder $f(a)$ .
Sign: $(x + a) \Rightarrow f(-a)$ .

See the full worked example for factors of polynomials →

Solving cubics

Find a root, divide, solve the quadratic.

Procedure.

Step 1: Find a root. Try small integer values: $\pm 1, \pm 2, \pm 3$ , plus rational candidates of the form $\pm \frac{p}{q}$ where $p$ divides the constant term and $q$ divides the leading coefficient.

Step 2: Divide. Once $f(a) = 0$ is confirmed, polynomial long division of $f(x)$ by $(x - a)$ gives a quadratic.

Step 3: Solve the quadratic. Factorise or use the quadratic formula.

Worked walkthrough. $f(x) = 2x^3 - 3x^2 - 11x + 6$ .

Test $x = 3$ : $54 - 27 - 33 + 6 = 0$ . ✓
Divide by $(x - 3)$ : get $2x^2 + 3x - 2$ .
Factorise: $(2x - 1)(x + 2)$ .
Roots: $x = 3, \frac{1}{2}, -2$ .

Cambridge tip. Don't stop after finding the first root — find all three.

Test integer/rational candidates.
Divide to reduce to a quadratic.
Solve the quadratic for remaining roots.

See the full worked example for factors of polynomials →

Finding unknown coefficients

Use given factors to set up equations.

Common exam question: 'Given that $(x + 2)$ is a factor of $f(x) = x^3 + kx^2 - 4x - 12$ , find $k$ .'

Procedure.

Apply factor theorem: $f(-2) = 0$ .
Substitute: $(-2)^3 + k(-2)^2 - 4(-2) - 12 = 0$ .
Simplify: $-8 + 4k + 8 - 12 = 0$ , so $4k = 12$ , $k = 3$ .

Two unknowns. If the polynomial has TWO unknown coefficients, you'll need TWO conditions (e.g. two given factors, or one factor plus a remainder).

Cambridge tip. Read the question carefully — distinguish 'is a factor' (set $f(a) = 0$ ) from 'leaves a remainder of $r$ ' (set $f(a) = r$ ).

$(x - a)$ factor → $f(a) = 0$ .
Remainder $r$ → $f(a) = r$ .
Two unknowns need two conditions.

See the full worked example for factors of polynomials →

How it’s examined

Polynomials appear on most Paper 1s. Most-tested: factorise a cubic given a root (5-7 marks), find unknown coefficient using factor/remainder theorem (4-5 marks), apply remainder theorem to get a system in two unknowns (5 marks).

Sources: Cambridge IGCSE Additional Mathematics 0606 syllabus (2025-2027); 0606 Examiner Reports 2022-2024; 0606/12 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-09.

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Step-by-step worked examples — Factors of Polynomials

Step-by-step solutions to past-paper-style questions on factors of polynomials, written exactly the way a tutor would explain them at the board.

1Use the remainder theorem (5 marks)
Extended• Adapted from 0606/12 May/Jun 2024 Q6• remainder-theorem
Question
When $f(x) = 2x^3 + ax^2 + bx - 3$ is divided by $(x - 1)$ the remainder is $-2$ , and by $(x + 2)$ the remainder is $-15$ . Find the values of $a$ and $b$ . (5 marks)
Step-by-step solution
1. Step 1
  Apply remainder theorem (1 mark). $f(1) = -2$ and $f(-2) = -15$ .
2. Step 2
  $f(1) = -2$ (2 marks). $2(1)^3 + a(1)^2 + b(1) - 3 = -2$ , so $2 + a + b - 3 = -2$ , giving $a + b = -1$ . (Equation 1)
3. Step 3
  $f(-2) = -15$ (1 mark). $2(-2)^3 + a(-2)^2 + b(-2) - 3 = -15$ , so $-16 + 4a - 2b - 3 = -15$ , giving $4a - 2b = 4$ , simplify to $2a - b = 2$ . (Equation 2)
4. Step 4
  Solve simultaneously (1 mark). Add (1) + (2): $3a = 1$ , $a = \frac{1}{3}$ . Then $b = -1 - \frac{1}{3} = -\frac{4}{3}$ .
Answer
$a = \dfrac{1}{3}$ , $b = -\dfrac{4}{3}$ .
2Factorise a cubic (6 marks)
Extended• factor-theorem
Question
Factorise $f(x) = 2x^3 - 3x^2 - 11x + 6$ completely. (6 marks)
Step-by-step solution
1. Step 1
  Test integer roots (2 marks). Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$ . Try $x = 3$ : $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ . ✓
2. Step 2
  $(x - 3)$ is a factor (1 mark). Divide $f(x)$ by $(x - 3)$ .
3. Step 3
  Polynomial long division (2 marks). $2x^3 - 3x^2 - 11x + 6 \div (x - 3) = 2x^2 + 3x - 2$ .
4. Step 4
  Factorise the quadratic (1 mark). $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ .
Answer
$f(x) = (x - 3)(2x - 1)(x + 2)$ . Roots: $x = 3, \frac{1}{2}, -2$ .
3Find unknown coefficient using factor theorem (4 marks)
Extended• factor-theorem
Question
If $(x + 2)$ is a factor of $g(x) = x^3 + kx^2 - 4x - 12$ , find $k$ . (4 marks)
Step-by-step solution
1. Step 1
  Apply factor theorem (2 marks). Since $(x + 2)$ is a factor, $g(-2) = 0$ .
2. Step 2
  Substitute (2 marks). $(-2)^3 + k(-2)^2 - 4(-2) - 12 = 0$ . So $-8 + 4k + 8 - 12 = 0$ , $4k = 12$ , $k = 3$ .
Answer
$k = 3$ .

Key Formulae — Factors of Polynomials

The formulae you need to memorise for factors of polynomials on the Cambridge IGCSE 0606 paper, with every variable defined in plain English and a note on when to use it.

Remainder theorem
$f(x) = (x - a)q(x) + f(a)$
$f(x)$
Polynomial being divided
$a$
Number giving the divisor (x - a)
$q(x)$
Quotient polynomial
$f(a)$
The remainder when f is divided by (x - a)
When to use
Finding the remainder when f(x) is divided by (x - a) without doing long division.
Factor theorem
$(x - a) \text{ is a factor of } f(x) \iff f(a) = 0$
$f(x)$
Polynomial
$a$
Candidate root
When to use
Identifying factors and roots of polynomials.
Example
If f(2) = 0 then (x - 2) is a factor of f(x).

Key Definitions and Keywords — Factors of Polynomials

Definitions to memorise and the exact keywords mark schemes credit for factors of polynomials answers — sharpened from recent examiner reports for the 2026 0606 sitting.

Polynomial
Examiner keyword
An expression $a_n x^n + a_{n-1}x^{n-1} + \ldots + a_0$ with non-negative integer powers.
Degree
The highest power of $x$ . Cubic = 3, quartic = 4, etc.
Root (zero)
Examiner keyword
A value of $x$ for which $f(x) = 0$ .
Factor theorem
Examiner keyword
$(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ .
Remainder theorem
Examiner keyword
When $f(x)$ is divided by $(x - a)$ , the remainder is $f(a)$ .

Common Mistakes and Misconceptions — Factors of Polynomials

The traps other students keep falling into on factors of polynomials questions — taken from recent Cambridge IGCSE 0606 examiner reports and mark schemes — and how to avoid them.

✕For factor $(x + 2)$ , testing $f(2)$ instead of $f(-2)$
0606 Examiner Reports 2022-2024
Why it happens
Sign confusion.
How to avoid it
$(x - a) \Rightarrow$ test $f(a)$ . $(x + 2) = (x - (-2)) \Rightarrow$ test $f(-2)$ . Always match the sign with the bracket.
✕Stopping after finding one root
Why it happens
Question seems answered.
How to avoid it
Cubics have up to 3 roots, quartics up to 4. Find one, divide, solve the resulting quadratic.
✕Arithmetic errors in polynomial long division
Why it happens
Many small steps.
How to avoid it
Check by multiplying back: divisor × quotient + remainder = original.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Factors of Polynomials — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Factors of Polynomials

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Use the remainder theorem (5 marks)

2Factorise a cubic (6 marks)

3Find unknown coefficient using factor theorem (4 marks)

Remainder theorem

Factor theorem

Polynomial

Degree

Root (zero)

Factor theorem

Remainder theorem

✕For factor (x+2)(x + 2)(x+2), testing f(2)f(2)f(2) instead of f(−2)f(-2)f(−2)

✕Stopping after finding one root

✕Arithmetic errors in polynomial long division

Practice questions

Past paper style quiz

Assess your understanding

Factors of Polynomials — frequently asked questions

✕For factor $(x + 2)$ , testing $f(2)$ instead of $f(-2)$