# Factor & Remainder Theorems and Cubic Equations

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u3-factor-remainder-theorems-and-cubic/

This guide teaches you to apply the remainder and factor theorems for cubic polynomials, factorise cubics fully, and solve cubic equations, aligned exactly to the CIE IGCSE Additional Mathematics 0606 2025–2027 syllabus.

**Prerequisites:** Expanding and simplifying algebraic expressions; Solving quadratic equations by factorisation and formula

## Learning objectives

- Apply the remainder theorem to find the remainder when a polynomial is divided by a linear factor
- Use the factor theorem to identify linear factors of cubic polynomials
- Factorise cubics fully into linear factors using inspection or polynomial long division
- Solve cubic equations by factorisation

## The Remainder Theorem

**Remainder Theorem** — For any polynomial $f(x)$, the remainder when $f(x)$ is divided by the linear term $(mx + c)$ is equal to $f\left(-\frac{c}{m}\right)$. For a divisor $(x - a)$, the remainder simplifies to $f(a)$.

*Example:* For $f(x) = x^3 + 2x - 3$, the remainder when divided by $(x - 1)$ is $f(1) = 1 + 2 - 3 = 0$.

The remainder theorem eliminates the need for time-consuming polynomial long division when you only need to find the remainder of a division operation. It is tested frequently in both calculator and non-calculator papers.

**Worked example:** Find the remainder when $f(x) = 2x^3 - 5x^2 + x - 4$ is divided by $(2x + 1)$.

1. 1. Solve the divisor equal to 0 to find the substitution value: $2x + 1 = 0 \implies x = -\frac{1}{2}$
2. 2. Substitute $x = -\frac{1}{2}$ into $f(x)$:
3. $$f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 5\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 4$$
4. 3. Simplify the expression:
5. $$= 2\left(-\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) - \frac{1}{2} - 4 = -\frac{1}{4} - \frac{5}{4} - \frac{2}{4} - \frac{16}{4} = -6$$
6. 4. The remainder is $-6$.

> **Exam tip:** Double check the sign of your substitution value: sign errors are the most common mistake on remainder theorem questions.

*Calculator:* allowed

## The Factor Theorem

**Factor Theorem** — A special case of the remainder theorem: if the remainder when $f(x)$ is divided by $(x - a)$ is 0, then $(x - a)$ is a factor of $f(x)$. The reverse is also true: if $(x - a)$ is a factor of $f(x)$, then $f(a) = 0$.

*Example:* For $f(x) = x^3 - 3x^2 - 4x + 12$, $f(2) = 0$, so $(x - 2)$ is a factor of $f(x)$.

For cubics with integer coefficients, the only possible integer roots are factors of the constant term, so test these values first to quickly find linear factors. When the leading coefficient $a \neq 1$, a root may be rational rather than a whole number: any rational root has the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient, so you test candidates $\frac{p}{q}$ (for example, a cubic beginning $36x^3$ can have a root such as $x = -\frac{1}{4}$, since $4$ divides $36$). In practice, 0606 questions of this type usually hand you a factor directly (for example, 'given that $(2x + 1)$ is a factor') or specify the coefficients, so once you have it you can divide by the stated factor immediately rather than searching for the root yourself.

**Worked example:** Show that $(x + 3)$ is a factor of $f(x) = 2x^3 + 3x^2 - 11x - 6$.

1. 1. Solve the divisor equal to 0 to find the substitution value: $x + 3 = 0 \implies x = -3$
2. 2. Substitute $x = -3$ into $f(x)$:
3. $$f(-3) = 2(-3)^3 + 3(-3)^2 -11(-3) -6 = 2(-27) + 3(9) + 33 -6$$
4. $$= -54 + 27 + 33 - 6 = 0$$
5. 3. Since $f(-3) = 0$, by the factor theorem, $(x + 3)$ is a factor of $f(x)$.

> **tip**
>
> Start testing with small integer factors ($\pm1, \pm2, \pm3$) first: these are the most common factors in 0606 exam questions.

*Calculator:* allowed

## Factorising Cubic Polynomials

Once you identify one linear factor of a cubic using the factor theorem, you can write the cubic as $(linear\ factor)(quadratic\ factor)$. You can find the quadratic factor using either coefficient matching (inspection) or polynomial long division, then factorise the quadratic further if possible.

**Worked example:** Factorise $f(x) = x^3 - 3x^2 - 4x + 12$ fully into linear factors.

1. 1. We already found $(x - 2)$ is a factor, so write: $f(x) = (x - 2)(ax^2 + bx + c)$
2. 2. Expand the right-hand side and match coefficients to the original cubic:
3. $$ax^3 + (b - 2a)x^2 + (c - 2b)x - 2c = x^3 - 3x^2 -4x +12$$
4. 3. Solve for coefficients: $a = 1$, $b - 2 = -3 \implies b = -1$, $-2c = 12 \implies c = -6$
5. 4. The quadratic factor is $x^2 - x - 6$, which factorises to $(x - 3)(x + 2)$
6. 5. Fully factorised cubic: $(x - 2)(x - 3)(x + 2)$

Sometimes a question asks you to write a polynomial in the form $(mx + c)q(x) + r$, where $q(x)$ is the quotient and $r$ is the remainder, even when $(mx + c)$ is not a factor. This is just polynomial division presented as an identity: divide as usual, and the number left at the end of the division is the remainder $r$.

**Worked example:** Express $f(x) = 2x^3 - 7x^2 + 11x - 11$ in the form $(2x - 5)q(x) + r$.

1. 1. Divide $2x^3 - 7x^2 + 11x - 11$ by $(2x - 5)$. First term: $2x^3 \div 2x = x^2$; then $x^2(2x - 5) = 2x^3 - 5x^2$, and subtracting leaves $-2x^2 + 11x$.
2. 2. Next term: $-2x^2 \div 2x = -x$; then $-x(2x - 5) = -2x^2 + 5x$, and subtracting leaves $6x - 11$.
3. 3. Last term: $6x \div 2x = 3$; then $3(2x - 5) = 6x - 15$, and subtracting leaves the remainder $4$.
4. 4. So the quotient is $q(x) = x^2 - x + 3$ and the remainder is $r = 4$:
5. $$2x^3 - 7x^2 + 11x - 11 = (2x - 5)(x^2 - x + 3) + 4$$
6. 5. Check with the remainder theorem: $f\left(\frac{5}{2}\right)$ should equal $r$, and indeed $f\left(\frac{5}{2}\right) = 4$.

> **Exam tip:** If you use long division, always write the cubic in descending powers of x, including 0 coefficients for any missing powers (e.g. write $x^3 + 2x + 1$ as $x^3 + 0x^2 + 2x +1$) to avoid subtraction errors.

*Calculator:* allowed

## Solving Cubic Equations

To solve a cubic equation $f(x) = 0$, first factorise $f(x)$ fully into linear factors, then set each factor equal to 0 to find all roots. Roots may be integers, rational, or irrational.

**Worked example:** Solve $2x^3 + 3x^2 - 11x - 6 = 0$.

1. 1. Test integer factors of the constant term: $f(2) = 16 + 12 - 22 - 6 = 0$, so $(x - 2)$ is a factor.
2. 2. Divide the cubic by $(x - 2)$ to get the quadratic factor $2x^2 +7x +3$
3. 3. Factorise the quadratic: $2x^2 +7x +3 = (2x +1)(x +3)$
4. 4. Set each linear factor equal to 0:
5. $$x - 2 = 0 \implies x = 2; \quad 2x +1 =0 \implies x = -\frac{1}{2}; \quad x +3 =0 \implies x = -3$$
6. 5. The full set of solutions is $x = -3, x = -\frac{1}{2}, x = 2$

**Check your understanding**

1. What is the first step to solve a cubic equation in the 0606 exam?

   - A. Use the quadratic formula immediately
   - B. Find one linear factor using the factor theorem
   - C. Differentiate the cubic
   - D. Guess all three roots

   *Why:* You first find one linear factor by testing integer factors of the constant term, then factorise the remaining quadratic to find the other roots.

*Calculator:* allowed

## Finding Unknown Coefficients from a Factor and a Remainder

The most frequently examined question type in this topic gives a cubic $p(x) = ax^3 + bx^2 + cx + d$ that contains two unknown constants, together with two conditions. Usually one condition states that a linear expression is a factor, and the other states the remainder after division by a different linear expression. Each condition gives one equation, so the two conditions form a pair of simultaneous equations that you solve for the two unknowns. Once the cubic is fully known, you factorise it or solve $p(x) = 0$.

Translate each condition into an equation before solving. '$(x - r)$ is a factor of $p(x)$' means $p(r) = 0$ by the factor theorem. '$p(x)$ divided by $(x - r)$ leaves remainder $k$' means $p(r) = k$ by the remainder theorem. For a divisor of the form $(mx + c)$, substitute $x = -\frac{c}{m}$.

**Worked example:** The cubic $p(x) = 2x^3 + ax^2 + bx - 12$ has $(x - 1)$ as a factor, and leaves a remainder of $6$ when divided by $(x + 2)$. Find the values of $a$ and $b$, factorise $p(x)$ fully, and solve $p(x) = 0$.

1. 1. Factor condition: since $(x - 1)$ is a factor, the factor theorem gives $p(1) = 0$.
2. $$p(1) = 2(1)^3 + a(1)^2 + b(1) - 12 = a + b - 10 = 0 \implies a + b = 10$$
3. 2. Remainder condition: dividing by $(x + 2)$ leaves remainder $6$, so by the remainder theorem $p(-2) = 6$.
4. $$p(-2) = 2(-2)^3 + a(-2)^2 + b(-2) - 12 = -16 + 4a - 2b - 12 = 4a - 2b - 28$$
5. $$4a - 2b - 28 = 6 \implies 4a - 2b = 34 \implies 2a - b = 17$$
6. 3. Solve the simultaneous equations. Add $a + b = 10$ and $2a - b = 17$ to eliminate $b$:
7. $$3a = 27 \implies a = 9, \quad b = 10 - a = 1$$
8. 4. So $p(x) = 2x^3 + 9x^2 + x - 12$. Since $(x - 1)$ is already known to be a factor, divide by it to find the quadratic factor:
9. $$2x^3 + 9x^2 + x - 12 = (x - 1)(2x^2 + 11x + 12)$$
10. 5. Factorise the quadratic: $2x^2 + 11x + 12 = (2x + 3)(x + 4)$, so the full factorisation is:
11. $$p(x) = (x - 1)(2x + 3)(x + 4)$$
12. 6. Solve $p(x) = 0$ by setting each linear factor equal to 0:
13. $$x - 1 = 0 \implies x = 1; \quad 2x + 3 = 0 \implies x = -\frac{3}{2}; \quad x + 4 = 0 \implies x = -4$$
14. 7. Check the roots and the given remainder: $p(1) = 2 + 9 + 1 - 12 = 0$, $p(-4) = -128 + 144 - 4 - 12 = 0$, and $p(-2) = -16 + 36 - 2 - 12 = 6$. The solutions are $x = -4, -\frac{3}{2}, 1$.

> **tip**
>
> Write both conditions as equations first, then eliminate one unknown by adding or subtracting them. After finding the constants, substitute back into both original conditions to confirm the factor gives 0 and the division gives the stated remainder before you factorise.

> **Exam tip:** The linear expression you were told is a factor is already one factor of the cubic, so divide by it immediately once you know the coefficients, rather than testing for a new root.

*Calculator:* allowed

## When a Cubic Has Only One Real Root

Not every cubic factorises into three linear factors. After you divide out the linear factor found with the factor theorem, the remaining quadratic factor may not factorise. Use the discriminant $b^2 - 4ac$ of that quadratic to decide how many real roots the cubic has: if the discriminant is negative the quadratic has no real roots, so the cubic has only one real root; if it is zero the quadratic gives a repeated root. Exam questions often ask you to 'show that the equation has only one real root', and this is exactly the reasoning they expect.

**Worked example:** Show that the equation $x^3 - 4x^2 + 9x - 10 = 0$ has only one real root.

1. 1. Test integer factors of the constant term $-10$: $f(2) = 8 - 16 + 18 - 10 = 0$, so $(x - 2)$ is a factor.
2. 2. Divide to obtain the quadratic factor:

   $$x^3 - 4x^2 + 9x - 10 = (x - 2)(x^2 - 2x + 5)$$
3. 3. Examine the discriminant of the quadratic factor $x^2 - 2x + 5$:
4. $$b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0$$
5. 4. Since the discriminant is negative, $x^2 - 2x + 5 = 0$ has no real roots. The only real root of the cubic comes from $(x - 2) = 0$, so $x = 2$ is the only real root.

> **tip**
>
> If the quadratic factor's discriminant is exactly $0$, the quadratic gives a repeated root, so the cubic has a repeated (double) root alongside the root from the linear factor. For example, $x^3 - 5x^2 + 8x - 4 = (x - 1)(x - 2)^2$ has a repeated root at $x = 2$. Only when the discriminant is positive does the cubic have three distinct real roots.

> **Exam tip:** To decide the number of real roots of a cubic, factor out the known linear factor first, then read off the discriminant of the remaining quadratic: negative means one real root, zero means a repeated root, positive means three distinct real roots.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Substituting $x = a$ for divisor $(x + a)$ instead of $x = -a$
  - Why it fails: The remainder theorem requires solving the divisor equal to 0, so $(x + a) = 0$ gives $x = -a$, not $x = a$
  - Correct: Always rearrange the linear divisor to solve for x before substituting into $f(x)$
- **Wrong:** Forgetting 0 coefficients for missing powers during long division
  - Why it fails: Missing terms lead to incorrect subtraction steps and wrong quadratic factors
  - Correct: Rewrite the cubic with all powers of x present, e.g. $x^3 + 2x + 1$ becomes $x^3 + 0x^2 + 2x + 1$ before dividing
- **Wrong:** Stopping factorisation at $(linear)(quadratic)$ even if the quadratic can be factorised further
  - Why it fails: Exam questions often ask for fully factorised form or all roots, so incomplete factorisation loses marks
  - Correct: Always check if the quadratic factor can be factorised into linear terms, use the quadratic formula for non-integer roots if needed
- **Wrong:** Testing random values instead of integer factors of the constant term
  - Why it fails: Only integer factors of the constant term can be integer roots, so testing random values wastes time
  - Correct: Test $\pm1, \pm2, \pm k$ where $k$ is a factor of the constant term first to find linear factors quickly

## Cheatsheet

| Concept | Rule | Exam Use Case |
| --- | --- | --- |
| Remainder Theorem | Remainder of $f(x) \div (mx + c) = f(-c/m)$ | Find remainder without long division |
| Factor Theorem | If $f(a) = 0$, $(x - a)$ is a factor of $f(x)$ | Identify linear factors of cubics |
| Cubic Factorisation | Cubic = (linear factor)(quadratic) → fully linear | Get factorised form for solving equations |
| Cubic Solution | Set each linear factor = 0 to find roots | Solve for all values of x in cubic equations |

## What's next

Now that you have mastered factor and remainder theorems and cubic equations, you can apply these foundational algebraic skills to more advanced topics in the CIE 0606 syllabus. Cubic factorisation is used extensively in polynomial graph sketching, differentiation of polynomial functions, and solving kinematics problems involving displacement, velocity, and acceleration. Regular practice of these skills will help you build speed and accuracy for your exam, as these concepts are tested in almost every 0606 paper.

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