Factor & Remainder Theorems and Cubic Equations

Q: Do I have to use polynomial long division to factorise cubics for 0606?

No, you can use inspection (matching coefficients) if you find it faster, but long division is a reliable fallback for trickier cubics. Both methods are fully accepted in the exam.

Q: Can I use the quadratic formula for the quadratic factor of a cubic?

Yes, if the quadratic does not factorise into integer linear terms, you can use the quadratic formula or completing the square to find the remaining real roots, including irrational roots.

CIE IGCSE Additional Mathematics· 3.1, 3.2, 3.3· 25 min read

1. The Remainder Theorem★★☆☆☆⏱ 5 min

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📘 Definition

Remainder Theorem

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

Example:

For $f (x) = x^{3} + 2 x - 3$ , 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 $f (x) = 2 x^{3} - 5 x^{2} + x - 4$ is divided by $(2 x + 1)$ .

1
1. Solve the divisor equal to 0 to find the substitution value: $2 x + 1 = 0 ⟹ x = - \frac{1}{2}$
2
1. 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
1. 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
1. 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.

2. The Factor Theorem★★☆☆☆⏱ 5 min

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📘 Definition

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 $f (x) = x^{3} - 3 x^{2} - 4 x + 12$ , $f (2) = 0$ , 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 $36 x^{3}$ 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 $(2 x + 1)$ 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 $f (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ .

1
1. Solve the divisor equal to 0 to find the substitution value: $x + 3 = 0 ⟹ x = - 3$
2
1. 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
1. Since $f (- 3) = 0$ , by the factor theorem, $(x + 3)$ is a factor of $f (x)$ .

3. Factorising Cubic Polynomials★★★☆☆⏱ 7 min

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Once you identify one linear factor of a cubic using the factor theorem, you can write the cubic as $(l in e a r f a c t or) (q u a d r a t i c f a c t or)$ . 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 $f (x) = x^{3} - 3 x^{2} - 4 x + 12$ fully into linear factors.

1
1. We already found $(x - 2)$ is a factor, so write: $f (x) = (x - 2) (a x^{2} + b x + c)$
2
1. 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
1. Solve for coefficients: $a = 1$ , $b - 2 = - 3 ⟹ b = - 1$ , $- 2 c = 12 ⟹ c = - 6$
5
1. The quadratic factor is $x^{2} - x - 6$ , which factorises to $(x - 3) (x + 2)$
6
1. Fully factorised cubic: $(x - 2) (x - 3) (x + 2)$

Sometimes a question asks you to write a polynomial in the form $(m x + c) q (x) + r$ , where $q (x)$ is the quotient and $r$ is the remainder, even when $(m x + c)$ 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 $f (x) = 2 x^{3} - 7 x^{2} + 11 x - 11$ in the form $(2 x - 5) q (x) + r$ .

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

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 $x^{3} + 2 x + 1$ as $x^{3} + 0 x^{2} + 2 x + 1$ ) to avoid subtraction errors.

4. Solving Cubic Equations★★★☆☆⏱ 6 min

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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 $2 x^{3} + 3 x^{2} - 11 x - 6 = 0$ .

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

✓ Quick check

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
Reveal answer
B —
You first find one linear factor by testing integer factors of the constant term, then factorise the remaining quadratic to find the other roots.

5. Finding Unknown Coefficients from a Factor and a Remainder★★★★☆⏱ 8 min

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The most frequently examined question type in this topic gives a cubic $p (x) = a x^{3} + b x^{2} + c x + d$ 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 $(m x + c)$ , substitute $x = - \frac{c}{m}$ .

📐 Worked Example

The cubic $p (x) = 2 x^{3} + a x^{2} + b x - 12$ 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
1. 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
1. Solve the simultaneous equations. Add $a + b = 10$ and $2 a - b = 17$ to eliminate $b$ :
7
$3a = 27 \implies a = 9, \quad b = 10 - a = 1$
8
1. So $p (x) = 2 x^{3} + 9 x^{2} + x - 12$ . 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
1. Factorise the quadratic: $2 x^{2} + 11 x + 12 = (2 x + 3) (x + 4)$ , so the full factorisation is:
11
$p(x) = (x - 1)(2x + 3)(x + 4)$
12
1. 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
1. 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 $x = - 4, - \frac{3}{2}, 1$ .

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.

6. When a Cubic Has Only One Real Root★★★★☆⏱ 6 min

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

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.

7. Common Pitfalls

Wrong move:

Substituting $x = a$ for divisor $(x + a)$ instead of $x = - a$

Why:

The remainder theorem requires solving the divisor equal to 0, so $(x + a) = 0$ gives $x = - a$ , not $x = a$

Correct move:

Always rearrange the linear divisor to solve for x before substituting into $f (x)$

Wrong move:

Forgetting 0 coefficients for missing powers during long division

Why:

Missing terms lead to incorrect subtraction steps and wrong quadratic factors

Correct move:

Rewrite the cubic with all powers of x present, e.g. $x^{3} + 2 x + 1$ becomes $x^{3} + 0 x^{2} + 2 x + 1$ before dividing

Wrong move:

Stopping factorisation at $(l in e a r) (q u a d r a t i c)$ even if the quadratic can be factorised further

Why:

Exam questions often ask for fully factorised form or all roots, so incomplete factorisation loses marks

Correct move:

Always check if the quadratic factor can be factorised into linear terms, use the quadratic formula for non-integer roots if needed

Wrong move:

Testing random values instead of integer factors of the constant term

Why:

Only integer factors of the constant term can be integer roots, so testing random values wastes time

Correct move:

Test $\pm 1, \pm 2, \pm k$ where $k$ is a factor of the constant term first to find linear factors quickly

8. Quick Reference Cheatsheet

Concept	Rule	Exam Use Case
Remainder Theorem	Remainder of $f (x) \div (m x + c) = f (- c / m)$	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

9. Frequently Asked

Do I have to use polynomial long division to factorise cubics for 0606?

No, you can use inspection (matching coefficients) if you find it faster, but long division is a reliable fallback for trickier cubics. Both methods are fully accepted in the exam.

Can I use the quadratic formula for the quadratic factor of a cubic?

Yes, if the quadratic does not factorise into integer linear terms, you can use the quadratic formula or completing the square to find the remaining real roots, including irrational roots.

Going deeper

question bankCIE 0606 Polynomials Past QuestionsTopic-sorted questions with mark schemes
practice worksheetCubic Factorisation Drills20+ problems with step-by-step solutions

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.