The factor and remainder theorems

Q: Do I need to use algebraic long division for these questions?

No, for 4PM1 S3T02, you can use substitution with the factor/remainder theorems directly for remainder calculations and factor checks. You only need coefficient matching or inspection to find the quadratic factor of a cubic once a linear factor is confirmed.

Q: What substitution value do I use for a factor of the form (ax + b)?

Set the linear expression equal to zero and solve for x: . Substitute this value into f(x); if the result is 0, (ax + b) is a factor, otherwise the result is the remainder when dividing by (ax + b).

Edexcel International GCSE Further Pure Mathematics· S3· 12 min read

1. The Remainder Theorem★★☆☆☆⏱ 3 min

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📘 Definition

Remainder Theorem

When a polynomial $f (x)$ is divided by a linear expression $(a x + b)$ , the remainder is equal to $f (- \frac{b}{a})$ . For division by $(a x - b)$ , the remainder is $f (\frac{b}{a})$ .

Example:

Dividing $f (x)$ by $(x - 3)$ gives a remainder equal to $f (3)$

This theorem eliminates the need for algebraic long division to find remainders for linear divisors, saving valuable time in exams. You only need to substitute the correct x-value into the polynomial to get the remainder directly.

📐 Worked Example

Find the remainder when $f (x) = 2 x^{3} + 5 x^{2} - x + 4$ is divided by $(2 x - 1)$ .

1
Identify $a = 2$ and $b = 1$ from the divisor $(2 x - 1)$ . The remainder equals $f (\frac{b}{a}) = f (\frac{1}{2})$ .
2
$f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 5\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 4$
3
Calculate each term: $2 \times \frac{1}{8} = \frac{1}{4}$ , $5 \times \frac{1}{4} = \frac{5}{4}$ , so total = $\frac{1}{4} + \frac{5}{4} - \frac{2}{4} + \frac{16}{4} = 5$ .
4
The remainder is 5.

Exam tip:

Always double-check the sign of your substitution value: for $(a x + b)$ use $x = - b / a$ , for $(a x - b)$ use $x = b / a$ . Sign errors are the most common mistake on these questions.

2. The Factor Theorem★★★☆☆⏱ 3 min

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📘 Definition

Factor Theorem

A special case of the remainder theorem: if $f (a) = 0$ , then $(x - a)$ is a factor of $f (x)$ . For a linear factor of the form $(a x + b)$ , if $f (- \frac{b}{a}) = 0$ , then $(a x + b)$ is a factor of $f (x)$ .

This theorem lets you verify if a given linear expression is a factor of a polynomial without expanding or performing long division. For 4PM1, you will almost always be given one linear factor for cubic factorisation questions, and you can use this theorem to confirm it first.

📐 Worked Example

Show that $(x + 2)$ is a factor of $f (x) = x^{3} + 3 x^{2} - 4$ .

1
Rewrite the divisor as $(x - (- 2))$ , so $a = - 2$ . We need to calculate $f (- 2)$ .
2
$f(-2) = (-2)^3 + 3(-2)^2 - 4 = -8 + 12 - 4 = 0$
3
Since $f (- 2) = 0$ , by the factor theorem, $(x + 2)$ is a factor of $f (x)$ .

Exam tip:

For questions that ask you to "Show that (x - a) is a factor", this substitution step is all you need to write for full marks, no extra working is required.

3. Factorising Cubics With a Provided Linear Factor★★★★☆⏱ 4 min

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Once you confirm a linear factor of a cubic, you can write the cubic as the product of the linear factor and a general quadratic expression of the form $A x^{2} + B x + C$ . You then find the values of A, B, and C by equating coefficients or using inspection.

📐 Worked Example

Given that $(x + 2)$ is a factor of $f (x) = x^{3} + 3 x^{2} - 4$ , factorise $f (x)$ fully.

1
Write the cubic as the product of the given linear factor and a general quadratic: $x^{3} + 3 x^{2} - 4 = (x + 2) (a x^{2} + b x + c)$ .
2
Equate coefficients for each power of x: For $x^{3}$ : $1 = 1 \times a ⟹ a = 1$ . For the constant term: $- 4 = 2 \times c ⟹ c = - 2$ . For $x^{2}$ : $3 = 1 \times b + 2 \times a$ , substitute $a = 1$ so $b = 1$ .
3
The quadratic factor is $x^{2} + x - 2$ . Factorise this further using inspection: $x^{2} + x - 2 = (x + 2) (x - 1)$ .
4
Combine factors to get the fully factorised form: $f (x) = (x + 2)^{2} (x - 1)$ .

Exam tip:

Always expand your final factorised form briefly to check it matches the original cubic. This takes 10 seconds and catches almost all coefficient matching errors.

4. Mixed Exam-style Applications★★★★☆⏱ 2 min

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📐 Worked Example

Given that $(2 x + 3)$ is a factor of $f (x) = 6 x^{3} + 11 x^{2} - x - 6$ , find the remainder when $f (x)$ is divided by $(x + 2)$ , then factorise $f (x)$ fully.

1
First calculate the remainder for division by $(x + 2)$ : substitute $x = - 2$ into $f (x)$ .
2
$f(-2) = 6(-2)^3 + 11(-2)^2 - (-2) - 6 = -48 + 44 + 2 - 6 = -8$
3
The remainder is -8. Next, write the cubic as $(2 x + 3) (a x^{2} + b x + c)$ . Equate coefficients: $a = 3$ , $c = - 2$ , $b = 1$ , so the quadratic is $3 x^{2} + x - 2$ .
4
Factorise the quadratic: $3 x^{2} + x - 2 = (3 x - 2) (x + 1)$ . The fully factorised form is $f (x) = (2 x + 3) (3 x - 2) (x + 1)$ .

✓ Quick check

What is the remainder when $f (x) = 3 x^{3} - 2 x^{2} + 5 x - 1$ is divided by $(x + 1)$ ?
- A) -11
- B) -3
- C) 9
- D) 11
Reveal answer
A —
Substitute $x = - 1$ into $f (x)$ : $3 (- 1)^{3} - 2 (- 1)^{2} + 5 (- 1) - 1 = - 3 - 2 - 5 - 1 = - 11$ .

5. Common Pitfalls

Wrong move:

Using $x = b / a$ as the substitution value for divisor $(a x + b)$ instead of $x = - b / a$

Why:

Sign error when rearranging the linear divisor to find the root, leading to incorrect remainder or factor check results

Correct move:

Always set the linear divisor equal to zero and solve for x directly: $a x + b = 0 ⟹ x = - b / a$ , to avoid sign mistakes

Wrong move:

Trying to find a linear factor of a cubic when no factor is provided

Why:

The 4PM1 specification explicitly provides one linear factor for all cubic factorisation questions in this subtopic

Correct move:

Use the provided factor directly, only test small integer roots if explicitly asked to find a factor with no guidance

Wrong move:

Leaving the quadratic factor of a cubic unfactorised when it can be split into linear terms

Why:

Exam questions almost always ask for full factorisation, so partial factorisation leads to lost marks

Correct move:

After finding the quadratic factor, always check if it can be factorised further using inspection or the quadratic formula

Wrong move:

Making coefficient matching errors when writing the quadratic factor

Why:

Forgetting that the product of the linear and quadratic factors must equal the original cubic, so coefficients for every power of x must align

Correct move:

Check the constant term first (it is the product of the constant terms of the linear and quadratic factors) then expand your final factorised form to verify

Wrong move:

Using algebraic long division for all factor/remainder checks

Why:

Long division takes more time and is more error-prone than simple substitution for these tasks

Correct move:

Use substitution for remainder/factor checks, only use coefficient matching or inspection to find the quadratic factor once you confirm the linear factor

6. Quick Reference Cheatsheet

Task	Rule	Substitution Value
Find remainder when $f (x)$ divided by $(x - a)$	Remainder = $f (a)$	$x = a$
Find remainder when $f (x)$ divided by $(a x + b)$	Remainder = $f (- b / a)$	$x = - \frac{b}{a}$
Check if $(x - a)$ is factor of $f (x)$	If $f (a) = 0$ , $(x - a)$ is a factor	$x = a$
Check if $(a x + b)$ is factor of $f (x)$	If $f (- b / a) = 0$ , $(a x + b)$ is a factor	$x = - \frac{b}{a}$
Factorise cubic given linear factor $(a x + b)$	Write cubic = $(a x + b) (A x^{2} + B x + C)$ , equate coefficients	Substitute to confirm factor first

7. Frequently Asked

Do I need to use algebraic long division for these questions?

No, for 4PM1 S3_T02, you can use substitution with the factor/remainder theorems directly for remainder calculations and factor checks. You only need coefficient matching or inspection to find the quadratic factor of a cubic once a linear factor is confirmed.

What substitution value do I use for a factor of the form (ax + b)?

Set the linear expression equal to zero and solve for x: $a x + b = 0 ⟹ x = - \frac{b}{a}$ . Substitute this value into f(x); if the result is 0, (ax + b) is a factor, otherwise the result is the remainder when dividing by (ax + b).

Going deeper

What's Next

Now that you have mastered the factor and remainder theorems, you are ready to progress to the next subtopics in the Identities and inequalities unit for Edexcel IGCSE Further Pure Maths. Next, you will learn to use these factorised cubic expressions to solve polynomial equations, and then apply your knowledge to solve algebraic inequalities. These theorems are also foundational for later topics including sketching cubic graphs and solving problems involving polynomial intersections. Make sure you practice past paper questions on this subtopic to build speed and accuracy, as these questions are often combined with other algebraic topics in longer exam questions. Always double-check your substitution and coefficient matching steps to avoid losing easy marks.