Algebraic expressions and factorising

Q: What is the key difference between an algebraic expression and an algebraic equation?

An expression is a collection of terms with no equals sign (e.g. \(3x + 5\)), while an equation contains an equals sign setting two expressions equal (e.g. \(3x + 5 = 11\)).

Q: How do you collect like terms to simplify an algebraic expression?

Combine only terms that have the exact same letters and powers , keeping the sign directly in front of each term attached to it.

Q: How do you multiply and divide single algebraic terms?

When multiplying, multiply the numbers and add the powers of matching letters. When dividing, divide the numbers and subtract the powers of matching letters.

Q: How do you expand a pair of double brackets like \((x + a)(x + b)\)?

Use the FOIL method (First, Outside, Inside, Last) to produce four terms, then collect like middle terms.

Q: What is the method for factorising an expression into a single bracket?

Find the Highest Common Factor (HCF) of the numbers and variables, write it outside the bracket, and divide each term by it to get the terms inside.

Q: How do you factorise a basic quadratic expression in the form \(x^2 + bx + c\)?

Find two numbers that multiply to \(c\) and add to \(b\) , then write the answer as \((x + p)(x + q)\).

Q: How do you factorise a harder quadratic \(ax^2 + bx + c\) where \(a > 1\)?

Use the AC method : find two numbers that multiply to \(ac\) and add to \(b\), split the middle term, and factorise by grouping in pairs.

Mathematics 2210 · CCEA GCSE · 10 min read · Updated: 31 Aug 2026

Welcome to Algebraic Expressions & Factorising

Welcome! Algebra is often thought of as a confusing world of random letters and numbers, but it is really just a powerful shorthand system for solving everyday problems. Think of algebra as a set of puzzle tools: when you do not know a exact value, you give it a letter name (like \(x\) or \(y\)) so you can still work with it.

In this chapter, you will master two opposite superpowers: expanding (opening brackets up) and factorising (putting expressions back into brackets). Don't worry if this seems tricky at first—once you spot the patterns, it becomes just like solving familiar number puzzles!

1. The Language of Algebra: Key Building Blocks

Before we jump into calculations, let's make sure we speak the same algebraic language. Here are the core terms you need to know:

• Variable: A letter representing an unknown number (e.g. \(x\), \(y\), \(n\)).
• Term: A single number, a single variable, or numbers and variables multiplied together (e.g. \(5\), \(x\), \(4y\), \(-3x^2\)).
• Coefficient: The number multiplying the variable. In \(7x\), the coefficient is \(7\). If you see \(-x\), the coefficient is \(-1\). If you see just \(x\), the coefficient is \(1\).
• Constant: A fixed number on its own with no variable attached (e.g. the \(+4\) in \(2x + 4\)).
• Expression: A collection of terms joined by \(+\) or \(-\) signs, with no equals sign (e.g. \(3x + 5\)).
• Equation: Two expressions set equal to each other, containing an equals sign (e.g. \(3x + 5 = 11\)).
• Formula: A rule showing the relationship between different quantities (e.g. \(A = lw\) for the area of a rectangle).
• Identity: A statement that is true for every possible value of the variable, written using the \(\equiv\) symbol (e.g. \(2(x + 3) \equiv 2x + 6\)).

Did You Know?

The word algebra comes from the Arabic word al-jabr, meaning "reunion of broken parts", introduced by the 9th-century mathematician Al-Khwarizmi!

Key Takeaway

An expression has no equals sign (e.g. \(4x - 7\)), while an equation always has an equals sign (e.g. \(4x - 7 = 13\)).

2. Simplifying & Collecting Like Terms

Simplifying an expression means writing it in its shortest, neatest possible form by combining terms that are the same type.

The Fruit Basket Analogy

Imagine you have \(3\) apples, \(2\) bananas, \(4\) more apples, and \(1\) banana.
You would naturally say: "\(7\) apples and \(3\) bananas."
In algebra, if \(a =\) apples and \(b =\) bananas:
\(3a + 2b + 4a + b = 7a + 3b\).
You cannot combine them into \(10ab\) because apples and bananas are different things!

Crucial Rule: Watch the Signs!

The plus or minus sign directly in front of a term belongs to that term. Always box or circle the sign with the term to avoid sign errors.

Important: Powers Make Terms Different

\(x\) and \(x^2\) are not like terms! Think of \(x\) as a straight line of length \(x\), and \(x^2\) as a flat square area of \(x \times x\). You cannot blend a line and an area together.

Worked Example: Simplifying

Simplify: \(5x + 3y - 2x + 7y - 4\)
Step 1: Group like \(x\) terms: \(5x - 2x = 3x\)
Step 2: Group like \(y\) terms: \(+3y + 7y = +10y\)
Step 3: Bring down the constant number: \(-4\)
Final Answer: \(3x + 10y - 4\)

Multiplying and Dividing Terms

When multiplying algebraic terms, multiply the numbers first, then add powers of matching letters:
• \(3x \times 4 = 12x\)
• \(2x \times 5x = 10x^2\)
• \(3a^2b \times 4ab^3 = 12a^3b^4\)
When dividing, divide numbers and subtract powers of matching letters:
• \(\frac{12x^5}{4x^2} = 3x^3\)

Key Takeaway

Only add and subtract terms that have the exact same letters and powers. Keep the sign that sits directly in front of each term.

3. Expanding Brackets

Expanding (or multiplying out) means removing the brackets by multiplying everything on the inside by whatever is on the outside.

Part A: Single Brackets

Think of the number outside the bracket as a friendly host at a party who must shake hands with every single guest inside the room.

Rule: \(a(b + c) = ab + ac\)

Example 1: Expand \(4(3x + 5)\)
• Multiply \(4 \times 3x = 12x\)
• Multiply \(4 \times (+5) = +20\)
• Answer: \(12x + 20\)

Example 2 (Watch the negative sign!): Expand \(-3(2x - 7)\)
• Multiply \(-3 \times 2x = -6x\)
• Multiply \(-3 \times (-7) = +21\) (negative \(\times\) negative \(=\) positive!)
• Answer: \(-6x + 21\)

Example 3 (Expanding and simplifying): Expand and simplify \(2(4x + 3) + 3(2x - 1)\)
• First bracket: \(2(4x + 3) = 8x + 6\)
• Second bracket: \(3(2x - 1) = 6x - 3\)
• Combine: \(8x + 6 + 6x - 3 = 14x + 3\)

Part B: Double Brackets (Multiplying Two Binomials)

When multiplying two brackets like \((x + 2)(x + 5)\), every term in the first bracket must multiply every term in the second bracket. A popular memory aid is FOIL:

• F - First terms in each bracket
• O - Outside terms
• I - Inside terms
• L - Last terms in each bracket

Worked Example: Standard Double Bracket

Expand and simplify \((x + 3)(x + 4)\):
• First: \(x \times x = x^2\)
• Outside: \(x \times 4 = +4x\)
• Inside: \(3 \times x = +3x\)
• Last: \(3 \times 4 = +12\)
Put it together: \(x^2 + 4x + 3x + 12\)
Combine the middle terms: \(4x + 3x = 7x\)
Final Answer: \(x^2 + 7x + 12\)

Worked Example: Mixed Signs

Expand and simplify \((2x - 3)(x + 5)\):
• First: \(2x \times x = 2x^2\)
• Outside: \(2x \times 5 = +10x\)
• Inside: \(-3 \times x = -3x\)
• Last: \(-3 \times 5 = -15\)
Combine like terms: \(+10x - 3x = +7x\)
Final Answer: \(2x^2 + 7x - 15\)

Common Pitfall Alert: Squaring a Bracket

Never do this: \((x + 5)^2 \neq x^2 + 25\). This is one of the most common mistakes in GCSE Maths!
Correct method: Always write it out as two brackets first: \((x + 5)(x + 5)\)
Then expand: \(x^2 + 5x + 5x + 25 = x^2 + 10x + 25\).

Key Takeaway

For double brackets, use FOIL to get four terms, then simplify by combining the middle like terms.

4. Factorising into a Single Bracket

Factorising is the exact opposite of expanding. You are looking for the Highest Common Factor (HCF) of all terms to pull outside a set of brackets.

Step-by-Step Method

1. Find the largest number that divides into all coefficients.
2. Find the highest power of any common letters.
3. Write this complete HCF outside the bracket.
4. Divide each original term by the HCF to see what stays inside.

Worked Examples

Example 1: Factorise \(6x + 15\)
• HCF of \(6\) and \(15\) is \(3\). (No common letter).
• Divide: \(6x \div 3 = 2x\) and \(15 \div 3 = 5\).
• Answer: \(3(2x + 5)\)

Example 2: Factorise fully \(8x^2 - 12x\)
• HCF of \(8\) and \(12\) is \(4\).
• Both terms share an \(x\), so common letter is \(x\).
• Total HCF outside \(= 4x\).
• Divide: \(8x^2 \div 4x = 2x\) and \(-12x \div 4x = -3\).
• Answer: \(4x(2x - 3)\)

Example 3: Factorise fully \(6a^2b + 9ab^2\)
• Number HCF: \(3\)
• Letter HCF: \(ab\)
• Outside: \(3ab\)
• Inside: \(6a^2b \div 3ab = 2a\) and \(9ab^2 \div 3ab = 3b\)
• Answer: \(3ab(2a + 3b)\)

Quick Check Trick

Whenever you finish factorising, quickly expand your answer in your head! If you get back your starting expression, you know you are \(100\%\) correct.

5. Factorising Quadratics into Two Brackets

A quadratic expression has the standard form \(ax^2 + bx + c\), where the highest power of \(x\) is \(2\).

Type A: Basic Quadratics (\(x^2 + bx + c\))

When the coefficient of \(x^2\) is \(1\), we factorise into two brackets: \((x + \text{?})(x + \text{?})\).

The Golden Rule: Find two numbers that multiply to give the end number (\(c\)) and add to give the middle number (\(b\)).

Sign Rules Helper:

• If \(c\) is positive, both numbers have the same sign (both \(+\) or both \(-\), matching \(b\)).
• If \(c\) is negative, the two numbers have different signs (one \(+\) and one \(-\)).

Worked Examples

Example 1 (Both positive): Factorise \(x^2 + 7x + 12\)
• Need two numbers that multiply to \(+12\) and add to \(+7\).
• Factor pairs of \(12\): \(1 \times 12\) (sum \(13\)), \(2 \times 6\) (sum \(8\)), \(3 \times 4\) (sum \(7\)).
• The winning pair is \(+3\) and \(+4\).
• Answer: \((x + 3)(x + 4)\)

Example 2 (Negative middle, positive end): Factorise \(x^2 - 8x + 15\)
• Need two numbers that multiply to \(+15\) and add to \(-8\).
• Since the product is positive and sum is negative, both numbers must be negative.
• Factor pairs: \(-1 \times -15\) (sum \(-16\)), \(-3 \times -5\) (sum \(-8\)).
• The winning pair is \(-3\) and \(-5\).
• Answer: \((x - 3)(x - 5)\)

Example 3 (Negative end number): Factorise \(x^2 + 2x - 24\)
• Need two numbers that multiply to \(-24\) and add to \(+2\).
• Different signs needed! Pairs: \(-4 \times 6 = -24\), and \(-4 + 6 = +2\).
• The winning pair is \(-4\) and \(+6\).
• Answer: \((x - 4)(x + 6)\)

6. The Difference of Two Squares (DOTS)

This is a special quadratic pattern that examiners love to test. It applies whenever you have two squared terms separated by a minus sign.

The Formula: \(a^2 - b^2 \equiv (a + b)(a - b)\)

Why does this work?

If you expand \((a + b)(a - b)\), the middle terms cancel out: \(a^2 - ab + ab - b^2 = a^2 - b^2\).

Worked Examples

Example 1: Factorise \(x^2 - 49\)
• Square root of \(x^2\) is \(x\).
• Square root of \(49\) is \(7\).
• Answer: \((x + 7)(x - 7)\)

Example 2: Factorise \(4x^2 - 81\)
• Square root of \(4x^2\) is \(2x\).
• Square root of \(81\) is \(9\).
• Answer: \((2x + 9)(2x - 9)\)

Example 3: Factorise \(25y^2 - 1\)
• Square root of \(25y^2\) is \(5y\).
• Square root of \(1\) is \(1\).
• Answer: \((5y + 1)(5y - 1)\)

Watch Out!

\(x^2 + 25\) CANNOT be factorised using this method because there is a plus sign, not a minus sign!

7. Higher Tier: Harder Quadratics (\(ax^2 + bx + c\))

When the number in front of \(x^2\) is greater than \(1\) (e.g. \(2x^2 + 7x + 3\)), we use the AC Method (Grouping).

Step-by-Step AC Method:

1. Multiply \(a \times c\).
2. Find two numbers that multiply to give \(ac\) and add to give \(b\).
3. Split the middle \(x\) term into two parts using these numbers.
4. Factorise the first two terms and the last two terms separately.
5. Factorise out the common bracket.

Worked Example: Harder Quadratic

Factorise: \(2x^2 + 7x + 3\)
• Step 1: Multiply \(a \times c = 2 \times 3 = 6\).
• Step 2: Find two numbers that multiply to \(6\) and add to \(7\). The numbers are \(1\) and \(6\).
• Step 3: Split the middle term \(7x\) into \(1x + 6x\):
\(2x^2 + 6x + 1x + 3\)
• Step 4: Factorise in pairs:
From \(2x^2 + 6x\), pull out \(2x \implies 2x(x + 3)\)
From \(+1x + 3\), pull out \(+1 \implies +1(x + 3)\)
• Step 5: Notice the shared bracket \((x + 3)\)! Pull it together:
Final Answer: \((2x + 1)(x + 3)\)

Quick Revision Checklist

Before sitting your exam, check that you can:

• Collect like terms by paying careful attention to signs.
• Expand a single bracket: \(a(b + c) = ab + ac\).
• Expand double brackets using FOIL and simplify the middle terms.
• Factorise into a single bracket by finding the HCF.
• Factorise a quadratic \(x^2 + bx + c\) by finding two numbers that multiply to \(c\) and add to \(b\).
• Spot the Difference of Two Squares: \(a^2 - b^2 = (a + b)(a - b)\).
• Factorise harder quadratics with \(a > 1\) using the AC method.
• Check your factorisation answers by expanding them back out!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the key difference between an algebraic expression and an algebraic equation?

An expression is a collection of terms with no equals sign (e.g. \(3x + 5\)), while an equation contains an equals sign setting two expressions equal (e.g. \(3x + 5 = 11\)).

How do you collect like terms to simplify an algebraic expression?

Combine only terms that have the exact same letters and powers, keeping the sign directly in front of each term attached to it.

How do you multiply and divide single algebraic terms?

When multiplying, multiply the numbers and add the powers of matching letters. When dividing, divide the numbers and subtract the powers of matching letters.

How do you expand a single bracket like \(a(b + c)\)?

Multiply every term inside the bracket by the term outside: \(a(b + c) = ab + ac\).

How do you expand a pair of double brackets like \((x + a)(x + b)\)?

Use the FOIL method (First, Outside, Inside, Last) to produce four terms, then collect like middle terms.

What is the method for factorising an expression into a single bracket?

Find the Highest Common Factor (HCF) of the numbers and variables, write it outside the bracket, and divide each term by it to get the terms inside.

How do you factorise a basic quadratic expression in the form \(x^2 + bx + c\)?

Find two numbers that multiply to \(c\) and add to \(b\), then write the answer as \((x + p)(x + q)\).

What is the Difference of Two Squares (DOTS) rule?

\(a^2 - b^2 \equiv (a + b)(a - b)\)

How do you factorise a harder quadratic \(ax^2 + bx + c\) where \(a > 1\)?

Use the AC method: find two numbers that multiply to \(ac\) and add to \(b\), split the middle term, and factorise by grouping in pairs.

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