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Collecting Like Terms and Basic Algebraic Operations Revision Notes | GCSE AQA Higher Maths

To expand and simplify, expand each bracket, write out all the terms, then collect like terms. Free notes with worked examples and 13 questions answered.

Updated 21 Mar 2026 · Revise Right Team

Collecting Like Terms and Basic Algebraic Operations is part of Notation, vocabulary and manipulation in GCSE AQA Higher Maths.

4 min

Reading time

39

Lessons in this topic

GCSE AQA Higher

Pathway

Reviewed by Revise Right

Content review

What this lesson covers

To expand and simplify, expand each bracket, write out all the terms, then collect like terms. Free notes with worked examples and 13 questions answered.

  • Identify and collect like terms in algebraic expressions
  • Expand expressions by multiplying a single term over a bracket
  • Factorise expressions by taking out common factors
  • Apply these skills to simplify and manipulate algebraic expressions in exam contexts

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Key ideas to keep in view

Use this lesson to stay inside Notation, vocabulary and manipulation while connecting the detail back to the wider Algebra topic.

  • Like terms: Terms in an algebraic expression that share exactly the same variable(s) raised to the same power(s). Like terms can be added or subtracted; unlike terms cannot.
  • Expanding brackets: The process of multiplying a term outside a bracket by every term inside the bracket, removing the bracket from the expression.
  • Factorising: The reverse of expanding: writing an expression as a product of its factors by identifying and extracting the highest common factor.
  • Highest Common Factor (HCF): The largest factor shared by all terms in an expression. Used when factorising to determine what is placed outside the bracket.

Revision notes

Read the core explanations from Collecting Like Terms and Basic Algebraic Operations before testing yourself from memory.

1. Why Algebra Needs Tidying Up

Algebra is the language of mathematics — but like any language, it gets messy. An expression like $3x + 5y - x + 2y$ carries the same information as $2x + 7y$, but the second form is far easier to work with. The skill of simplifying expressions sits at the heart of almost every GCSE algebra topic that follows.

Three techniques unlock most of that tidying work: collecting like terms, expanding brackets, and factorising. Master these and you have the toolkit for equations, quadratics, and beyond. This lesson builds each one from scratch.

2. Collecting Like Terms

Think of algebra like a shopping bag. You can count apples with apples and oranges with oranges, but you cannot add apples to oranges and call the result a single thing. In algebra, $3x$ and $5x$ are both "$x$-apples" — they can be combined. But $3x$ and $5x^2$ are different types entirely.

To collect like terms, group the matching types together and then add or subtract their coefficients (the numbers in front). The variable part stays unchanged — you are only combining the coefficients.

Worked example: Collecting Like Terms: Single Variable

Simplify $5x + 3 - 2x + 7$

  1. $x$-terms: $5x$ and $-2x$ Number terms: $3$ and $7$ — Colour-coding or underlining each group helps you avoid mixing types. [M1] for correct identification.
  2. $5x + (-2x) = 5x - 2x = 3x$ — Subtract the coefficients: $5 - 2 = 3$. The variable $x$ stays as $x$.
  3. $3 + 7 = 10$ — Constants (plain numbers) are like terms with each other.
  4. $3x + 10$ — [A1] for the correct simplified form. No further simplification is possible.

Answer: $3x + 10$

Now try a slightly more involved example with two different variables. The process is identical — you just have more groups to manage.

Worked example: Visualising Like Terms

Simplify $4a + 3b - a + 5b - 2$

  1. $4a$ and $-a$ — Note that $-a$ means $-1a$. The coefficient is $-1$, not zero.
  2. $4a + (-1a) = 4a - a = 3a$ — $4 - 1 = 3$, so the result is $3a$. [M1]
  3. $3b + 5b = 8b$ — $3 + 5 = 8$, giving $8b$.
  4. $-2$ (no other constant to combine with) — There is only one constant here, so it stays as $-2$.
  5. $3a + 8b - 2$ — [A1] Three separate unlike terms remain. This is fully simplified.

Answer: $3a + 8b - 2$

3. Multiplying a Single Term Over a Bracket

Brackets in algebra are a way of packaging terms together. When a term sits directly outside a bracket — like $3(x + 4)$ — it is multiplying everything inside. Expanding means distributing that outside term to each term inside, one at a time.

The key discipline here is to multiply the outside term by every term inside — not just the first one. Missing the second term is the single most common error on this topic in GCSE papers.

Worked example: Expanding: Numerical Multiplier

Expand $4(3x - 2)$

  1. Outside term: $4$ Inside terms: $3x$ and $-2$ — The outside term multiplies every term inside the bracket.
  2. $4 \times 3x = 12x$ — Multiply the coefficients: $4 \times 3 = 12$. The variable $x$ is carried through. [M1]
  3. $4 \times (-2) = -8$ — Positive times negative gives negative. Do not drop the sign.
  4. $12x - 8$ — [A1] The bracket has been removed. No like terms remain to collect here.

Answer: $12x - 8$

Worked example: Expanding: Variable Multiplier

Expand $3x(2x + 5)$

  1. Outside term: $3x$ Inside terms: $2x$ and $5$ — The outside term here contains a variable, so the result will include $x^2$ terms.
  2. $3x \times 2x = 6x^2$ — Multiply coefficients: $3 \times 2 = 6$. Multiply variables: $x \times x = x^2$. [M1]
  3. $3x \times 5 = 15x$ — The number $5$ has no variable, so the result is $15x$, not $15x^2$.
  4. $6x^2 + 15x$ — [A1] These two terms are unlike ($x^2$ vs $x$), so no further collecting is possible.

Answer: $6x^2 + 15x$

Sometimes expanding is followed immediately by collecting like terms. This is where both skills combine — and where exam questions earn their marks.

Worked example: Expanding Then Collecting

Expand and simplify $2(3x + 4) + 5(x - 1)$

  1. $2(3x + 4) = 6x + 8$ — $2 \times 3x = 6x$ and $2 \times 4 = 8$. [M1]
  2. $5(x - 1) = 5x - 5$ — $5 \times x = 5x$ and $5 \times (-1) = -5$. [M1]
  3. $6x + 8 + 5x - 5$ — Do not collect yet — write out all four terms first to avoid sign errors.
  4. $6x + 5x = 11x$ — $6 + 5 = 11$.
  5. $8 + (-5) = 8 - 5 = 3$ — Keep the sign attached to the number when grouping.
  6. $11x + 3$ — [A1] Fully simplified. Two unlike terms remain.

Answer: $11x + 3$

4. Taking Out Common Factors (Factorising)

Factorising is expanding in reverse. Where expanding removes brackets, factorising introduces them. The goal is to rewrite an expression as a product — something multiplied by something else.

The method relies on spotting what every term in the expression shares. That shared part is the Highest Common Factor (HCF), and it moves outside the bracket. What remains inside the bracket is what each original term becomes after dividing by the HCF.

Worked example: Factorising: Numerical HCF

Factorise $12x + 8$

  1. Factors of $12$: $1, 2, 3, 4, 6, 12$ Factors of $8$: $1, 2, 4, 8$ HCF $= 4$ — The largest number dividing both $12$ and $8$ is $4$. [M1]
  2. $12x \div 4 = 3x$ $8 \div 4 = 2$ — These results go inside the bracket.
  3. $4(3x + 2)$ — [A1] The HCF sits outside; the divided terms sit inside.
  4. $4 \times 3x = 12x$ and $4 \times 2 = 8$ So $4(3x + 2) = 12x + 8$ ✓ — Always verify by expanding back. If you recover the original expression, the factorisation is correct.

Answer: $4(3x + 2)$

Worked example: Factorising: Numerical and Variable HCF

Factorise $6x^2 + 9x$

  1. Factors of $6$: $1, 2, 3, 6$ Factors of $9$: $1, 3, 9$ HCF of coefficients $= 3$ — The largest number dividing both $6$ and $9$ is $3$.
  2. Terms: $x^2$ and $x$ Highest power of $x$ in both: $x^1 = x$ — $x^2$ contains $x$ as a factor; $x$ contains $x$ as a factor. The HCF of the variable parts is $x$. [M1]
  3. HCF $= 3 \times x = 3x$ — Both the numerical and variable parts of the HCF are combined.
  4. $6x^2 \div 3x = 2x$ $9x \div 3x = 3$ — For $6x^2 \div 3x$: divide coefficients $6 \div 3 = 2$, divide variables $x^2 \div x = x$. Result: $2x$.
  5. $3x(2x + 3)$ — [A1] Check: $3x \times 2x = 6x^2$ and $3x \times 3 = 9x$. Correct.

Answer: $3x(2x + 3)$

5. Retrieval Practice

You have now seen all three core techniques. Before moving to exam strategy, test whether the methods have stuck. Work through each problem before checking the answer.

6. Exam Technique and Common Mistakes

These three skills appear across the entire GCSE paper — in equations, inequalities, proof, and beyond. Getting them right under exam conditions comes down to a handful of disciplined habits.

One particular trap appears regularly in questions that combine expanding with a subtracted bracket. Look at $4(x + 3) - 2(x + 1)$. The $-2$ multiplies both $x$ and $1$ inside the second bracket, giving $-2x - 2$, not $-2x + 1$. Writing out the expansion in full before collecting is the safest approach.

7. Bringing It Together: A Mixed Exam Question

Exam questions rarely test one skill in isolation. A typical question might ask you to expand, collect, and then factorise the result — all in one part. Here is how that looks in practice.

Worked example: Mixed Skills: Expand, Simplify, Factorise

Show that $3(2x + 4) + 6(x - 1)$ can be written in the form $a(bx + c)$ where $a$, $b$, and $c$ are integers. State the values of $a$, $b$, and $c$.

  1. $3(2x + 4) = 6x + 12$ — $3 \times 2x = 6x$ and $3 \times 4 = 12$. [M1]
  2. $6(x - 1) = 6x - 6$ — $6 \times x = 6x$ and $6 \times (-1) = -6$. [M1]
  3. $6x + 12 + 6x - 6$ — List every term before collecting to avoid sign errors.
  4. $x$-terms: $6x + 6x = 12x$ Constants: $12 - 6 = 6$ Result: $12x + 6$ — [A1] for the correct simplified expression $12x + 6$.
  5. HCF of $12$ and $6$ is $6$. No shared variable factor (the constant $6$ has no $x$). — The HCF is $6$. [M1]
  6. $12x + 6 = 6(2x + 1)$ — $12x \div 6 = 2x$ and $6 \div 6 = 1$. Check: $6 \times 2x = 12x$ and $6 \times 1 = 6$. ✓ [A1]
  7. $a = 6$, $b = 2$, $c = 1$ — Read directly from $6(2x + 1)$. The question asks for integers, and all three values are integers. [A1]

Answer: $6(2x + 1)$, where $a = 6$, $b = 2$, $c = 1$

Frequently asked questions

How do you expand and simplify brackets?

To expand and simplify, expand each bracket, write out all the terms, then collect like terms. For example, 2(3x + 4) + 5(x - 1) = 6x + 8 + 5x - 5 = 11x + 3; with a subtracted bracket, -2(x + 1) gives -2x - 2.

How do you expand a single bracket?

To expand a single bracket, multiply the outside term by every term inside the bracket, not just the first one. For example, 4(3x - 2) = 12x - 8 and 3x(2x + 5) = 6x² + 15x.

How do you collect like terms?

To collect like terms, group terms with identical variable parts and add or subtract their coefficients, keeping the variable part unchanged. For example, 4a + 3b - a + 5b - 2 simplifies to 3a + 8b - 2.

What does fully factorise mean?

Fully factorise means taking the highest common factor of every term outside the bracket, including any shared variable. For 6x² + 9x the full HCF is 3x, giving 3x(2x + 3), because both the number and variable parts are taken out.

How do you factorise by taking out a common factor?

To factorise by taking out a common factor, find the HCF of every term, put it outside a bracket, and divide each term by it. For example, 12x + 8 = 4(3x + 2) and 6x² + 9x = 3x(2x + 3).

How do you simplify 5x + 3 - 2x + 7?

5x + 3 - 2x + 7 simplifies to 3x + 10. The x-terms give 5x - 2x = 3x and the number terms give 3 + 7 = 10, and no further simplification is possible.

How do you expand 4(3x - 2)?

4(3x - 2) expands to 12x - 8. The outside 4 multiplies every term inside: 4 × 3x = 12x and 4 × (-2) = -8, keeping the negative sign.

How do you expand 3x(2x + 5)?

3x(2x + 5) expands to 6x² + 15x. Multiplying 3x × 2x gives 6x² and 3x × 5 gives 15x, and these are unlike terms so no further collecting is possible.

How do you factorise 12x + 8?

12x + 8 factorises to 4(3x + 2). The HCF of 12 and 8 is 4, and dividing each term by 4 gives 3x and 2; expanding 4(3x + 2) back to 12x + 8 checks the answer.

How do you factorise 6x² + 9x?

6x² + 9x factorises to 3x(2x + 3). The HCF of the coefficients is 3 and the HCF of x² and x is x, so the full HCF is 3x, and dividing gives 2x and 3.

How do you expand brackets with a negative number outside?

When a negative number sits outside a bracket, it multiplies every term inside, including the sign. In 4(x + 3) - 2(x + 1), the -2 gives -2x - 2, not -2x + 1, so writing out the expansion in full before collecting is safest.

Are x and x² like terms?

x and x² are not like terms, because like terms must have identical variable parts with the same letters raised to the same powers. So 4x and -7x are like terms, but 4x and 4x² are not and cannot be combined.

What is the coefficient of -a?

The coefficient of -a is -1, because -a means -1a. So 4a - a = 4a + (-1a) = 3a when collecting like terms.

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