Expanding & Factorising Brackets

Master expanding single and double brackets, and learn to reverse the process through factorisation

GCSE Algebra
Expanding & Factorising Brackets – cover
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Introduction

Brackets are one of the most powerful and essential tools in algebra. They allow us to group parts of an expression together, to control the order in which operations are carried out, and to build more complicated structures step by step. Whenever you see a bracket in algebra, it is a signal that something inside must be multiplied, simplified, or carefully managed. Two of the most important skills that every GCSE student needs are expanding brackets and factorising brackets. These two skills are opposite processes: expanding removes the brackets by multiplying, while factorising introduces brackets by collecting terms together. Understanding both directions is vital, because mathematics often requires moving back and forth between expanded and factorised forms.

For example, suppose you are asked to expand: \[ 3(x + 4) = 3x + 12 \] Here the bracket disappears and the expression becomes longer. Later, you might meet an expression such as: \[ 12x + 18 = 6(2x + 3) \] This is factorised form: we have turned a long expression back into a product of simpler parts. These two skills are not only useful in isolation but also play a central role in many GCSE topics including solving equations, simplifying algebraic fractions, and working with quadratic graphs.

The importance of brackets extends far beyond school mathematics. In engineering, algebraic expressions are used to calculate stresses, forces, and loads, often requiring expansion to simplify results. In computer science, algorithms and formulas frequently involve expressions that need to be rewritten for efficiency. In finance, compound interest formulas can be expanded or factorised depending on what we are solving for. And in physics, the equations of motion often involve quadratics which must be factorised to find meaningful solutions. Brackets are, quite literally, everywhere in mathematics and its applications.

In this tutorial we will:

  • Define the key vocabulary needed to understand expansion and factorisation.
  • Explain in detail how to expand single and double brackets, including negative signs and fractional coefficients.
  • Show step-by-step methods for factorising expressions, including quadratics and special cases like the difference of two squares.
  • Provide worked examples to illustrate the techniques clearly.
  • Highlight common mistakes and how to avoid them.
  • Offer a wide range of practice and challenge questions to consolidate learning.

By the end of this tutorial you should feel confident in moving between expanded and factorised forms of algebraic expressions, understanding why the two skills are connected, and applying them successfully in both exam-style questions and real-life contexts.

Key Vocabulary

Before we dive into the techniques of expanding and factorising, it is essential to understand the key words and phrases that appear again and again. Having clear definitions helps avoid confusion and ensures that you know exactly what each question is asking.

  • Expression: A mathematical statement made up of numbers, variables (letters), and operation symbols, but without an equals sign. For example, \(3x + 7\) and \((x + 2)(x - 5)\) are expressions.
  • Term: A part of an expression separated by + or − signs. In \(4x^2 + 3x - 5\), there are three terms: \(4x^2\), \(3x\), and \(-5\).
  • Coefficient: The number multiplying a variable. In \(7y\), the coefficient is 7. If there is no visible number, the coefficient is 1 (as in \(x = 1x\)).
  • Bracket: A grouping symbol, usually round brackets \(()\), used to show that terms inside should be multiplied or treated as one unit. Example: \(2(x + 5)\).
  • Expand: To remove brackets by multiplying out. Example: \(3(x + 4) = 3x + 12\).
  • Factorise: To put an expression back into brackets by finding common factors. Example: \(12x + 18 = 6(2x + 3)\).
  • Common factor: A number or expression that divides evenly into all terms. Example: in \(15x + 20\), the common factor is 5.
  • Quadratic: An expression containing \(x^2\) (a square term). Example: \(x^2 + 5x + 6\).
  • Difference of two squares: A special quadratic pattern of the form \(a^2 - b^2\), which factorises as \((a - b)(a + b)\).

We will use these words repeatedly throughout the tutorial. Make sure you are comfortable with them, as they form the language of algebra.

Expanding Brackets

Expanding (or “multiplying out”) brackets means using the distributive law to remove brackets and write an equivalent expression without them. The distributive law works for any numbers or variables: \[ a(b + c) = ab + ac \quad\text{and}\quad a(b - c) = ab - ac. \] You must multiply the term outside the bracket by every term inside the bracket, then simplify by collecting like terms.

1) Single Brackets (Distributive Law)

When a single term (a “monomial”) sits outside a bracket, multiply it by each term inside.

  • Example A (positive outside): \(3(x + 5)\)
    \(= 3\cdot x + 3\cdot 5 = 3x + 15\).
  • Example B (negative outside): \(-2(4x - 7)\)
    \(= -2\cdot 4x + (-2)\cdot(-7) = -8x + 14\).
  • Example C (variable outside): \(3x(2x - 5)\)
    \(= 3x\cdot 2x + 3x\cdot(-5) = 6x^2 - 15x\).
  • Example D (fraction/decimal outside): \(\tfrac{1}{2}x(6x + 8)\)
    \(= \tfrac{1}{2}x\cdot 6x + \tfrac{1}{2}x\cdot 8 = 3x^2 + 4x\).
    Or \(0.4(5x - 3) = 2x - 1.2\).
  • Example E (minus sign only): \(-(x - 4)\)
    Think of “\(-1\)” outside: \((-1)\cdot x + (-1)\cdot(-4) = -x + 4\).

A common exam pattern is a number or variable before a bracket that starts with a minus: \[ 7 - (2x - 3) = 7 + \big[-(2x - 3)\big] = 7 - 2x + 3 = 10 - 2x. \] Here the minus in front flips the signs of everything inside.

Sign rules reminder: \(+\times+=+\), \(+\times-= -\), \(-\times+= -\), \(-\times-=+\).

2) Double Brackets (Two Binomials)

When two brackets are multiplied, such as \((ax + b)(cx + d)\), each term in the first bracket multiplies each term in the second. Many students use FOIL (First, Outer, Inner, Last), but it’s simply the distributive law done twice. You can also use the grid (area) method to organise the products.

Method 1: Distributive (term-by-term)

\((x + 3)(x + 5)\):

  1. Multiply \(x\) by the second bracket: \(x(x + 5) = x^2 + 5x\).
  2. Multiply \(+3\) by the second bracket: \(3(x + 5) = 3x + 15\).
  3. Add and collect like terms: \(x^2 + 5x + 3x + 15 = x^2 + 8x + 15\).

Method 2: FOIL (a memory aid)

  • First: \(x\cdot x = x^2\)
  • Outer: \(x\cdot 5 = 5x\)
  • Inner: \(3\cdot x = 3x\)
  • Last: \(3\cdot 5 = 15\)

Sum: \(x^2 + 8x + 15\).

Method 3: Grid (area) model

Lay out a 2×2 grid with the terms of each bracket labelling rows/columns. Fill each cell with the product, then add them:

  • Cells: \(x\cdot x = x^2\), \(x\cdot 5 = 5x\), \(3\cdot x = 3x\), \(3\cdot 5 = 15\).
  • Add: \(x^2 + 5x + 3x + 15 = x^2 + 8x + 15\).

Careful, signs!

  • Example F: \((2x - 3)(x + 4)\)
    \(= 2x\cdot x + 2x\cdot 4 + (-3)\cdot x + (-3)\cdot 4\)
    \(= 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12\).
  • Example G: \((x - 6)(x - 2)\)
    \(= x^2 - 2x - 6x + 12 = x^2 - 8x + 12\).
  • Example H (mixed signs/coefficients): \((3x + 2)(-2x + 7)\)
    \(= 3x\cdot(-2x) + 3x\cdot 7 + 2\cdot(-2x) + 2\cdot 7\)
    \(= -6x^2 + 21x - 4x + 14 = -6x^2 + 17x + 14\).

Fractions and decimals inside brackets

  • Example I (fractions): \(\big(\tfrac{1}{3}x + 2\big)(6x - 3)\)
    \(=\tfrac{1}{3}x\cdot 6x + \tfrac{1}{3}x\cdot(-3) + 2\cdot 6x + 2\cdot(-3)\)
    \(= 2x^2 - x + 12x - 6 = 2x^2 + 11x - 6\).
  • Example J (decimals): \((0.5x - 0.2)(4x + 6)\)
    \(= 0.5x\cdot 4x + 0.5x\cdot 6 + (-0.2)\cdot 4x + (-0.2)\cdot 6\)
    \(= 2x^2 + 3x - 0.8x - 1.2 = 2x^2 + 2.2x - 1.2\).

Why FOIL works (and when it doesn’t)

FOIL is not a new rule: it’s the distributive law applied to two-term brackets. It only applies directly to binomials (two-term brackets). For trinomials or more complicated brackets, revert to full distribution or the grid method. The grid scales easily (2×3, 3×3, etc.) and prevents missed terms.

3) Expand then Simplify: Collecting Like Terms

After multiplying, always combine like terms (same variable and power) and put the result in descending powers of the variable. This gives a neat, standard form of the expanded expression.

  • Example K: \((x + 4)(2x + 1)\)
    \(= x\cdot 2x + x\cdot 1 + 4\cdot 2x + 4\cdot 1\)
    \(= 2x^2 + x + 8x + 4 = 2x^2 + 9x + 4\).
  • Example L (multiple like terms): \((2x - 3)(x - 1)\)
    \(= 2x^2 - 2x - 3x + 3 = 2x^2 - 5x + 3\).

4) Quality Checks (catch errors quickly)

  • Degree check: Multiplying two linear brackets should give a quadratic (highest power \(x^2\)). If you don’t see an \(x^2\) term, something went wrong.
  • Symmetry check: If the brackets are identical \((x+a)(x+a)\), the middle term should be \(2ax\) and the constant \(a^2\).
  • Substitution check: Pick an easy value (e.g. \(x=1\)) and evaluate both the original product and your expansion. They must match.

5) Typical Pitfalls (and fixes)

  • Missing a term: Students often forget the inner or outer product in double brackets. Use the grid or a checklist (F–O–I–L) to ensure all products are included.
  • Sign slips: Treat minus signs as “stickers” on the terms they precede; carry them into the product. Write intermediate steps rather than doing it all in your head.
  • Not simplifying: Always collect like terms. Leaving \(x^2 + 3x + 5x + 4\) instead of \(x^2 + 8x + 4\) will cost marks.
  • Overusing FOIL: FOIL is fine for binomials, but use distribution or a grid for anything more complex.

With careful distribution, organised working (FOIL or grid), and a final tidy-up (collecting like terms), you can expand any bracketed expression accurately and confidently.

Factorising Expressions

Factorising is the reverse process of expanding. Instead of multiplying brackets out, we rewrite an expression by putting brackets back in. It is one of the most important algebraic skills because it allows us to simplify, to solve equations, and to reveal the structure of expressions. Whenever you solve a quadratic equation by setting factors equal to zero, you are using factorisation. Think of it as “compressing” an expression into a more compact, product form.

1) Taking Out a Common Factor

The simplest type of factorisation is removing a common factor shared by all terms.

  • Example A: \(12x + 18\). The highest common factor (HCF) of 12 and 18 is 6. Factorise: \(12x + 18 = 6(2x + 3)\).
  • Example B: \(15y^2 - 20y\). HCF is 5y. Factorise: \(15y^2 - 20y = 5y(3y - 4)\).
  • Example C: \(7a^3b - 14a^2b^2\). Common factor is \(7a^2b\). Factorise: \(= 7a^2b(a - 2b)\).

Always check that every term inside the bracket has no remaining common factor; if so, factorise further.

2) Factorising Quadratics (x² + bx + c)

Quadratics of the form \(x^2 + bx + c\) are very common. The method is to look for two numbers that multiply to \(c\) and add to \(b\).

  • Example D: \(x^2 + 5x + 6\). Numbers that multiply to 6 and add to 5 → 2 and 3. Factorise: \((x + 2)(x + 3)\).
  • Example E: \(x^2 - 7x + 10\). Numbers that multiply to 10 and add to -7 → -5 and -2. Factorise: \((x - 5)(x - 2)\).

If no such pair exists, the quadratic may not factorise neatly with integers. In those cases, we use the quadratic formula or completing the square (topics in higher algebra).

3) Factorising Quadratics (ax² + bx + c)

When the coefficient of \(x^2\) is not 1, the process is slightly harder. A systematic way is to use the “product-sum” method:

  1. Multiply the coefficient of \(x^2\) (a) by the constant (c).
  2. Find two numbers that multiply to \(a \times c\) and add to \(b\).
  3. Split the middle term using these numbers.
  4. Group into two pairs and factorise each group.
  5. Take out the common bracket.
  • Example F: \(2x^2 + 7x + 3\). Step 1: \(a \times c = 2 \times 3 = 6\). Step 2: Numbers that multiply to 6 and add to 7 → 6 and 1. Step 3: Rewrite as \(2x^2 + 6x + x + 3\). Step 4: Group: \((2x^2 + 6x) + (x + 3)\). Step 5: Factorise: \(2x(x + 3) + 1(x + 3)\). Step 6: Final: \((2x + 1)(x + 3)\).
  • Example G: \(3x^2 - 11x - 4\). Step 1: \(a \times c = 3 \times -4 = -12\). Step 2: Numbers that multiply to -12 and add to -11 → -12 and 1. Step 3: Rewrite as \(3x^2 - 12x + x - 4\). Step 4: Group: \((3x^2 - 12x) + (x - 4)\). Step 5: Factorise: \(3x(x - 4) + 1(x - 4)\). Step 6: Final: \((3x + 1)(x - 4)\).

4) Special Cases

Some quadratics have patterns you can learn to recognise immediately.

  • Perfect square trinomial: Expressions like \(x^2 + 6x + 9\). Factorises as \((x + 3)(x + 3) = (x + 3)^2\). General form: \(x^2 + 2ax + a^2 = (x + a)^2\).
  • Difference of two squares: Expressions like \(x^2 - 16\). Recognise as \(a^2 - b^2 = (a - b)(a + b)\). Example: \(x^2 - 16 = (x - 4)(x + 4)\).

5) Factorising by Grouping

Sometimes four-term expressions can be factorised by grouping into two pairs.

  • Example H: \(ab + ac + xb + xc\). Group: \((ab + ac) + (xb + xc)\). Factorise: \(a(b + c) + x(b + c)\). Final: \((a + x)(b + c)\).
  • Example I: \(2y^2 + 6y + yx + 3x\). Group: \((2y^2 + 6y) + (yx + 3x)\). Factorise: \(2y(y + 3) + x(y + 3)\). Final: \((2y + x)(y + 3)\).

6) Advanced Cases (with negatives and fractions)

Factorising can involve more than whole numbers. Be careful with negatives and fractions.

  • Example J (negative common factor): \(-12x^2 - 18x\). Factor out -6x: \(-6x(2x + 3)\).
  • Example K (fractional coefficients): \(\tfrac{1}{2}x^2 + \tfrac{3}{2}x\). Factor out \(\tfrac{1}{2}x\): \(\tfrac{1}{2}x(x + 3)\).

7) Why Factorise?

Factorisation is not just an exercise: it has important purposes.

  • Solving equations: If \(x^2 + 5x + 6 = 0\), factorising gives \((x + 2)(x + 3) = 0\). Therefore \(x = -2\) or \(x = -3\).
  • Simplifying fractions: \(\frac{x^2 - 9}{x - 3}\) simplifies to \(x + 3\) once factorised.
  • Graphing: Factorised form shows the roots (x-intercepts) of a quadratic graph.
  • Recognising patterns: Difference of squares and perfect squares appear frequently in algebra and beyond.

Factorisation is therefore a gateway skill: it opens the door to solving real problems in equations, functions, and modelling.

Worked Examples

Now that we understand the methods for expanding and factorising, let’s look at some fully worked examples. Each solution shows the reasoning step by step. Pay close attention to the structure, because in exams you are often awarded marks for the method, not just the final answer.

Expanding Brackets

Example 1: Single bracket
Expand \(4(x + 7)\).
Solution: Multiply 4 by each term inside the bracket: \(4 \times x = 4x\), \(4 \times 7 = 28\). Final answer: \(4x + 28\).

Example 2: With a negative
Expand \(-3(2x - 5)\).
Solution: Multiply -3 by each term: \(-3 \times 2x = -6x\), \(-3 \times -5 = +15\). Final answer: \(-6x + 15\).

Example 3: Double brackets
Expand \((x + 3)(x + 8)\).
Solution: First: \(x \times x = x^2\). Outer: \(x \times 8 = 8x\). Inner: \(3 \times x = 3x\). Last: \(3 \times 8 = 24\). Add together: \(x^2 + 8x + 3x + 24 = x^2 + 11x + 24\).

Example 4: Double brackets with negatives
Expand \((2x - 5)(x - 4)\).
Solution: First: \(2x \imes x = 2x^2\). Outer: \(2x \times -4 = -8x\). Inner: \(-5 \times x = -5x\). Last: \(-5 \times -4 = +20\). Add together: \(2x^2 - 8x - 5x + 20 = 2x^2 - 13x + 20\).

Example 5: Fractional coefficients
Expand \(\tfrac{1}{2}(x - 6)(2x + 3)\).
Solution: First expand the brackets: \((x - 6)(2x + 3) = 2x^2 + 3x - 12x - 18 = 2x^2 - 9x - 18\). Now multiply everything by \(\tfrac{1}{2}\): \(\tfrac{1}{2}(2x^2 - 9x - 18) = x^2 - 4.5x - 9\).

Factorising Expressions

Example 6: Taking out a common factor
Factorise \(18x + 24\).
Solution: The HCF of 18 and 24 is 6. Factor out 6: \(18x + 24 = 6(3x + 4)\).

Example 7: Quadratic with coefficient 1
Factorise \(x^2 + 9x + 20\).
Solution: Find two numbers that multiply to 20 and add to 9 → 5 and 4. So, \(x^2 + 9x + 20 = (x + 5)(x + 4)\).

Example 8: Quadratic with coefficient > 1
Factorise \(2x^2 + 11x + 5\).
Solution: Multiply \(a \times c = 2 \times 5 = 10\). Numbers that multiply to 10 and add to 11 → 10 and 1. Rewrite: \(2x^2 + 10x + x + 5\). Group: \((2x^2 + 10x) + (x + 5)\). Factorise: \(2x(x + 5) + 1(x + 5)\). Final: \((2x + 1)(x + 5)\).

Example 9: Difference of two squares
Factorise \(x^2 - 49\).
Solution: Recognise as \(a^2 - b^2\). Here \(a = x\), \(b = 7\). So, \(x^2 - 49 = (x - 7)(x + 7)\).

Example 10: Grouping method
Factorise \(3ab + 6a + 2b + 4\).
Solution: Group terms: \((3ab + 6a) + (2b + 4)\). Factorise each: \(3a(b + 2) + 2(b + 2)\). Final: \((3a + 2)(b + 2)\).

These examples show the main types of problems you will face. By practising them regularly, you will gain confidence in deciding which technique to use and how to apply it step by step.

Common Mistakes

Even when students understand the rules, there are common traps that lead to lost marks. Recognising these mistakes and learning how to avoid them is just as important as learning the correct methods.

1) Forgetting to Multiply Every Term

Mistake: \(2(x + 3) = 2x + 3\) Correct: \(2(x + 3) = 2x + 6\). Remember: the 2 multiplies every term in the bracket, not just the first one.

2) Sign Errors

Mistake: \(-3(x - 4) = -3x - 4\) Correct: \(-3(x - 4) = -3x + 12\). The negative must multiply through correctly: \(-3 \times -4 = +12\).

3) Mixing Up Expanding and Factorising

Mistake: Writing \(3x + 6\) as \(3x(x + 2)\). Correct: \(3x + 6 = 3(x + 2)\). Here the factor taken out should be 3, not 3x. Always check by expanding back — if it doesn’t match, something is wrong.

4) Wrong Middle Term in Quadratics

Mistake: \(x^2 + 7x + 12 = (x + 3)(x + 4)\). Correct: \((x + 3)(x + 4) = x^2 + 7x + 12\) is correct, but many students guess factors wrongly. For example, writing \((x + 2)(x + 6) = x^2 + 8x + 12\) gives the wrong middle term. Always check by expanding to verify.

5) Incorrect Difference of Squares

Mistake: \(x^2 + 9 = (x + 3)(x - 3)\). Correct: \(x^2 - 9 = (x + 3)(x - 3)\). The formula only works for subtraction, not addition.

6) Not Simplifying After Expansion

Mistake: \((x + 2)(x + 3) = x^2 + 2x + 3x + 6\). Correct: \(= x^2 + 5x + 6\). Always collect like terms for a clean final answer.

The safest habit is to double-check your result by re-expanding the factorised form or by substituting a small value for \(x\) into both the original and your answer. If they don’t match, an error has crept in.

Practice Questions

Try these practice questions to check your understanding. Work carefully and always check your answers by re-expanding or re-factorising. Do not look at the answers straight away — attempt them first to build confidence.

A) Expanding Brackets

  1. Expand: \(6(x + 4)\)
  2. Expand: \(-5(y - 7)\)
  3. Expand: \(2x(3x - 8)\)
  4. Expand: \((x + 5)(x + 9)\)
  5. Expand: \((x - 6)(x + 2)\)
  6. Expand: \((2x + 3)(x - 4)\)
  7. Expand: \((3x - 1)(x - 7)\)
  8. Expand: \((4x + 2)(2x + 5)\)
  9. Expand: \((-x + 3)(x - 4)\)
  10. Expand: \((x + 10)(x - 10)\)

B) Factorising Expressions

  1. Factorise: \(14x + 21\)
  2. Factorise: \(20y^2 - 15y\)
  3. Factorise: \(x^2 + 11x + 24\)
  4. Factorise: \(x^2 - 13x + 42\)
  5. Factorise: \(2x^2 + 7x + 3\)
  6. Factorise: \(3x^2 + 10x + 7\)
  7. Factorise: \(x^2 - 36\)
  8. Factorise: \(4x^2 - 25\)
  9. Factorise: \(x^2 + 16x + 64\)
  10. Factorise: \(5x^2 - 20x\)

Challenge Questions

These are more demanding problems that require careful reasoning. They connect expansion and factorisation to real contexts or more complex algebra.

  1. A rectangle has sides \((x + 4)\) and \((x + 7)\). Write an expression for its area and expand it.
  2. Expand and simplify: \((x + 3)(x + 2)(x - 5)\).
  3. Factorise completely: \(6x^2 + 9x - 15\).
  4. Expand: \((2x - 3)^2\).
  5. Factorise: \(9x^2 - 16\).
  6. Simplify: \(\frac{x^2 - 25}{x - 5}\).
  7. The product of two consecutive integers is \(x(x+1)\). Expand this expression.
  8. The difference between the squares of two numbers is written as \(a^2 - b^2\). Factorise it.
  9. Solve by factorising: \(x^2 + 7x + 10 = 0\).
  10. Solve by factorising: \(2x^2 - 5x - 3 = 0\).

Conclusion

Expanding and factorising brackets are two of the most important and widely used algebraic skills in GCSE Mathematics. They are exact opposites: expanding multiplies out brackets to create a longer expression, while factorising compresses a long expression back into bracket form. Understanding both directions is essential, because many exam questions require you to move from one form to the other with confidence.

Expansion relies on the distributive law: each term outside a bracket must multiply every term inside. By practising single brackets, double brackets, and more challenging cases with negatives or fractions, you develop accuracy and fluency. Factorisation is the reverse process, where we look for common factors or apply special patterns such as quadratics or the difference of two squares. Both skills work hand in hand — checking one process against the other is one of the best ways to avoid errors.

Beyond exams, these techniques are used in real applications: from calculating areas in geometry to solving equations in science and engineering. By mastering them now, you are building a foundation that will support further studies in mathematics, physics, computing, and many other fields. Keep practising until the steps feel natural, and always double-check your results.

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