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Maths · Algebra

Expanding and factorising

Expanding multiplies brackets out; factorising puts them back. They are the same skill run in opposite directions, and between them they turn up in almost every algebra question on the paper.

  • 6 key terms
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Teacher resources

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Student handouts

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Last Lesson and Before

Answer each one, then check.

  1. 1

    Simplify \(3a + 5a - 2a\).

    Show answerHide answer

    \(6a\)

  2. 2

    Work out \(-3 \times -4\).

    Show answerHide answer

    \(12\)

  3. 3

    What is the HCF of 12 and 18?

    Show answerHide answer

    \(6\)

  4. 4

    Last lesson: simplify \(x \times x\).

    Show answerHide answer

    \(x^2\)

  5. 5

    Last lesson: simplify \(2x \times 5x^2\).

    Show answerHide answer

    \(10x^3\)

Learning Objectives

  1. 1Expand single brackets and simplify.
  2. 2Factorise expressions by taking out the highest common factor.
  3. 3Expand double brackets.
  4. 4Factorise quadratic expressions of the form \(x^2 + bx + c\).
  5. 5Factorise the difference of two squares, \(x^2 - a^2\).

The Words You Need

  • Term

    One part of an expression: \(3x\), \(-5\) and \(x^2\) are terms.

  • Expression

    Terms added or subtracted, with no equals sign: \(3x^2 - 5x + 2\).

  • Coefficient

    The number in front of a letter: the 3 in \(3x^2\).

  • Like terms

    Same letters and powers: \(4x\) and \(-7x\), but not \(4x\) and \(4x^2\).

  • Expand

    Multiply out the brackets.

  • Factorise

    Put back into brackets - the opposite of expanding.

Expanding Single Brackets

Multiply every term inside the bracket by the term outside.

  • Numbers

    \(4(3x - 7) = 12x - 28\).

  • Letters

    \(x(x + 5) = x^2 + 5x\).

  • Negative outside

    \(-3(x - 4) = -3x + 12\): a negative times a negative is positive.

  • Expand and simplify

    Expand every bracket, then collect like terms.

Expand and Simplify

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

Show the solutionHide the solution
  1. 1 Expand the first bracket \(5(2x + 3) = 10x + 15\)
  2. 2 Expand the second: the \(-3\) multiplies both terms \(-3(x - 4) = -3x + 12\)
  3. 3 Write it all out \(10x + 15 - 3x + 12\)
  4. 4 Collect like terms \(10x - 3x = 7x\) and \(15 + 12 = 27\)

Answer\(7x + 27\)

Factorising Fully

Factorise fully \(12x^2y - 18xy^2\)

Show the solutionHide the solution
  1. 1 Find the HCF of the numbers HCF of 12 and 18 \(= 6\)
  2. 2 Find the HCF of the letters Both terms have \(x\) and \(y\): \(xy\)
  3. 3 The HCF goes outside \(6xy(\ \ldots\ )\)
  4. 4 Divide each term by the HCF \(12x^2y \div 6xy = 2x\) and \(18xy^2 \div 6xy = 3y\)

Answer\(6xy(2x - 3y)\)

The Grid Method for (x + 4)(x - 3)

Multiply each row by each column, then collect like terms: \(x^2 - 3x + 4x - 12 = x^2 + x - 12\).

  • \(x\)

    x: \(x^2\). −3: \(-3x\)

  • \(+4\)

    x: \(+4x\). −3: \(-12\)

Squaring a Bracket

Expand and simplify \((2x - 5)^2\)

Show the solutionHide the solution
  1. 1 Write it as two brackets \((2x - 5)(2x - 5)\)
  2. 2 First times first \(2x \times 2x = 4x^2\)
  3. 3 Outer and inner \(2x \times (-5) = -10x\) and \((-5) \times 2x = -10x\)
  4. 4 Last times last \((-5) \times (-5) = +25\)
  5. 5 Collect like terms \(4x^2 - 20x + 25\)

Answer\(4x^2 - 20x + 25\) (not \(4x^2 + 25\))

Factorising x² + bx + c

Find two numbers that MULTIPLY to give \(c\) and ADD to give \(b\).

  • The brackets

    \(x^2 + bx + c = (x + p)(x + q)\), where \(pq = c\) and \(p + q = b\).

  • \(c\) positive

    Both numbers have the same sign as \(b\): \(x^2 - 7x + 12 = (x - 3)(x - 4)\).

  • \(c\) negative

    The numbers have opposite signs: \(x^2 + x - 12 = (x + 4)(x - 3)\).

  • Check

    Expand your answer: it should give the expression you started with.

Factorising a Quadratic

Factorise \(x^2 - 2x - 15\)

Show the solutionHide the solution
  1. 1 List pairs that multiply to \(-15\) \(1\) and \(-15\), \(-1\) and \(15\), \(3\) and \(-5\), \(-3\) and \(5\)
  2. 2 Pick the pair that adds to \(-2\) \(3 + (-5) = -2\)
  3. 3 Write the brackets \((x + 3)(x - 5)\)
  4. 4 Check by expanding \(x^2 - 5x + 3x - 15 = x^2 - 2x - 15\)

Answer\((x + 3)(x - 5)\)

The Difference of Two Squares

One square number (or square term) minus another factorises straight away.

  • The rule

    \(x^2 - a^2 = (x + a)(x - a)\).

  • Examples

    \(x^2 - 49 = (x + 7)(x - 7)\). \(x^2 - 1 = (x + 1)(x - 1)\).

  • Why it works

    \((x + 7)(x - 7) = x^2 - 7x + 7x - 49\): the middle terms cancel.

  • Spot it

    Two terms, both square, with a MINUS between them. \(x^2 + 49\) does not factorise.

Match the Expression to Its Factors

  • \(x^2 + 5x + 6\)

    \((x + 2)(x + 3)\)

  • \(x^2 - 9\)

    \((x + 3)(x - 3)\)

  • \(x^2 + x - 6\)

    \((x + 3)(x - 2)\)

  • \(x^2 - 5x + 6\)

    \((x - 2)(x - 3)\)

  • \(4x + 12\)

    \(4(x + 3)\)

  • \(x^2 + 3x\)

    \(x(x + 3)\)

Expand, Then Undo

Work in pairs. Partner A expands (a) \((x + 5)(x + 2)\) (b) \((x - 4)(x + 6)\) (c) \((x - 3)^2\). Partner B factorises (d) \(x^2 + 9x + 14\) (e) \(x^2 - 4x - 21\) (f) \(x^2 - 100\). Then swap papers and check each other's answers by doing the opposite process.

1. Partner A: expand.

2. Partner B: factorise.

3. Swap and check by reversing.

A good answer shows: (a) \(x^2 + 7x + 10\) (b) \(x^2 + 2x - 24\) (c) \(x^2 - 6x + 9\) (d) \((x + 2)(x + 7)\) (e) \((x - 7)(x + 3)\) (f) \((x + 10)(x - 10)\)

Can I...?

  1. 1Expand a single bracket.
  2. 2Expand and simplify two single brackets.
  3. 3Factorise by taking out the HCF.
  4. 4Factorise fully.
  5. 5Expand double brackets.
  6. 6Square a bracket.
  7. 7Factorise \(x^2 + bx + c\).
  8. 8Factorise the difference of two squares.

Summary & Exam Focus

  • Expand: multiply every term inside by every term outside.
  • Factorise: take out the HCF of the numbers AND the letters.
  • Double brackets: four multiplications, then collect like terms.
  • \(x^2 + bx + c\): find two numbers that multiply to \(c\) and add to \(b\).
  • \(x^2 - a^2 = (x + a)(x - a)\).

Exam focus

Factorise \(x^2 + 9x + 20\) (2 marks) (2 marks)

Always check a factorisation by expanding it again. It takes ten seconds and catches sign errors that would cost the mark.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Expand
Multiply out brackets.
Factorise
Write an expression as a product of factors, using brackets.
Factorise fully
Take out the highest common factor, so no common factor is left inside.
Quadratic expression
An expression whose highest power is \(x^2\), such as \(x^2 + 5x + 6\).
Difference of two squares
An expression of the form \(x^2 - a^2\), which factorises to \((x + a)(x - a)\).
Identity
Two expressions that are equal for every value of \(x\), written with \(\equiv\).

Questions and answers

10 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Non-calculator 1 mark Easier

Expand \(4(3x - 7)\)

Mark scheme — 1 mark available

  • \(12x - 28\) — B1

Model answer

\(12x - 28\)

2. Exam question Non-calculator 2 marks Easier

Expand and simplify \(3(2y + 5) - 2(y - 4)\)

Mark scheme — 2 marks available

  • \(6y + 15\) or \(-2y + 8\) correct — M1
  • \(4y + 23\) — A1

Model answer

\(6y + 15 - 2y + 8 = 4y + 23\)

3. Exam question Non-calculator 2 marks Easier

Factorise fully \(15y^2 + 10y\)

Mark scheme — 2 marks available

  • \(5y(3y + 2)\) — B2
  • A correct partial factorisation, e.g. \(5(3y^2 + 2y)\) or \(y(15y + 10)\) — B1

Model answer

\(5y(3y + 2)\)

4. Exam question Non-calculator 2 marks Easier

Expand and simplify \((x + 6)(x - 2)\)

Mark scheme — 2 marks available

  • At least 3 of the 4 terms correct: \(x^2\), \(-2x\), \(+6x\), \(-12\) — M1
  • \(x^2 + 4x - 12\) — A1

Model answer

\(x^2 - 2x + 6x - 12 = x^2 + 4x - 12\)

5. Exam question Non-calculator 2 marks Easier

Factorise \(x^2 + 9x + 20\)

Mark scheme — 2 marks available

  • \((x + 4)(x + 5)\) — B2
  • \((x \pm 4)(x \pm 5)\) — B1

Model answer

\((x + 4)(x + 5)\)

6. Exam question Non-calculator 1 mark Easier

Factorise \(x^2 - 64\)

Mark scheme — 1 mark available

  • \((x + 8)(x - 8)\) — B1

Model answer

\((x + 8)(x - 8)\)

7. Exam question Non-calculator · Show that 3 marks Easier

The diagram shows a rectangle. The length is \((x + 5)\) cm and the width is \((x - 2)\) cm. Show that the area of the rectangle, in cm², is \(x^2 + 3x - 10\).

A rectangle with length x plus 5 cm and width x minus 2 cm.

Mark scheme — 3 marks available

  • Area \(= (x + 5)(x - 2)\) — M1
  • At least 3 of the 4 terms correct — M1
  • Fully correct expansion reaching \(x^2 + 3x - 10\) — A1

Model answer

Area \(= (x + 5)(x - 2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10\)

8. Multiple choice 1 mark Easier

Expand \(3(x - 5)\)

  1. A \(3x - 5\)
  2. B \(3x - 15\) Correct
  3. C \(x - 15\)
  4. D \(3x + 15\)

Why: Multiply both terms by 3: \(3 \times x = 3x\) and \(3 \times -5 = -15\).

9. Multiple choice 1 mark Core

Factorise \(x^2 + 7x + 10\)

  1. A \((x + 2)(x + 5)\) Correct
  2. B \((x + 1)(x + 10)\)
  3. C \((x + 7)(x + 10)\)
  4. D \((x - 2)(x - 5)\)

Why: \(2 \times 5 = 10\) and \(2 + 5 = 7\).

10. Multiple choice 1 mark Stretch

Expand and simplify \((x - 4)^2\)

  1. A \(x^2 + 16\)
  2. B \(x^2 - 16\)
  3. C \(x^2 - 8x - 16\)
  4. D \(x^2 - 8x + 16\) Correct

Why: \((x - 4)(x - 4) = x^2 - 4x - 4x + 16 = x^2 - 8x + 16\).