OpenRevise GCSE

Viewing as

Student Teacher

Teacher view: planning notes, the answers to every question, and the teacher copies of the files.

Maths · Algebra

More expanding and factorising

Three brackets, quadratics that start with a number in front of the squared term, and the difference of two squares with coefficients - plus using factorising to simplify algebraic fractions and to prove facts about every number at once.

  • Higher
  • 5 key terms
  • All boards
Download the teacher pack · 5 files

Teacher resources

The teacher copies: slides with the questions built in, the answers, and anything else attached to this lesson for whoever is teaching it.

Student handouts

The same files the students see, to print or hand out.

From Earlier Lessons

Answer each one, then check.

  1. 1

    Expand and simplify \((x + 3)(x + 5)\).

    Show answerHide answer

    \(x^2 + 8x + 15\)

  2. 2

    Factorise \(x^2 + 5x + 6\).

    Show answerHide answer

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

  3. 3

    Factorise \(x^2 - 25\).

    Show answerHide answer

    \((x + 5)(x - 5)\)

  4. 4

    Simplify \(6x^2 \div 2x\).

    Show answerHide answer

    \(3x\)

  5. 5

    Work out \(7^2 - 3^2\).

    Show answerHide answer

    \(49 - 9 = 40\)

Learning Objectives

  1. 1Expand the product of three binomials.
  2. 2Factorise quadratics of the form \(ax^2 + bx + c\).
  3. 3Factorise the difference of two squares with coefficients, such as \(9x^2 - 16y^2\).
  4. 4Simplify algebraic fractions by factorising.
  5. 5Use expanding and factorising in algebraic proof.

Expanding Three Brackets

Expand and simplify \((x + 1)(x + 2)(x + 3)\)

Show the solutionHide the solution
  1. 1 Expand the first two brackets \((x + 1)(x + 2) = x^2 + 3x + 2\)
  2. 2 Multiply every term by \(x\) \(x(x^2 + 3x + 2) = x^3 + 3x^2 + 2x\)
  3. 3 Multiply every term by \(+3\) \(3(x^2 + 3x + 2) = 3x^2 + 9x + 6\)
  4. 4 Collect like terms \(x^3 + 6x^2 + 11x + 6\)

Answer\(x^3 + 6x^2 + 11x + 6\)

Three Brackets with Negatives

Expand and simplify \((x - 2)(x + 3)(2x - 1)\)

Show the solutionHide the solution
  1. 1 Expand the first two \((x - 2)(x + 3) = x^2 + x - 6\)
  2. 2 Multiply by \(2x\) \(2x^3 + 2x^2 - 12x\)
  3. 3 Multiply by \(-1\) \(-x^2 - x + 6\)
  4. 4 Collect like terms \(2x^3 + x^2 - 13x + 6\)

Answer\(2x^3 + x^2 - 13x + 6\) (check \(x = 1\): \((-1)(4)(1) = -4\) and \(2 + 1 - 13 + 6 = -4\))

The Difference of Two Squares, Again

\(a^2 - b^2 = (a + b)(a - b)\) works for any two squares, not just \(x^2\) and a number.

  • With coefficients

    \(9x^2 - 16y^2 = (3x)^2 - (4y)^2 = (3x + 4y)(3x - 4y)\).

  • Take out a factor first

    \(2x^2 - 18 = 2(x^2 - 9) = 2(x + 3)(x - 3)\).

  • With numbers

    \(101^2 - 99^2 = (101 + 99)(101 - 99) = 200 \times 2 = 400\) - no squaring needed.

  • Watch for it

    \(50 - 2x^2 = 2(25 - x^2) = 2(5 + x)(5 - x)\).

Splitting the Middle Term

A reliable method for any quadratic \(ax^2 + bx + c\).

  1. 1 Multiply

    Work out \(a \times c\).
  2. 2 Find the pair

    Two numbers that multiply to \(ac\) and add to \(b\).
  3. 3 Split

    Write \(bx\) as the sum of those two \(x\) terms.
  4. 4 Pairs

    Factorise the first two terms and the last two terms.
  5. 5 Common bracket

    Take out the bracket they share.

Factorising a Quadratic with a Coefficient

Factorise \(6x^2 + 11x - 10\)

Show the solutionHide the solution
  1. 1 \(a \times c\) \(6 \times (-10) = -60\)
  2. 2 Two numbers that multiply to \(-60\) and add to 11 \(15\) and \(-4\)
  3. 3 Split the middle term \(6x^2 + 15x - 4x - 10\)
  4. 4 Factorise in pairs \(3x(2x + 5) - 2(2x + 5)\)
  5. 5 Take out the common bracket \((2x + 5)(3x - 2)\)

Answer\((2x + 5)(3x - 2)\) (check: \(6x^2 - 4x + 15x - 10\))

Simplifying Algebraic Fractions

Factorise the top and the bottom, then cancel any bracket they share.

  • Cancel brackets, not terms

    In \(\dfrac{x + 3}{x + 5}\), nothing cancels: the 3 and the 5 are not factors.

  • Factorise everything

    Take out common factors, use the difference of two squares, factorise quadratics.

  • Then cancel

    A bracket on the top and the bottom divides to 1.

Simplifying an Algebraic Fraction

Simplify fully \(\dfrac{x^2 - 9}{2x^2 + 5x - 3}\)

Show the solutionHide the solution
  1. 1 Top: difference of two squares \(x^2 - 9 = (x + 3)(x - 3)\)
  2. 2 Bottom: \(ac = -6\); 6 and \(-1\) multiply to \(-6\) and add to 5 \(2x^2 + 6x - x - 3 = (x + 3)(2x - 1)\)
  3. 3 Write it factorised \(\dfrac{(x + 3)(x - 3)}{(x + 3)(2x - 1)}\)
  4. 4 Cancel \((x + 3)\) \(\dfrac{x - 3}{2x - 1}\)

Answer\(\dfrac{x - 3}{2x - 1}\)

Algebraic Proof

Algebra can prove something is true for every number at once.

  • Even and odd

    An even number is \(2n\); an odd number is \(2n + 1\), where \(n\) is an integer.

  • Consecutive numbers

    \(n\), \(n + 1\), \(n + 2\), ...

  • The method

    Write the expression, expand and simplify, then factorise to show the property.

  • Example

    \((n + 3)^2 - (n + 1)^2 = n^2 + 6n + 9 - n^2 - 2n - 1 = 4n + 8 = 4(n + 2)\), which is always a multiple of 4.

Match the Quadratic to Its Factors

  • \(2x^2 + 7x + 3\)

    \((2x + 1)(x + 3)\)

  • \(3x^2 - 5x - 2\)

    \((3x + 1)(x - 2)\)

  • \(4x^2 - 25\)

    \((2x + 5)(2x - 5)\)

  • \(2x^2 - 8\)

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

  • \(6x^2 + x - 1\)

    \((3x - 1)(2x + 1)\)

  • \(x^3 - 4x\)

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

Factorise and Cancel

(a) Factorise \(2x^2 + 9x + 4\). (b) Factorise \(3x^2 - 10x + 8\). (c) Factorise fully \(3x^2 - 75\). (d) Simplify \(\dfrac{x^2 + 3x - 10}{x^2 - 4}\). (e) Expand \((x + 2)^3\). (f) Prove that the sum of three consecutive integers is always a multiple of 3.

1. Look for a common factor first.

2. Then the difference of two squares or splitting the middle term.

3. Check by expanding.

A good answer shows: (a) \((2x + 1)(x + 4)\) (b) \((3x - 4)(x - 2)\) (c) \(3(x + 5)(x - 5)\) (d) \(\dfrac{(x + 5)(x - 2)}{(x + 2)(x - 2)} = \dfrac{x + 5}{x + 2}\) (e) \(x^3 + 6x^2 + 12x + 8\) (f) \(n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)\)

Can I...?

  1. 1Expand three brackets.
  2. 2Factorise \(ax^2 + bx + c\) by splitting the middle term.
  3. 3Factorise the difference of two squares with coefficients.
  4. 4Take out a common factor before factorising.
  5. 5Simplify algebraic fractions.
  6. 6Use \(2n\) and \(2n + 1\) for even and odd numbers.
  7. 7Write an algebraic proof.
  8. 8Check a factorisation by expanding.

Summary & Exam Focus

  • Three brackets: expand two, then multiply every term by the third.
  • \(ax^2 + bx + c\): find two numbers that multiply to \(ac\) and add to \(b\), split, factorise in pairs.
  • \(a^2 - b^2 = (a + b)(a - b)\), with any squares; take out common factors first.
  • Algebraic fractions: factorise top and bottom, then cancel brackets.

Exam focus

Factorise \(3x^2 + 10x + 8\) (2 marks) (2 marks)

In a proof, finish with a sentence: "\(4(n + 2)\) is a multiple of 4, so the expression is always a multiple of 4". The last mark is for saying what you have shown.

Key terms

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

Binomial
An expression with two terms, such as \((x + 3)\).
Product of three binomials
Three brackets multiplied, such as \((x + 1)(x + 2)(x + 3)\).
Splitting the middle term
Writing \(bx\) as two terms so that \(ax^2 + bx + c\) can be factorised in pairs.
Algebraic fraction
A fraction with algebra on the top, the bottom or both.
Proof
An argument showing something is true in every case, not just the ones tested.

Questions and answers

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

1. Exam question Non-calculator 2 marks Easier

Factorise \(3x^2 + 10x + 8\)

Mark scheme — 2 marks available

  • \((3x \pm 4)(x \pm 2)\), or a correct split of the middle term — M1
  • \((3x + 4)(x + 2)\) — A1

Model answer

\(ac = 24\); 6 and 4 multiply to 24 and add to 10. \(3x^2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (3x + 4)(x + 2)\)

2. Exam question Non-calculator 3 marks Easier

Expand and simplify \((x + 4)(x - 1)(x + 2)\)

Mark scheme — 3 marks available

  • Two brackets expanded correctly — M1
  • At least 6 correct terms from multiplying by the third bracket — M1
  • \(x^3 + 5x^2 + 2x - 8\) — A1

Model answer

\((x + 4)(x - 1) = x^2 + 3x - 4\). \((x^2 + 3x - 4)(x + 2) = x^3 + 2x^2 + 3x^2 + 6x - 4x - 8 = x^3 + 5x^2 + 2x - 8\)

3. Exam question Non-calculator 2 marks Easier

Factorise fully \(50 - 2x^2\)

Mark scheme — 2 marks available

  • \(2(25 - x^2)\), or \((10 + 2x)(5 - x)\) — M1
  • \(2(5 + x)(5 - x)\) — A1

Model answer

\(2(25 - x^2) = 2(5 + x)(5 - x)\)

4. Exam question Non-calculator 3 marks Easier

Simplify fully \(\dfrac{x^2 + 3x - 10}{x^2 - 4}\)

Mark scheme — 3 marks available

  • \((x + 5)(x - 2)\) — M1
  • \((x + 2)(x - 2)\) — M1
  • \(\dfrac{x + 5}{x + 2}\) — A1

Model answer

\(x^2 + 3x - 10 = (x + 5)(x - 2)\) and \(x^2 - 4 = (x + 2)(x - 2)\). Cancelling \((x - 2)\): \(\dfrac{x + 5}{x + 2}\).

5. Exam question Non-calculator · Prove 3 marks Easier

Prove that \((2n + 1)^2 - (2n - 1)^2\) is a multiple of 8 for all positive integer values of \(n\).

Mark scheme — 3 marks available

  • One bracket expanded correctly — M1
  • \(8n\) — A1
  • A concluding statement: \(8n\) is a multiple of 8 — C1

Model answer

\((2n + 1)^2 = 4n^2 + 4n + 1\) and \((2n - 1)^2 = 4n^2 - 4n + 1\). Subtracting: \(8n\). \(8n = 8 \times n\), so it is always a multiple of 8.

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

The diagram shows a square of side \((2x + 3)\) cm with a square of side \((x + 1)\) cm cut from one corner. Show that the shaded area, in cm², can be written as \((3x + 4)(x + 2)\).

A square of side 2x plus 3 cm with a square of side x plus 1 cm cut from one corner, leaving an L-shape.

Mark scheme — 3 marks available

  • \((2x + 3)^2 - (x + 1)^2\) — M1
  • Uses the difference of two squares, or expands both to reach \(3x^2 + 10x + 8\) — M1
  • Correctly reaches \((3x + 4)(x + 2)\) — A1

Model answer

Shaded area \(= (2x + 3)^2 - (x + 1)^2\). This is a difference of two squares: \(\big((2x + 3) + (x + 1)\big)\big((2x + 3) - (x + 1)\big) = (3x + 4)(x + 2)\).

7. Multiple choice 1 mark Easier

Factorise \(4x^2 - 9\)

  1. A \((4x + 3)(x - 3)\)
  2. B \((2x - 3)^2\)
  3. C \((2x + 3)(2x - 3)\) Correct
  4. D \((4x + 9)(x - 1)\)

Why: \(4x^2 = (2x)^2\) and \(9 = 3^2\), so \((2x + 3)(2x - 3)\).

8. Multiple choice 1 mark Core

Factorise \(2x^2 + 5x + 3\)

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

Why: \(ac = 6\); 2 and 3 add to 5. \(2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)\).

9. Multiple choice 1 mark Stretch

Simplify \(\dfrac{x^2 - 1}{x + 1}\)

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

Why: \(x^2 - 1 = (x + 1)(x - 1)\). Cancel \((x + 1)\) to leave \(x - 1\).