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Maths · Algebra

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Simplifying and Expanding Expressions

Algebraic notation, collecting like terms, and expanding single brackets, double brackets and squared brackets.

  • 10 key terms
  • All boards

Learning Objectives

  1. 1Use algebraic notation correctly and tell apart an expression, an equation, a formula and an identity.
  2. 2Simplify expressions by collecting like terms and using the laws of indices.
  3. 3Expand single brackets and pairs of brackets, including squared brackets.
  4. 4Expand and simplify expressions such as \((3x - 2)(x + 5)\) without making sign errors.

Algebra is arithmetic with letters

A letter in algebra stands for a number that is unknown or that can change, so every rule of arithmetic still applies. The skills in this lesson, simplifying and expanding, are used in almost every later topic: solving equations, factorising, sequences, graphs and proof. Most of the marks lost in algebra come from sign errors and from missing terms, so the habit of working one careful line at a time pays off straight away.

Algebraic notation

Algebra is written in a shorthand that you must read and write fluently.

  • Multiplication

    The multiplication sign is left out: \(3 \times a = 3a\), \(a \times b = ab\) and \(a \times a = a^2\). The number always goes in front of the letter.

  • Division

    Division is written as a fraction: \(a \div b = \dfrac{a}{b}\) and \(\dfrac{12x}{3} = 4x\).

  • Terms

    A term is a single part of an expression, and it includes the sign in front of it. The expression \(5x^2 - 3x + 7\) has three terms: \(5x^2\), \(-3x\) and \(+7\).

  • Coefficient

    The number multiplying a letter is its coefficient. In \(-3x\) the coefficient of \(x\) is \(-3\).

Expression, equation, formula or identity?

Four words that look similar and mean different things.

  • Expression

    A collection of terms with no equals sign, such as \(3x + 5\). You can simplify or expand it, but you cannot solve it.

  • Equation

    Two expressions joined by an equals sign that is true only for particular values, such as \(3x + 5 = 20\). You solve it to find the unknown.

  • Formula

    A rule that links two or more quantities, such as \(A = lw\) for the area of a rectangle.

  • Identity

    An equation that is true for every value of the letter, such as \(2(x + 3) \equiv 2x + 6\). It is written with the symbol \(\equiv\).

Collecting like terms

Like terms have exactly the same letters with exactly the same powers.

  • Spotting like terms

    \(3x\) and \(5x\) are like terms, and so are \(4x^2\) and \(-x^2\). But \(3x\) and \(3x^2\) are not, and neither are \(3x\) and \(3y\).

  • Collecting

    Add or subtract the coefficients and keep the letters, taking the sign in front of each term with it. \(4a + 3b - a + 5b = 3a + 8b\).

  • Multiplying terms

    Multiply the numbers and add the powers of the same letter. \(3x \times 4x^2 = 12x^3\).

  • Dividing terms

    Divide the numbers and subtract the powers. \(12x^5 \div 3x^2 = 4x^3\).

Collecting like terms

Simplify \(5x^2 + 3x - 2x^2 + 7 - 4x\).

Show the solutionHide the solution
  1. 1 Group the \(x^2\) terms \(5x^2 - 2x^2 = 3x^2\).
  2. 2 Group the \(x\) terms \(3x - 4x = -x\).
  3. 3 Keep the number \(+7\) has no like term, so it stays as it is.
  4. 4 Write the answer \(3x^2 - x + 7\).

Answer\(3x^2 - x + 7\)

Expanding single brackets

Expanding means removing the brackets by multiplying every term inside by the term outside.

  • The basic rule

    \(3(2x - 5) = 3 \times 2x + 3 \times (-5) = 6x - 15\). Every term inside the bracket must be multiplied.

  • Negative outside

    A negative outside changes every sign inside: \(-2(x - 4) = -2x + 8\), not \(-2x - 8\).

  • Letters outside

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

  • Expand and simplify

    Expand each bracket, then collect like terms: \(3(x + 2) + 2(x - 5) = 3x + 6 + 2x - 10 = 5x - 4\).

Brackets with a negative number

Expand and simplify \((x + 7)(x - 4)\).

Show the solutionHide the solution
  1. 1 Multiply every pair \(x \times x = x^2\), \(x \times (-4) = -4x\), \(7 \times x = 7x\) and \(7 \times (-4) = -28\).
  2. 2 Write all four terms \(x^2 - 4x + 7x - 28\).
  3. 3 Collect like terms \(-4x + 7x = 3x\).
  4. 4 Answer \(x^2 + 3x - 28\).

Answer\(x^2 + 3x - 28\)

Brackets with a coefficient

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

Show the solutionHide the solution
  1. 1 Multiply every pair \(2x \times x = 2x^2\), \(2x \times (-5) = -10x\), \(3 \times x = 3x\) and \(3 \times (-5) = -15\).
  2. 2 Write all four terms \(2x^2 - 10x + 3x - 15\).
  3. 3 Collect like terms \(-10x + 3x = -7x\).
  4. 4 Answer \(2x^2 - 7x - 15\).

Answer\(2x^2 - 7x - 15\)

Squaring a bracket

A squared bracket is two identical brackets multiplied, so it has a middle term.

  • Write it out twice

    \((x + 3)^2 = (x + 3)(x + 3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9\).

  • The middle term

    The \(6x\) comes from \(2 \times x \times 3\). Forgetting it, and writing \(x^2 + 9\), is the most common mistake in this topic.

  • With a coefficient

    \((2x - 1)^2 = (2x - 1)(2x - 1) = 4x^2 - 2x - 2x + 1 = 4x^2 - 4x + 1\).

  • Check with a number

    Put \(x = 1\) into both sides: \((1 + 3)^2 = 16\) and \(1 + 6 + 9 = 16\).

Common mistakes when expanding

What students write

  • \((x + 3)^2 = x^2 + 9\)
  • \(-2(x - 4) = -2x - 8\)
  • \(3x \times 4x = 12x\)

What is correct

  • \((x + 3)^2 = x^2 + 6x + 9\)
  • \(-2(x - 4) = -2x + 8\)
  • \(3x \times 4x = 12x^2\)

Expressions in context: perimeter and area

Many exam questions use algebraic expressions for the sides of shapes.

  • Perimeter

    Add all the sides and collect like terms. A rectangle \((3x + 1)\) by \((x + 4)\) has perimeter \(2(3x + 1 + x + 4) = 2(4x + 5) = 8x + 10\).

  • Area

    Multiply the length by the width and expand. \((3x + 1)(x + 4) = 3x^2 + 12x + x + 4 = 3x^2 + 13x + 4\).

  • Compound shapes

    Add or subtract the areas of simple shapes, then simplify.

  • "Show that" questions

    Start from one side and work to the stated result, showing every line. Do not start from the answer.

Expanding three terms

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

Show the solutionHide the solution
  1. 1 Expand the first bracket \(x(x + 2) = x^2 + 2x\).
  2. 2 Expand the second bracket \(-3(x - 4) = -3x + 12\), because \(-3 \times (-4) = +12\).
  3. 3 Write both together \(x^2 + 2x - 3x + 12\).
  4. 4 Collect like terms \(2x - 3x = -x\), so the answer is \(x^2 - x + 12\).

Answer\(x^2 - x + 12\)

A coefficient in both brackets

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

Show the solutionHide the solution
  1. 1 Multiply every pair \(2x \times 3x = 6x^2\), \(2x \times 4 = 8x\), \(-3 \times 3x = -9x\) and \(-3 \times 4 = -12\).
  2. 2 Write all four terms \(6x^2 + 8x - 9x - 12\).
  3. 3 Collect like terms \(8x - 9x = -x\).
  4. 4 Answer \(6x^2 - x - 12\).

Answer\(6x^2 - x - 12\)

A special case: no middle term

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

Show the solutionHide the solution
  1. 1 Multiply every pair \(x \times x = x^2\), \(x \times (-3) = -3x\), \(3 \times x = 3x\) and \(3 \times (-3) = -9\).
  2. 2 Collect like terms \(-3x + 3x = 0\), so the middle terms cancel out.
  3. 3 Answer \(x^2 - 9\). This pattern, a difference of two squares, is used a lot in factorising.

Answer\(x^2 - 9\)

Expanding three brackets (Higher tier)

Expand two brackets first, and then multiply the result by the third.

  • Two at a time

    Expand the first pair of brackets and tidy it up before bringing in the third.

  • Example

    \((x + 1)(x + 2)(x + 3)\). First \((x + 1)(x + 2) = x^2 + 3x + 2\).

  • Multiply by the third bracket

    \((x^2 + 3x + 2)(x + 3) = x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 = x^3 + 6x^2 + 11x + 6\).

  • Check

    Put \(x = 1\) into both forms. \(2 \times 3 \times 4 = 24\) and \(1 + 6 + 11 + 6 = 24\).

Using algebra to prove results

Algebra lets you show that something is true for every number, which examples alone cannot do.

  • Even and odd numbers

    An even number can be written \(2n\) and an odd number can be written \(2n + 1\), where \(n\) is an integer.

  • Example

    Two consecutive integers are \(n\) and \(n + 1\), so their sum is \(n + n + 1 = 2n + 1\), which is always odd.

  • Say what the letters mean

    Write "let \(n\) be an integer" at the start of the proof.

  • Finish with a sentence

    End by stating what you have shown, such as "so the sum is always odd".

Test yourself

  1. 1

    Simplify \(3x + 2y - x - 5y\).

    Show answerHide answer

    \(2x - 3y\).

  2. 2

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

    Show answerHide answer

    \(4x - 12\).

  3. 3

    Expand and simplify \((x + 2)(x + 5)\).

    Show answerHide answer

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

  4. 4

    Simplify \(2a \times 3a^2\).

    Show answerHide answer

    \(6a^3\).

  5. 5

    Expand and simplify \((x - 1)^2\).

    Show answerHide answer

    \(x^2 - 2x + 1\).

Exam technique: expanding

Expanding is quick, but each of the usual errors costs marks.

  • Write every term before you collect

    Four products from two brackets means four terms to show, even if two of them cancel.

  • Watch every sign

    A negative times a negative is a positive, which is the most common slip.

  • Check by substituting

    Put \(x = 1\) or \(x = 2\) into the original and your answer. If they differ, there is an error.

  • "Expand and simplify" needs both

    Expanding without collecting like terms usually loses the last mark.

Summary and exam focus

  • Terms include their sign, and only like terms with identical letters and powers can be collected.
  • When multiplying terms, multiply the numbers and add the powers; when dividing, divide the numbers and subtract the powers.
  • Expanding a single bracket means multiplying every term inside, taking care of negative signs.
  • For two brackets, multiply every term in the first by every term in the second and then collect like terms.
  • A squared bracket is two brackets, and it always has a middle term.

Exam focus

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

Write out all four products on one line before you collect like terms, so you can see that you have four terms. A grid is fine if you are asked to show your working, and a correct four-term expansion earns the method marks even if the final sign is wrong.

Key terms

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

Consecutive
Following one after another in order, such as the consecutive integers 7, 8 and 9.
Term
A single part of an expression, with the sign in front of it, such as \(-3x\).
Expression
A collection of terms with no equals sign, such as \(3x + 5\).
Equation
A statement that two expressions are equal, which is true only for certain values.
Formula
A rule that links quantities using letters, such as \(A = lw\).
Identity
An equation that is true for all values of the letters, written with the symbol \(\equiv\).
Like terms
Terms with the same letters raised to the same powers.
Coefficient
The number that multiplies a letter in a term.
Expand
Multiply out brackets to remove them.
Simplify
Write an expression in its shortest form by collecting like terms.

Questions and answers

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

1. Exam question Simplify 2 marks Easier

Simplify \(5a + 3b - 2a + 4b\).

Mark scheme — 2 marks available

  • \(3a\) or \(7b\) correct — M1
  • \(3a + 7b\) — A1

Model answer

Collect the \(a\) terms: \(5a - 2a = 3a\). Collect the \(b\) terms: \(3b + 4b = 7b\). The answer is \(3a + 7b\).

2. Exam question Expand 3 marks Core

Expand and simplify \(4(x + 3) - 2(x - 1)\).

Mark scheme — 3 marks available

  • \(4x + 12\) or \(-2x + 2\) correct — M1
  • \(4x + 12 - 2x + 2\) — M1
  • \(2x + 14\) — A1

Model answer

\(4(x + 3) = 4x + 12\) and \(-2(x - 1) = -2x + 2\). So the total is \(4x + 12 - 2x + 2 = 2x + 14\).

3. Exam question Simplify 3 marks Core

Simplify (a) \(3a \times 4a^2\) (1 mark) (b) \((2x^2y)^3\) (2 marks)

Mark scheme — 3 marks available

  • (a) \(12a^3\) — B1
  • (b) Any two of \(8\), \(x^6\), \(y^3\) correct — M1
  • (b) \(8x^6y^3\) — A1

Model answer

(a) Multiply the numbers and add the powers: \(3 \times 4 = 12\) and \(a \times a^2 = a^3\), so \(12a^3\). (b) Cube every part: \(2^3 \times (x^2)^3 \times y^3 = 8x^6y^3\).

4. Exam question Expand 2 marks Core

Expand and simplify \((x + 5)(x - 2)\).

Mark scheme — 2 marks available

  • Four correct terms, \(x^2 - 2x + 5x - 10\), or three of the four correct — M1
  • \(x^2 + 3x - 10\) — A1

Model answer

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

5. Exam question Expand 3 marks Core

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

Mark scheme — 3 marks available

  • At least three of the four terms correct — M1
  • \(2x^2 - 8x + 3x - 12\) — M1
  • \(2x^2 - 5x - 12\) — A1

Model answer

\(2x \times x = 2x^2\), \(2x \times (-4) = -8x\), \(3 \times x = 3x\), \(3 \times (-4) = -12\). So the expansion is \(2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12\).

6. Exam question Show that 3 marks Core

A rectangle has length \((x + 3)\) cm and width \((2x - 1)\) cm. Show that the area of the rectangle is \((2x^2 + 5x - 3)\) cm\(^2\).

Mark scheme — 3 marks available

  • Area \(= (x + 3)(2x - 1)\) — M1
  • At least three of the four terms \(2x^2\), \(-x\), \(6x\), \(-3\) correct — M1
  • \(2x^2 - x + 6x - 3\) leading to \(2x^2 + 5x - 3\) — C1

Model answer

Area \(= (x + 3)(2x - 1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\), as required.

7. Exam question Explain 3 marks Core

Dan says that \((x + 3)^2 = x^2 + 9\). (a) Explain why Dan is wrong. (1 mark) (b) Expand and simplify \((x + 3)^2\). (2 marks)

Mark scheme — 3 marks available

  • (a) States that the middle term is missing, or that he has only squared each term — C1
  • (b) \(x^2 + 3x + 3x + 9\) — M1
  • (b) \(x^2 + 6x + 9\) — A1

Model answer

(a) Dan has not included the middle term. Squaring a bracket means multiplying it by itself, so there are four terms, not two. (b) \((x + 3)(x + 3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9\).

8. Multiple choice 1 mark Core

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

  1. A \(x^2 - 6\)
  2. B \(x^2 + 9\)
  3. C \(x^2 - 6x - 9\)
  4. D \(x^2 - 9\) Correct

Why: \(x^2 - 3x + 3x - 9 = x^2 - 9\), because the middle terms cancel.

9. Multiple choice 1 mark Core

\(n\) is an integer. Which of these expressions is always an odd number?

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

Why: \(2n\) is always even, so \(2n + 1\) is always odd.

10. Multiple choice 1 mark Easier

Simplify \(4x + 3y - x + 2y\).

  1. A \(3x + 5y\) Correct
  2. B \(5x + 5y\)
  3. C \(3x + y\)
  4. D \(8xy\)

Why: Collect the \(x\) terms: \(4x - x = 3x\). Collect the \(y\) terms: \(3y + 2y = 5y\).

11. Multiple choice 1 mark Core

Simplify \(3x \times 4x^2\).

  1. A \(12x^3\) Correct
  2. B \(12x^2\)
  3. C \(7x^3\)
  4. D \(7x^2\)

Why: Multiply the numbers (\(3 \times 4 = 12\)) and add the powers (\(x^1 \times x^2 = x^3\)).

12. Multiple choice 1 mark Easier

Expand \(3(2x - 5)\).

  1. A \(6x - 5\)
  2. B \(6x + 15\)
  3. C \(5x - 15\)
  4. D \(6x - 15\) Correct

Why: Multiply both terms inside by 3: \(3 \times 2x = 6x\) and \(3 \times (-5) = -15\).

13. Multiple choice 1 mark Core

Expand \(-2(x - 4)\).

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

Why: \(-2 \times x = -2x\) and \(-2 \times (-4) = +8\), so the answer is \(-2x + 8\).

14. Multiple choice 1 mark Core

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

  1. A \(x^2 + 8x + 8\)
  2. B \(2x + 8\)
  3. C \(x^2 + 15\)
  4. D \(x^2 + 8x + 15\) Correct

Why: \(x^2 + 5x + 3x + 15 = x^2 + 8x + 15\).

15. Multiple choice 1 mark Core

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

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

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

16. Multiple choice 1 mark Core

Which of these is an identity, true for every value of \(x\)?

  1. A \(x^2 = 4\)
  2. B \(3x + 1 = 10\)
  3. C \(2(x + 3) \equiv 2x + 6\) Correct
  4. D \(x + 5 = 2x\)

Why: Expanding the bracket shows \(2(x + 3)\) is always equal to \(2x + 6\). The others are true only for certain values.

17. Multiple choice 1 mark Stretch

Simplify \((2x^3)^2\).

  1. A \(4x^5\)
  2. B \(8x^6\)
  3. C \(4x^6\) Correct
  4. D \(2x^6\)

Why: Square the 2 to get 4 and multiply the powers: \((x^3)^2 = x^6\).