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Expanding bracketsThis topic is relevant for:

Here is everything you need to know about **algebraic expressions** for GCSE maths (Edexcel, AQA and OCR). You’ll learn what **algebraic expressions** are, how to **simplify algebraic expressions**, and the different methods for using algebraic expressions.

Look out for the algebraic expression worksheets, word problems and exam questions at the end.

**More from this topic: **Algebra

An **algebraic expression** is a set of terms with letters and numbers that are combined using addition

An expression that contains two terms is called a binomial.

\[E.g. 2x+3y\quad or \quad 2-5y^{2}\quad etc.\]

An expression that contains three terms is called a trinomial.

\[E.g. 2x+3y-5\quad or \quad 2-5y^{2}+6xy\quad etc.\]

Let’s define some of the keywords when using algebraic notation:

A **variable** is a symbol (often a letter) that is used to represent an unknown quantity.

\[E.g. x\quad or \quad y\quad or\quad a\quad etc.\]

Variables can also have **exponents** (be raised to a certain **power**).

\[E.g. x^{2}\quad or \quad y^{3}\]

A **coefficient** is the value that is before a variable. It tells us how many lots of the variable there is.

\[ \begin{aligned}
E.g. 5x&=x+x+x+x+x\\
&=5\times x
\end{aligned}\]

Here **coefficient** and **variable**.

A **term** is a number by itself, a variable by itself, or a combination of numbers and letters. If the term includes a variable it is called an algebraic term.

\[E.g. 2\quad or\quad 5xy\quad or\quad 12x^{2}\quad or \quad 12xy\quad etc.\]

An expression that contains one term is called a monomial.

A polynomial expression consists of two or more algebraic terms.

Get your free algebraic expressions worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free algebraic expressions worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEThere are various methods of using algebraic expressions and performing algebraic operations. These are summarised below.

For examples, practice questions and worksheets on each one follow the links to the step by step guides below or go straight to factorising and simplifying expressions.

Example of collecting like terms

\[5x+3y-3x-7y\]

\[5 x-3 x=2 x \quad 3 y-7 y=-4 y\]

So

\[5x+3y-3x-7y = 2x-4y\]

**Step-by-step guide:**Collecting like terms

Example of multiplying and dividing algebra

\[5 x y \times 4 x z
\]

\[5 \times 4 = 20\]

\[x \times x=x^{2}\
\]

\[y \times z=y z
\]

So

\[5 x y \times 4 x z=20 x^{2} y z\]

Example of expanding brackets

\[3(2x+5)\]

Multiply every term outside the bracket by every term inside the bracket.

✕ | 2x | + 5 |

3 | 6x | + 15 |

\[3 \times 2 x=6 x \\
3 \times 5=15 x \\
3(2x+5)=6x+15\]

**Step-by-step guide:**Expanding brackets**See also:**Expand and simplify

Example of algebraic fractions

\[\frac{14 x y}{8 x}\]

Divide the numerator and the denominator by

\[\frac{7y}{4}\]

**Step-by-step guide:**Algebraic fractions**See also:**Simplifying algebraic fractions

Example of writing and simplifying algebraic expressions

Write an expression for the perimeter of the shape.

\[\begin{aligned}
Perimeter&=2x+3+x-2+2x+3+x-2\\
&=6x+2
\end{aligned}
\]

**Step-by-step guide:**Simplifying expressions

Example of factorising in single brackets

\[6 x+3=3(2 x+1)\]

**Step-by-step guide:**Factorising single brackets

Example of factorising quadratic expressions into two brackets

\[2 x^{2}+5 x-3=(2 x-1)(x+3)\]

**Step-by-step guide:**Factorising quadratics

Example of difference of two squares

\[4 x^{2}-36=(2 x+6)(2 x-6)\]

**Step-by-step guide:**Difference of two squares

**Expressions, Equations and Formulas**

An algebraic **expression **is different from an **equation **and a** formula**.

An **expression** is a set of terms that are combined using arithmetic operations:

\[2x+5\]

An **equation** is an expression that equals something.

\[2x+5=15\]

We can solve equations to work out the value of the unknown variable

**Algebraic formulae** are a set of instructions which give a desired result.

E.g. ^{2}

Circumference of a Circle = πd

**Brackets terminology**

Brackets are sometimes referred to as parentheses.

1. Simplify:

6x^{2}y-2x^{2}y+4x^{2}-5x^{2}

10x^{2}y-7x^{2}

3x^{2}y

10x^{2}y+3x^{2}

4x^{2}y-x^{2}

By considering like terms in the expression 6x^{2}y-2x^{2}+4x^{2}-5x^{2} we have 6x^{2}y+4x^{2}=10x^{2}y and -2x^{2}-5x^{2}=-7x^{2} which simplifies to 10x^{2}y-7x^{2} .

2. Write an expression for the area of the parallelogram:

6x^{2}+17x+5

6x^{2}+5x+1

7x+7

10x+12

For the area, we need to multiply the length (3x+1) of the base and the perpendicular height (2x+1) .

Area =(3x+1)(2x+1)=6x^{2}+5x+13. Simplify:

\frac{9x^{2}y}{15x^{3}}

\frac{3y}{5x}

\frac{9y}{15x}

\frac{3x^{2}y}{5x}

\frac{3}{5}

The numerator and denominator of \frac{9x^{2}y}{15x^{3}} have a highest common factor of 3x^{2} . Therefore, we divide through by 3x^{2} resulting in the simplified algebraic fraction.

4. Simplify:

\frac{x^{2}-3x-10}{x^2-25}

\frac{x-5}{x-5}

\frac{x+2}{x-5}

\frac{x+2}{x+5}

x^{2}-3x-35

The numerator can be factorised as the product of two brackets

x^{2}-3x-10=(x+2)(x-5)The denominator can be factorised as it is the difference of two squares

x^{2}-25=(x+5)(x-5)This means we can write the fraction as

\frac{(x+2)(x-5)}{(x+5)(x-5)}The numerator and denominator have a common factor of x-5 , which we cancel, leaving the simplified fraction.

5. Simplify:

3x(4-5x+2y)

12-15x+6y

12x-15x^{2}+6xy

7x-9x^{2}+5xy

12x-15x+6y

We need to multiply each term inside the bracket by 3x

3x\times4=12x\\ 3x\times(-5x)=-15x^{2}\\ 3x\times2y=6xywhich we can combine into the expression we need as required.

6. Expand and simplify:

4(2x-1)-3(x+6)

5x-22

11x-22

x-5

11x+14

We can expand each bracket one at a time

4(2x-1)=8x-4and

-3(x+6)=-3x-18This can be further simplified by collecting like terms

8x-4-3x-18=5x-227. Fully factorise:

2{x}^2+x-6

(2 x+2)(x-3)

2\left(x^{2}+0.5 x-3\right)

(2 x-3)(x+2)

(x-3)(x+2)

Correct answer:

=(2x-3)(x+2)

To factorise the quadratic expression, we are looking for numbers that multiply to -12 and sum to +1. By considering factor pairs, we conclude that we need to use +4 and -3.

We can rewrite

2x^{2}+x-6as 2x^{2}+4x-3x-6

which can be factorised as

2x(x+2)-3(x+2)or more concisely

(2x-3)(x+2)8. Fully factorise:

4{x}^2-25

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

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

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

(3 x+20)(x+5)

Correct answer:

(2 x+5)(2 x-5)This is a special case (difference of two squares), which means we can take square roots of the coefficient of x and the constant term, then write one bracket with a + sign and the other bracket with a – sign.

1. Simplify: 4f – 2e + 3f + 5e

Show answer

7f + 3e

(2 marks)

2. Expand and simplify: 4x(2x – 7)

Show answer

8x^{2} – 28x

(2 marks)

3. Simplify:

\[\frac{15x^{3}y^{2}}{5xy^{3}}\]

Show answer

\[=\frac{3x^{2}}{y}\]

(2 marks)

- simplify expressions
- use language and properties precisely to analyse algebraic expressions
- simplify and manipulate algebraic expressions to maintain equivalence by:

– collecting like terms

– multiplying a single term over a bracket

– taking out common factors - translate simple situations or procedures into algebraic expressions

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