Algebraic Expressions

What you'll learn

How to simplify expressions involving powers, roots and negative indices.
How to rewrite exponential expressions using a common base.
How to factorise, expand and simplify algebraic expressions accurately.
How to rationalise surd denominators and give answers in exact form.

Expressions, terms and simplification

An algebraic expression is a mathematical phrase containing numbers, letters and operations, but no equals sign to solve.

For example, $3x^2 - 5x + 1$ is an expression. The separate parts $3x^2$ , $- 5 x$ and 1 are called terms.

Definition

Key vocabulary

A term is one part of an expression, separated by addition or subtraction.
A coefficient is the number multiplying a variable, such as 3 in $3x^2$ .
Like terms have exactly the same variable part, such as $4x^2$ and $7x^2$ .

To simplify means to rewrite an expression in an equivalent but cleaner form. In this topic, that often means using index laws, expanding brackets, factorising, or rationalising surds.

Index laws: powers and roots

An index or exponent tells you how many times a base is used as a factor. In $a^n$ , the base is $a$ and the index is $n$ .

A labelled form of a^n showing the base and index, with a small repeated-factor interpretation.

The most useful index laws are:

$am×an=am+na^m \times a^n = a^{m+n}$
$aman=am−n\frac{a^m}{a^n} = a^{m-n}$ , where $\neq 0$
$a^m)^n = a^{mn}$
$a−n=1ana^{-n} = \frac{1}{a^n}$
$a1n=ana^{\frac{1}{n}} = \sqrt[n]{a}$

Key Idea

Fractional powers

A fractional power combines a root and a power: $amna^{\frac{m}{n}}$ means take the nth root and the mth power.

Fractional powers can be interpreted as taking an nth root and an mth power in either order.

Example

Simplifying a fractional power

Simplify $(343x6125)13\left(\frac{343x^6}{125}\right)^{\frac{1}{3}}$ .

Recognise that a power of $13\frac{1}{3}$ means a cube root.
$(343x6125)13=343x61253\left(\frac{343x^6}{125}\right)^{\frac{1}{3}} = \sqrt[3]{\frac{343x^6}{125}}$
Take the cube root of the numerator and denominator.
$343x61253=343x631253\sqrt[3]{\frac{343x^6}{125}} = \frac{\sqrt[3]{343x^6}}{\sqrt[3]{125}}$
Use $343 = 7^3$ , $125 = 5^3$ and $x^6 = (x^2)^3$ .
$343x631253=7x25\frac{\sqrt[3]{343x^6}}{\sqrt[3]{125}} = \frac{7x^2}{5}$

Example

Simplifying with a negative fractional power

Simplify $(125x68y3)−23\left(\frac{125x^6}{8y^3}\right)^{-\frac{2}{3}}$ .

A negative index means take the reciprocal first.
$(125x68y3)−23=(8y3125x6)23\left(\frac{125x^6}{8y^3}\right)^{-\frac{2}{3}} = \left(\frac{8y^3}{125x^6}\right)^{\frac{2}{3}}$
A power of $23\frac{2}{3}$ means take the cube root, then square.
$(8y3125x6)23=(2y5x2)2\left(\frac{8y^3}{125x^6}\right)^{\frac{2}{3}} = \left(\frac{2y}{5x^2}\right)^2$
Square the numerator and denominator.
$(2y5x2)2=4y225x4\left(\frac{2y}{5x^2}\right)^2 = \frac{4y^2}{25x^4}$

Common Mistake

Even roots and variables

Strictly, $x2=∣x∣\sqrt{x^2} = |x|$ . In many index-law simplifications, variables are treated as positive unless a domain is stated. If a domain is given, use it.

Rewriting powers using a common base

Many exponential questions become easier once all powers are written with the same base.

Common bases reveal how numbers like 4, 8, 9 and 27 are related as powers of 2 or 3.

For example:

$4 = 2^2$
$8 = 2^3$
$9 = 3^2$
$27 = 3^3$

Example

Writing an expression as a power of 3

Express $27^{2x-1}$ in the form $3^y$ , giving $y$ in the form $a x + b$ .

Rewrite 27 as a power of 3.
$27^{2x-1} = (3^3)^{2x-1}$
Use $a^m)^n = a^{mn}$ .
$3^3)^{2x-1} = 3^{3(2x-1)}$
Expand the index.
$3^{3(2x-1)} = 3^{6x-3}$
Therefore $y = 6 x - 3$ .

Common Mistake

Multiplying the base instead of the index

When simplifying $3^3)^{2x-1}$ , multiply the indices to get $3^{6x-3}$ . Do not write $9^{2x-1}$ or $3^{2x+2}$ .

Solving exponential equations

An exponential equation is an equation where the unknown appears in an index, such as $2^x$ or $8^{x+1}$ .

If both sides can be written with the same base, equate the indices. If the equation contains terms like $4^x$ and $2^x$ , use a substitution such as $y = 2^x$ .

A decision flowchart for choosing between common-base methods and substitution in exponential equations.

Here is the decision process for exponential equations.

Flowchart for solving exponential equations

Example

Solving by using a common base

Find $x$ if $8^{x+1} = 4^{2x-3}$ .

Rewrite both sides as powers of 2.
$2^3)^{x+1} = (2^2)^{2x-3}$
Multiply the indices.
$2^{3x+3} = 2^{4x-6}$
Since the bases are equal, equate the indices.
$3 x + 3 = 4 x - 6$
Solve the linear equation.
$x = 9$

Example

Solving by substitution

Solve $4^x - 10(2^x) + 16 = 0$ .

Let $y = 2^x$ . Then rewrite $4^x$ .
$4^x = (2^2)^x = (2^x)^2 = y^2$
Substitute into the equation.
$y^2 - 10y + 16 = 0$
Factorise the quadratic.
$(y - 2) (y - 8) = 0$
Solve for $y$ .
$\quad \text{or} \quad y = 8$
Convert back to $x$ .
$2x=2or2x=82^x = 2 \quad \text{or} \quad 2^x = 8$
Therefore $x = 1$ or $x = 3$ .

Tip

Substitution clue

If you see both $2^x$ and $2^{2x}$ , try letting $y = 2^x$ . Then $2^{2x} = (2^x)^2 = y^2$ , which creates a quadratic.

The substitution y=2^x turns related exponential terms into powers of y.

Common Mistake

Accepting impossible substitution values

If $y = 2^x$ , then $y$ is always positive. So if your quadratic gives a negative value of $y$ , reject it.

“Hence” equations using a known quadratic

Sometimes you first solve a quadratic, then reuse its structure in a disguised equation.

Example

Using a previous quadratic result

Solve $u^2 - 35u + 216 = 0$ . Hence solve $t3−35t32+216=0t^3 - 35t^{\frac{3}{2}} + 216 = 0$ .

Factorise the quadratic.
$u^2 - 35u + 216 = (u - 8)(u - 27)$
So the quadratic has roots $u = 8$ and $u = 27$ .
In the second equation, let $t^{\frac{3}{2}}$ . Then $u^2 = t^3$ .
The equation becomes the same quadratic as before.
$u^2 - 35u + 216 = 0$
Therefore $t32=8t^{\frac{3}{2}} = 8$ or $t32=27t^{\frac{3}{2}} = 27$ .
Raise both sides to the power $23\frac{2}{3}$ .
$8^{\frac{2}{3}} \quad \text{or} \quad t = 27^{\frac{2}{3}}$
Simplify the answers.
$\quad \text{or} \quad t = 9$

Factorising and expanding

To factorise means to put an expression into brackets. To expand means to remove brackets by multiplying out.

Factorising completely usually means:

A checklist flowchart helps learners decide the order of factorising steps.

Take out the highest common factor.
Look for a difference of two squares.
Check whether any bracket can be factorised further.

Example

Factorising completely

Factorise $6x - 24x^3$ completely.

Take out the highest common factor, which is $6 x$ .
$6x - 24x^3 = 6x(1 - 4x^2)$
Recognise a difference of two squares.
$1 - 4x^2 = 1^2 - (2x)^2$
Factorise the bracket.
$1 - 4x^2 = (1 - 2x)(1 + 2x)$
Write the final answer.
$6x - 24x^3 = 6x(1 - 2x)(1 + 2x)$

Example

Expanding and simplifying brackets

Expand and simplify $2x - 3)(x + 1)^2$ .

An area/grid model shows every product formed when expanding the brackets.

Square the repeated bracket first.
$x + 1)^2 = x^2 + 2x + 1$
Multiply by the remaining bracket.
$2x - 3)(x^2 + 2x + 1) = 2x^3 + 4x^2 + 2x - 3x^2 - 6x - 3$
Collect like terms.
$2x^3 + x^2 - 4x - 3$

Surds and rationalising denominators

A surd is an irrational root left in exact form, such as $2\sqrt{2}$ or $5\sqrt{5}$ .

To rationalise the denominator means to rewrite a fraction so there is no surd on the bottom. For denominators like $\sqrt{b}$ , multiply by the conjugate $\sqrt{b}$ .

Multiplying by the conjugate removes the surd term by using the difference of two squares.

Definition

Conjugate

The conjugate of $\sqrt{b}$ is $\sqrt{b}$ . Multiplying conjugates uses the difference of two squares.

Example

Rationalising a denominator

Simplify $4+23−2\frac{4 + \sqrt{2}}{3 - \sqrt{2}}$ .

Multiply the top and bottom by the conjugate of the denominator.
$4+23−2×3+23+2\frac{4 + \sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}$
Expand the numerator.
$\sqrt{2})(3 + \sqrt{2}) = 14 + 7\sqrt{2}$
Expand the denominator.
$\sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7$
Divide by 7.
$14+727=2+2\frac{14 + 7\sqrt{2}}{7} = 2 + \sqrt{2}$

Example

Solving and giving the answer in surd form

Find $x$ if $2+xx=5\frac{2 + x}{x} = \sqrt{5}$ , giving your answer in the form $b\sqrt{5}$ .

Multiply both sides by $x$ , where $\neq 0$ .
$x\sqrt{5}$
Collect the $x$ terms on one side.
$x\sqrt{5} - x$
Factorise the right-hand side.
$x(\sqrt{5} - 1)$
Divide by $5−1\sqrt{5} - 1$ .
$\frac{2}{\sqrt{5} - 1}$
Rationalise the denominator.
$\frac{2}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{2(\sqrt{5} + 1)}{4}$
Write in the required form.
$\frac{1}{2} + \frac{1}{2}\sqrt{5}$

Exam technique

In the exam

Look for the structure first: common base, quadratic substitution, common factor, or surd denominator.
Show enough working to make your method clear, especially when changing bases or rationalising.
Check restrictions: denominators cannot be zero, and substitutions like $y = 2^x$ must be positive.

Self review

Check yourself

Can you simplify an expression with a negative fractional index without skipping the reciprocal step?
When should you use a substitution such as $y = 2^x$ ?
Can you rationalise a denominator using the correct conjugate?

What you'll learn

How to simplify expressions involving powers, roots and negative indices.
How to rewrite exponential expressions using a common base.
How to factorise, expand and simplify algebraic expressions accurately.
How to rationalise surd denominators and give answers in exact form.

Expressions, terms and simplification

An algebraic expression is a mathematical phrase containing numbers, letters and operations, but no equals sign to solve.

For example, $3x^2 - 5x + 1$ is an expression. The separate parts $3x^2$ , $- 5 x$ and 1 are called terms.

Definition

Key vocabulary

A term is one part of an expression, separated by addition or subtraction.
A coefficient is the number multiplying a variable, such as 3 in $3x^2$ .
Like terms have exactly the same variable part, such as $4x^2$ and $7x^2$ .

To simplify means to rewrite an expression in an equivalent but cleaner form. In this topic, that often means using index laws, expanding brackets, factorising, or rationalising surds.

Index laws: powers and roots

An index or exponent tells you how many times a base is used as a factor. In $a^n$ , the base is $a$ and the index is $n$ .

A labelled form of a^n showing the base and index, with a small repeated-factor interpretation.

The most useful index laws are:

$am×an=am+na^m \times a^n = a^{m+n}$
$aman=am−n\frac{a^m}{a^n} = a^{m-n}$ , where $\neq 0$
$a^m)^n = a^{mn}$
$a−n=1ana^{-n} = \frac{1}{a^n}$
$a1n=ana^{\frac{1}{n}} = \sqrt[n]{a}$

Key Idea

Fractional powers

A fractional power combines a root and a power: $amna^{\frac{m}{n}}$ means take the nth root and the mth power.

Fractional powers can be interpreted as taking an nth root and an mth power in either order.

Example

Simplifying a fractional power

Simplify $(343x6125)13\left(\frac{343x^6}{125}\right)^{\frac{1}{3}}$ .

Recognise that a power of $13\frac{1}{3}$ means a cube root.
$(343x6125)13=343x61253\left(\frac{343x^6}{125}\right)^{\frac{1}{3}} = \sqrt[3]{\frac{343x^6}{125}}$
Take the cube root of the numerator and denominator.
$343x61253=343x631253\sqrt[3]{\frac{343x^6}{125}} = \frac{\sqrt[3]{343x^6}}{\sqrt[3]{125}}$
Use $343 = 7^3$ , $125 = 5^3$ and $x^6 = (x^2)^3$ .
$343x631253=7x25\frac{\sqrt[3]{343x^6}}{\sqrt[3]{125}} = \frac{7x^2}{5}$

Example

Simplifying with a negative fractional power

Simplify $(125x68y3)−23\left(\frac{125x^6}{8y^3}\right)^{-\frac{2}{3}}$ .

A negative index means take the reciprocal first.
$(125x68y3)−23=(8y3125x6)23\left(\frac{125x^6}{8y^3}\right)^{-\frac{2}{3}} = \left(\frac{8y^3}{125x^6}\right)^{\frac{2}{3}}$
A power of $23\frac{2}{3}$ means take the cube root, then square.
$(8y3125x6)23=(2y5x2)2\left(\frac{8y^3}{125x^6}\right)^{\frac{2}{3}} = \left(\frac{2y}{5x^2}\right)^2$
Square the numerator and denominator.
$(2y5x2)2=4y225x4\left(\frac{2y}{5x^2}\right)^2 = \frac{4y^2}{25x^4}$

Common Mistake

Even roots and variables

Strictly, $x2=∣x∣\sqrt{x^2} = |x|$ . In many index-law simplifications, variables are treated as positive unless a domain is stated. If a domain is given, use it.

Rewriting powers using a common base

Many exponential questions become easier once all powers are written with the same base.

Common bases reveal how numbers like 4, 8, 9 and 27 are related as powers of 2 or 3.

For example:

$4 = 2^2$
$8 = 2^3$
$9 = 3^2$
$27 = 3^3$

Example

Writing an expression as a power of 3

Express $27^{2x-1}$ in the form $3^y$ , giving $y$ in the form $a x + b$ .

Rewrite 27 as a power of 3.
$27^{2x-1} = (3^3)^{2x-1}$
Use $a^m)^n = a^{mn}$ .
$3^3)^{2x-1} = 3^{3(2x-1)}$
Expand the index.
$3^{3(2x-1)} = 3^{6x-3}$
Therefore $y = 6 x - 3$ .

Common Mistake

Multiplying the base instead of the index

When simplifying $3^3)^{2x-1}$ , multiply the indices to get $3^{6x-3}$ . Do not write $9^{2x-1}$ or $3^{2x+2}$ .

Solving exponential equations

An exponential equation is an equation where the unknown appears in an index, such as $2^x$ or $8^{x+1}$ .

If both sides can be written with the same base, equate the indices. If the equation contains terms like $4^x$ and $2^x$ , use a substitution such as $y = 2^x$ .

A decision flowchart for choosing between common-base methods and substitution in exponential equations.

Here is the decision process for exponential equations.

Flowchart for solving exponential equations

Example

Solving by using a common base

Find $x$ if $8^{x+1} = 4^{2x-3}$ .

Rewrite both sides as powers of 2.
$2^3)^{x+1} = (2^2)^{2x-3}$
Multiply the indices.
$2^{3x+3} = 2^{4x-6}$
Since the bases are equal, equate the indices.
$3 x + 3 = 4 x - 6$
Solve the linear equation.
$x = 9$

Example

Solving by substitution

Solve $4^x - 10(2^x) + 16 = 0$ .

Let $y = 2^x$ . Then rewrite $4^x$ .
$4^x = (2^2)^x = (2^x)^2 = y^2$
Substitute into the equation.
$y^2 - 10y + 16 = 0$
Factorise the quadratic.
$(y - 2) (y - 8) = 0$
Solve for $y$ .
$\quad \text{or} \quad y = 8$
Convert back to $x$ .
$2x=2or2x=82^x = 2 \quad \text{or} \quad 2^x = 8$
Therefore $x = 1$ or $x = 3$ .

Tip

Substitution clue

If you see both $2^x$ and $2^{2x}$ , try letting $y = 2^x$ . Then $2^{2x} = (2^x)^2 = y^2$ , which creates a quadratic.

The substitution y=2^x turns related exponential terms into powers of y.

Common Mistake

Accepting impossible substitution values

If $y = 2^x$ , then $y$ is always positive. So if your quadratic gives a negative value of $y$ , reject it.

“Hence” equations using a known quadratic

Sometimes you first solve a quadratic, then reuse its structure in a disguised equation.

Example

Using a previous quadratic result

Solve $u^2 - 35u + 216 = 0$ . Hence solve $t3−35t32+216=0t^3 - 35t^{\frac{3}{2}} + 216 = 0$ .

Factorise the quadratic.
$u^2 - 35u + 216 = (u - 8)(u - 27)$
So the quadratic has roots $u = 8$ and $u = 27$ .
In the second equation, let $t^{\frac{3}{2}}$ . Then $u^2 = t^3$ .
The equation becomes the same quadratic as before.
$u^2 - 35u + 216 = 0$
Therefore $t32=8t^{\frac{3}{2}} = 8$ or $t32=27t^{\frac{3}{2}} = 27$ .
Raise both sides to the power $23\frac{2}{3}$ .
$8^{\frac{2}{3}} \quad \text{or} \quad t = 27^{\frac{2}{3}}$
Simplify the answers.
$\quad \text{or} \quad t = 9$

Factorising and expanding

To factorise means to put an expression into brackets. To expand means to remove brackets by multiplying out.

Factorising completely usually means:

A checklist flowchart helps learners decide the order of factorising steps.

Take out the highest common factor.
Look for a difference of two squares.
Check whether any bracket can be factorised further.

Example

Factorising completely

Factorise $6x - 24x^3$ completely.

Take out the highest common factor, which is $6 x$ .
$6x - 24x^3 = 6x(1 - 4x^2)$
Recognise a difference of two squares.
$1 - 4x^2 = 1^2 - (2x)^2$
Factorise the bracket.
$1 - 4x^2 = (1 - 2x)(1 + 2x)$
Write the final answer.
$6x - 24x^3 = 6x(1 - 2x)(1 + 2x)$

Example

Expanding and simplifying brackets

Expand and simplify $2x - 3)(x + 1)^2$ .

An area/grid model shows every product formed when expanding the brackets.

Square the repeated bracket first.
$x + 1)^2 = x^2 + 2x + 1$
Multiply by the remaining bracket.
$2x - 3)(x^2 + 2x + 1) = 2x^3 + 4x^2 + 2x - 3x^2 - 6x - 3$
Collect like terms.
$2x^3 + x^2 - 4x - 3$

Surds and rationalising denominators

A surd is an irrational root left in exact form, such as $2\sqrt{2}$ or $5\sqrt{5}$ .

To rationalise the denominator means to rewrite a fraction so there is no surd on the bottom. For denominators like $\sqrt{b}$ , multiply by the conjugate $\sqrt{b}$ .

Multiplying by the conjugate removes the surd term by using the difference of two squares.

Definition

Conjugate

The conjugate of $\sqrt{b}$ is $\sqrt{b}$ . Multiplying conjugates uses the difference of two squares.

Example

Rationalising a denominator

Simplify $4+23−2\frac{4 + \sqrt{2}}{3 - \sqrt{2}}$ .

Multiply the top and bottom by the conjugate of the denominator.
$4+23−2×3+23+2\frac{4 + \sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}$
Expand the numerator.
$\sqrt{2})(3 + \sqrt{2}) = 14 + 7\sqrt{2}$
Expand the denominator.
$\sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7$
Divide by 7.
$14+727=2+2\frac{14 + 7\sqrt{2}}{7} = 2 + \sqrt{2}$

Example

Solving and giving the answer in surd form

Find $x$ if $2+xx=5\frac{2 + x}{x} = \sqrt{5}$ , giving your answer in the form $b\sqrt{5}$ .

Multiply both sides by $x$ , where $\neq 0$ .
$x\sqrt{5}$
Collect the $x$ terms on one side.
$x\sqrt{5} - x$
Factorise the right-hand side.
$x(\sqrt{5} - 1)$
Divide by $5−1\sqrt{5} - 1$ .
$\frac{2}{\sqrt{5} - 1}$
Rationalise the denominator.
$\frac{2}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{2(\sqrt{5} + 1)}{4}$
Write in the required form.
$\frac{1}{2} + \frac{1}{2}\sqrt{5}$

Exam technique

In the exam

Look for the structure first: common base, quadratic substitution, common factor, or surd denominator.
Show enough working to make your method clear, especially when changing bases or rationalising.
Check restrictions: denominators cannot be zero, and substitutions like $y = 2^x$ must be positive.

Self review

Check yourself

Can you simplify an expression with a negative fractional index without skipping the reciprocal step?
When should you use a substitution such as $y = 2^x$ ?
Can you rationalise a denominator using the correct conjugate?

What you'll learn

Expressions, terms and simplification

Index laws: powers and roots

Rewriting powers using a common base

Solving exponential equations

“Hence” equations using a known quadratic

Factorising and expanding

Surds and rationalising denominators

Recap questions

Review Algebraic Expressions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Algebraic Expressions

What you'll learn

Expressions, terms and simplification

Index laws: powers and roots

Rewriting powers using a common base

Solving exponential equations

“Hence” equations using a known quadratic

Factorising and expanding

Surds and rationalising denominators

Recap questions

Review Algebraic Expressions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Algebraic Expressions

What you'll learn

Expressions, terms and simplification

Index laws: powers and roots

Rewriting powers using a common base

Solving exponential equations

“Hence” equations using a known quadratic

Factorising and expanding

Surds and rationalising denominators

Recap questions

Review Algebraic Expressions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Algebraic Expressions Revision Guide

Algebraic Expressions

What you'll learn

Expressions, terms and simplification

Index laws: powers and roots

Rewriting powers using a common base

Solving exponential equations

“Hence” equations using a known quadratic

Factorising and expanding

Surds and rationalising denominators

Recap questions

Review Algebraic Expressions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Algebraic Expressions Revision Guide