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?

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?

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

You've reached the end

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

You've reached the end

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

You've reached the end

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

You've reached the end

Algebraic Expressions Revision Guide