No. It is 1x^2, which is positive for any non-zero x.

J560 · 6.01–6.02Algebraic ExpressionsFoundation & HigherGrades 3–8

Index Laws

Q: Why can’t I add indices when the bases are different?

Index laws count repeated multiplication of the same base. x^2 y^3 contains two different things being multiplied, so it cannot become one power.

You have used index laws with numbers. The same laws work with letters, and that is where they earn their keep: simplifying $\dfrac{12x^{5}y^{2}}{4x^{2}y}$ or $(2a^{3})^{4}$ in a line or two. The one trap is forgetting that the number in front obeys ordinary arithmetic while only the powers follow the index laws.

45 min Algebraic Expressions

The big picture

An expression like $6a^{3}b$ has two kinds of information: a coefficient (6), which you multiply and divide normally, and powers of each letter, which you add, subtract or multiply using index laws. Handling those two parts separately makes every simplification routine.

Higher-tier questions combine the laws with negative and fractional indices and ask for answers in a particular form, such as $kx^{n}$ . That skill is used directly in A-Level differentiation, where every term must first be written as a power of $x$ .

GCSE Maths - Introduction to Powers/Indices & 3 Basic Rules (Powers Part 1/6) (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Before you start

Try a short calculation from the preceding topic without the notes.
Check the tier and grade band shown in this lesson.

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Simplify products and quotients of algebraic terms

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Simplify $a^{5} \times a^{3}$ .

Show answer and reasoning

$a^{8}$ . Add indices.

After watching 2: Simplify $\dfrac{20x^{6}}{5x^{2}}$ .

Show answer and reasoning

$4x^{4}$ . $20 \div 5 = 4$ ; $6 - 2 = 4$ .

What you'll learn

Simplify products and quotients of algebraic terms
Raise algebraic terms and brackets to powers
Use zero and negative indices with letters
Work with fractional indices and write expressions in the form $kx^{n}$ (Higher)

Multiplying and dividing terms

Multiply the coefficients, then add the powers of each letter separately: $4x^{3}y \times 5x^{2}y^{4} = 20x^{5}y^{5}$ .

Divide the coefficients, then subtract the powers: $\dfrac{18a^{7}b^{3}}{6a^{2}b^{3}} = 3a^{5}b^{0} = 3a^{5}$ .

A letter with no written power has power 1: $x = x^{1}$ .

Example 1Simplify

\dfrac{15p^{4}q \times 2p^{3}q^{2}}{10p^{5}q}

1Top:

30p^{7}q^{3}

2Divide by

10p^{5}q

: coefficients

3

, powers

p^{2}q^{2}

Answer

3p^{2}q^{2}

Your turnSame method, new numbers — try it before moving on.

Simplify $6a^5b^2 \times 3ab^4$ .

Your answer

Type it like 12a^3b^2.

Tip — Deal with numbers first, then each letter in turn. Writing the letters in a fixed order avoids missing one.

Powers of terms and brackets

To raise a term to a power, raise every factor inside: $(3x^{2}y)^{3} = 3^{3} \times x^{6} \times y^{3} = 27x^{6}y^{3}$ .

The coefficient is raised to the power as a number; the letters have their indices multiplied.

This only applies to products inside the bracket. $(x + y)^{2}$ is not $x^{2} + y^{2}$ — that needs expanding.

(ab)^{n} = a^{n}b^{n}, \qquad (a^{m})^{n} = a^{mn}

Powers of products and powers.

Example 1Simplify

(2m^{4})^{3} \div (4m^{2})^{2}

(2m^{4})^{3} = 8m^{12}

(4m^{2})^{2} = 16m^{4}

3Divide:

\dfrac{8}{16}m^{8}

Answer

\tfrac12 m^{8}

Your turnSame method, new numbers — try it before moving on.

Simplify $(3x^2y)^3$ .

Your answer

Type it like 8x^6y^3.

Why this works

A very common error is $(2m^{4})^{3} = 6m^{12}$ — multiplying the 2 by 3 instead of cubing it. Treat the coefficient as its own factor: $2^{3} = 8$ .

Zero and negative indices

$x^{0} = 1$ for any non-zero $x$ , so terms that cancel completely become 1.

$x^{-n} = \dfrac{1}{x^{n}}$ . Only the letter with the negative power moves to the denominator: $5x^{-2} = \dfrac{5}{x^{2}}$ , but $(5x)^{-2} = \dfrac{1}{25x^{2}}$ .

To write a fraction as a single term, move letters from the denominator by making their powers negative: $\dfrac{4}{x^{3}} = 4x^{-3}$ .

Example 1Write

\dfrac{3}{2x^{4}}

in the form

kx^{n}

1Split:

\dfrac{3}{2} \times \dfrac{1}{x^{4}}

\dfrac{1}{x^{4}} = x^{-4}

Answer

\tfrac32 x^{-4}

Fractional indices

$x^{1/2} = \sqrt{x}$ and $x^{1/3} = \sqrt[3]{x}$ ; in general $x^{m/n} = \left(\sqrt[n]{x}\right)^{m}$ .

Roots of algebraic terms follow the same rule: $\sqrt{16x^{6}} = (16x^{6})^{1/2} = 4x^{3}$ .

These are combined with the other laws exactly as integer powers are.

Example 1Simplify

\left(\dfrac{27x^{9}}{8y^{3}}\right)^{-2/3}

1Negative power: flip to

\left(\dfrac{8y^{3}}{27x^{9}}\right)^{2/3}

2Cube root:

\dfrac{2y}{3x^{3}}

3Square:

\dfrac{4y^{2}}{9x^{6}}

Answer

\dfrac{4y^{2}}{9x^{6}}

Your turnSame method, new numbers — try it before moving on.

Simplify $(9x^4)^{\frac{1}{2}}$ .

Your answer

Tip — Apply the root to the coefficient as a number and divide the letters’ powers by the root’s index — $\sqrt[3]{x^{9}} = x^{3}$ .

Think like an examiner

Show the coefficient and each letter handled separately.
Give the answer in the form requested, such as $kx^{n}$ .
For "show that" questions, write each index law step on its own line.

Laws of indices

x^{a} \times x^{b} = x^{a+b}

Multiply: add.

x^{a} \div x^{b} = x^{a-b}

Divide: subtract.

(x^{a})^{b} = x^{ab}

Power of a power.

x^{-a} = \tfrac{1}{x^{a}},\ x^{1/n} = \sqrt[n]{x}

Negative and fractional.

Watch out for these

Writing

(3x^{2})^{2} = 3x^{4}

Square the coefficient too:

9x^{4}

Writing

x^{3} \times x^{4} = x^{12}

Add the indices:

x^{7}

Writing

4x^{-1} = \dfrac{1}{4x}

Only

x

has the negative power:

\dfrac{4}{x}

Writing

(a + b)^{2} = a^{2} + b^{2}

Index laws apply to products, not sums; expand the brackets.

Stretch yourself

Given that $\dfrac{(x^{3})^{k} \times x^{5}}{\sqrt{x^{6}}} = x^{14}$ , find the value of $k$ .

Hint — Write every part as a single power of $x$ , including the square root.

Questions students ask

Why can’t I add indices when the bases are different?

Index laws count repeated multiplication of the same base.

x^{2} \times y^{3}

contains two different things being multiplied, so it cannot become one power.

What happens to a letter that cancels completely?

Its power becomes 0, and anything to the power 0 is 1, so it disappears from the answer.

x^{-2}

negative?

No. It is

\dfrac{1}{x^{2}}

, which is positive for any non-zero

x

Key takeaways

Coefficients use arithmetic; indices use the laws.
Add indices to multiply, subtract to divide, multiply for powers of powers.
Raise every factor in a bracket to the power, coefficient included.
Negative indices mean reciprocals of the letter they belong to.
Fractional indices are roots: root first, then power.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 33 marks

(a)Simplify $y^4 \times y^5$ .[1]
(b)Simplify $(2x^3)^4$ .[2]

Question 2Grade 62 marks

Simplify $(16a^8)^{\frac{3}{4}}$ .[2]

Question 3Grade 73 marks

Find the value of $n$ such that $5^{2n + 1} = 125^{n - 1}$ .[3]

Unlimited practice

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An external explanation or interactive activity to use alongside the lesson. It may include extension material.

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Simplify $(5y^{3})^{2}$ .
Write $\dfrac{7}{x^{2}}$ in the form $kx^{n}$ .
Simplify $(64x^{12})^{1/3}$ .
Simplify $(2x)^{-3}$ .

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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Related lessons

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Index Laws

You have used index laws with numbers. The same laws work with letters, and that is where they earn their keep: simplifying

\dfrac{12x^{5}y^{2}}{4x^{2}y}

(2a^{3})^{4}

in a line or two. The one trap is forgetting that the number in front obeys ordinary arithmetic while only the powers follow the index laws.

45 min Algebraic Expressions

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Simplify products and quotients of algebraic terms

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Simplify $a^{5} \times a^{3}$ .

Show answer and reasoning

$a^{8}$ . Add indices.

After watching 2: Simplify $\dfrac{20x^{6}}{5x^{2}}$ .

Show answer and reasoning

$4x^{4}$ . $20 \div 5 = 4$ ; $6 - 2 = 4$ .

Practise it the way it is examined

Question 1Grade 33 marks

(a)Simplify $y^4 \times y^5$ .[1]
(b)Simplify $(2x^3)^4$ .[2]

Question 2Grade 62 marks

Simplify $(16a^8)^{\frac{3}{4}}$ .[2]

Question 3Grade 73 marks

Find the value of $n$ such that $5^{2n + 1} = 125^{n - 1}$ .[3]