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

1MA1 · A4AlgebraFoundation & 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 $\frac{12 x ^{5} y ^{2}}{4 x ^{2} y}$ or $(2 a^{3})^{4}$ 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 $6 a^{3} b$ 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 $k x^{n}$ . That skill is used directly in A-Level differentiation, where every term must first be written as a power of $x$ .

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 $k x^{n}$ (Higher)

Multiplying and dividing terms

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

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

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

1Top:

30 p^{7} q^{3}

2Divide by

10 p^{5} q

: coefficients

3

, powers

p^{2} q^{2}

Answer

3 p^{2} q^{2}

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

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

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: $(3 x^{2} y)^{3} = 3^{3} \times x^{6} \times y^{3} = 27 x^{6} y^{3}$ .

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}, (a^{m})^{n} = a^{mn}

Powers of products and powers.

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

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

3Divide:

\frac{8}{16} m^{8}

Answer

\frac{1}{2} m^{8}

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

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

Your answer

Type it like 8x^6y^3.

Why this works

A very common error is $(2 m^{4})^{3} = 6 m^{12}$ — 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.

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

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

1Split:

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

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

Answer

\frac{3}{2} x^{- 4}

Fractional indices

$x^{1/2} = x$ and $x^{1/3} = 3 x$ ; in general $x^{m / n} = (n x)^{m}$ .

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

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

1Negative power: flip to

(\frac{8 y ^{3}}{27 x ^{9}})^{2/3}

2Cube root:

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

3Square:

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

Answer

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

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

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

Your answer

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

Think like an examiner

Show the coefficient and each letter handled separately.
Give the answer in the form requested, such as $k x^{n}$ .
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} = \frac{1}{x ^{a}}, x^{1/ n} = n x

Negative and fractional.

Watch out for these

Writing

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

Square the coefficient too:

9 x^{4}

Writing

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

Add the indices:

x^{7}

Writing

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

Only

x

has the negative power:

\frac{4}{x}

Writing

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

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

Stretch yourself

Given that $\frac{( x ^{3} ) ^{k} \times x ^{5}}{x ^{6}} = x^{14}$ , 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

\frac{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 $(2 x^{3})^{4}$ .[2]

Question 2Grade 62 marks

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

Question 3Grade 73 marks

Find the value of $n$ such that $5^{2 n + 1} = 12 5^{n - 1}$ .[3]

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Index Laws

Multiplying and dividing terms

Powers of terms and brackets

Zero and negative indices

Fractional indices

Practise it the way it is examined

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Related topics

Index Laws

Multiplying and dividing terms

Powers of terms and brackets

Zero and negative indices

Fractional indices

Practise it the way it is examined

Skill drill

Practise this topic

Related topics