Year 10 - Index laws with algebra - Free practice

Year 10 Index Laws with Algebra Worksheets & Practice

Simplify algebraic products and quotients using exponent laws, with worked solutions every time.

Quick answer

How do you simplify algebraic expressions using index laws?

Apply the index laws to the coefficients and each base separately. Product law: add indices when multiplying (aᵐ × aⁿ = aᵐ⁺ⁿ). Quotient law: subtract when dividing. Power law: multiply when raising a power to a power. For example, 2x³ × 3x² = 6x⁵.

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Worked examples

Year 10 index laws with algebra: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Multiplying algebraic terms using the product law
Simplify $3a^4 \times 5a^3$ .
Show worked solution
Key idea: When multiplying terms with the same base, add the exponents: $a^m \times a^n = a^{m+n}$ .
1. Multiply the coefficients: $3 \times 5 = 15$ .
2. Apply the product law to the powers of $a$ : $a^4 \times a^3 = a^{4+3} = a^7$ .
3. Write the result: $15a^7$ .
Answer: $15a^7$
Example 2 · Combining product, quotient and power laws
Simplify $\dfrac{(3x^2y)^3}{9x^4y^2}$ , giving your answer with positive indices only.
Show worked solution
Key idea: Expand any brackets using the power law first, then apply the quotient law to simplify.
1. Expand the numerator using the power law: $(3x^2y)^3 = 3^3 \cdot x^{2 \times 3} \cdot y^3 = 27x^6y^3$ .
2. Write the full expression: $\dfrac{27x^6y^3}{9x^4y^2}$ .
3. Divide the coefficients: $\dfrac{27}{9} = 3$ .
4. Apply the quotient law to each base: $x^{6-4} = x^2$ and $y^{3-2} = y^1 = y$ .
5. Write the result: $3x^2y$ .
Answer: $3x^2y$
Example 3 · Applying index laws in a real-life context
A square solar panel has a side length of $2x^3$ metres. Four of these panels are laid flat in a single row end to end, and the combined area is divided equally among $8x^2$ households. Write an expression for each household's share of the total panel area in square metres, giving your answer with positive indices only.
Show worked solution
Key idea: Find the total area of the four panels, then divide by the number of households using the quotient law.
1. Find the area of one panel: $(2x^3)^2 = 4x^6$ square metres.
2. Find the total area of four panels: $4 \times 4x^6 = 16x^6$ square metres.
3. Divide by the number of households: $\dfrac{16x^6}{8x^2}$ .
4. Divide the coefficients: $\dfrac{16}{8} = 2$ .
5. Apply the quotient law: $x^{6-2} = x^4$ .
Answer: $2x^4$ square metres per household

Common mistakes

Where students usually lose marks

Multiplying the exponents instead of adding them

$a^3 \times a^4 = a^{12}$

The product law says to add the exponents, not multiply them. The correct result is $a^{3+4} = a^7$ . Save multiplying exponents for the power law, $(a^m)^n = a^{mn}$ , where there is a bracket raised to a power.

Forgetting to apply the power to the coefficient

$(3x^2)^3 = 3x^6$

Every factor inside the bracket must be raised to the outside power. The coefficient 3 must also be cubed: $(3x^2)^3 = 3^3 \cdot x^{2 \times 3} = 27x^6$ .

Subtracting exponents the wrong way in a quotient

$\dfrac{a^3}{a^7} = a^4$

Always subtract the denominator's exponent from the numerator's exponent: $a^{3-7} = a^{-4}$ . If a positive index is required, write this as $\dfrac{1}{a^4}$ .

Treating a zero index as zero instead of one

$5a^0 = 0$

Any non-zero base raised to the power of zero equals 1, so $a^0 = 1$ . Therefore $5a^0 = 5 \times 1 = 5$ .

Try one yourself

Your turn

Simplify $\dfrac{(2x^3)^4}{x^5}$ , giving your answer as a coefficient times a power of $x$ .

What is the exponent of $x$ in the simplified expression?

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About this worksheet

Why we built it

Index laws with algebra extend the rules students first met with numbers to expressions involving pronumerals, coefficients and multiple bases. These worksheets work through simplifying products, quotients and powers of algebraic terms, with full worked solutions so students can check each step of their thinking and spot where errors creep in.

What's covered

Sub-skills your student will practise

✓Multiplying algebraic terms using the product law (aᵐ × aⁿ = aᵐ⁺ⁿ)
✓Dividing algebraic terms using the quotient law (aᵐ ÷ aⁿ = aᵐ⁻ⁿ)
✓Raising a power to a power using the power law ((aᵐ)ⁿ = aᵐⁿ)
✓Applying the zero index and negative index rules to algebraic terms
✓Simplifying expressions with multiple bases and mixed coefficients
✓Combining two or more index laws in a single simplification

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Frequently asked

Questions parents ask about index laws with algebra

What index laws do Year 10 students need to know for algebra?

Students work with the product law, quotient law, power of a power law, zero index and negative index rules. They apply these to algebraic terms with pronumerals and coefficients, not just numbers.

What is the difference between the product law and the power law?

The product law applies when multiplying two terms with the same base: you add the indices. The power law applies when a power is raised to another power: you multiply the indices. Mixing these up is one of the most common errors at Year 10.

Why do students lose marks simplifying index expressions in algebra?

The most frequent mistakes are forgetting to apply the index law to the coefficient as well as the pronumeral, and subtracting indices in the wrong order when dividing. Worked solutions make these slip-ups easy to catch and correct.

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