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Algebraic manipulation and laws of indices - Mathematics Year 9

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Introduction

What must stay equivalent when you expand brackets, cancel a factor or rearrange a formula? Track structure and domain conditions instead of applying a rule because an expression looks familiar.

Complete eight manipulation questions and rearrange two formulae. Connect an area model, exact algebra and substitution checks, and explain why checking one value can refute a claim but cannot prove every identity.

Distribute, collect and distinguish identities

An expression represents a value depending on its variables. An identity states two expressions agree for every value in their common domain; an equation may hold only for particular values. Collecting, expanding and factorising are ways to write an equivalent expression, not instructions to set it equal to zero or solve for an unspecified unknown.

Like terms have the same variable part, including exponents. Thus 3 x²−x²=2 x² and 2 x+5 x=7 x, butx² andx cannot be merged as one like term. In particularx²+x²=2 x², notx⁴. The fourth power arises in the different operationx²×x²=x⁴. State whether the terms are being added or multiplied before using index rules.

Original two-bracket expansion. Distribute each term ofx+3 across each term ofx+5. The four products arex², 5 x, 3 x, 15. Collect the two linear terms to get $(x + 3) (x + 5) = x^{2} + 8 x + 15$ . At x=2, the product 5×7=35 agrees with 4+16+15=35. This checks the arithmetic at that value; the distributive derivation establishes the identity for all real x.

An area model uses a rectangle with sidesx+3 andx+5, partitioned at lengthsx, 3 andx, 5. The four subareas arex², 3 x, 5 x, 15. As a literal positive-length diagram this requiresx>0. The algebraic distributive identity remains valid at other real values even when a negative variable cannot be interpreted as a physical side length.

For(2 x −1)(x+4), the four algebraic products are 2 x², 8 x, −x, −4, so the result is 2 x²+7 x −4. A signed term is still distributed to both terms in the second bracket. The incorrect 2 x²+8 x −1 omits products and does not preserve the expression.

Factorisation reverses distributivity. In 6 x²+9 x, both terms share 3 x, so write 3 x(2 x+3). Expanding back gives the original two terms. A greatest common numerical/variable factor is efficient, but a factorisation need not be unique in presentation. Keep the variable powers and all terms when checking; do not divide one term and forget the other.

At a single substitution a false expression can accidentally match the correct one. For examplex²+x equals 2 x at x=1 but not x=2. One counterexample can disprove a universal identity; one successful example does not establish it. Use an algebraic derivation for the general claim and a substitution as an additional error check.

Complete eight transformations with domain checks

For nonzero base a and integer indices, multiplication of like-base powers adds exponents and division subtracts them. Raising a power to a power multiplies exponents. For the positive-exponent producta³×a⁴=a⁷, the same identity also holds at a=0. Quotients with a in the denominator require a≠0. A cancelled factor does not retroactively make an original zero denominator valid.

Eight manipulation tasks. Cover the result column, derive each form and retain its domain. The table uses adjacency for multiplication and superscripts for integer powers.

Task	Original expression	Equivalent result	Domain retained
1	(x+3)(x+5)	x²+8 x+15	All real x
2	(2 x −1)(x+4)	2 x²+7 x −4	All real x
3	3 x²+2 x−x²+5 x	2 x²+7 x	All real x
4	6 x²+9 x	3 x(2 x+3)	All real x
5	a³×a⁴	a⁷	All real a
6	12 x²÷(3 x)	4 x	x≠0
7	(2 x³)²	4 x⁶	All real x
8	x⁵÷x²	x³	x≠0

In task 6, 12 x²÷(3 x)=(12÷3)×x²⁻¹=4 x, for x≠0. At x=2, the original 48÷6=8 and result 4×2=8 agree. At x=0, the simplified expression 4 x could be evaluated, but the original quotient 0/0 is undefined. The equality as a rewrite is limited to the original domain. Task 8 similarly cancelsx² from x⁵ only for nonzero x.

In task 7, square the whole product: (2 x³)²=2²×x⁶=4 x⁶. The coefficient must be squared too. At x=−1, the original(−2)²=4 and 4×(−1)⁶=4. Writing 2 x⁶ would fail this check. Brackets also distinguish(−x)²=x² from−x², where the exponent applies to x before the leading minus sign.

Compare Corbettmaths’ original grid method for(x+5)(x −3) with our different(x+3)(x+5) expression. Its four cells show why signed cross-products must be retained. Predict the middle coefficient before viewing the collection step. Original: corbettmaths, Expanding Two Brackets using a Grid - Corbettmaths. The written explanations and native tasks work without sound or the external player.

A useful workflow is to copy the expression, identify its operation, write a complete transformation and independently check a selected value. Expansion and factorisation should give the same polynomial coefficients when returned to a common form. A calculator substitution helps locate errors, but preserve an algebraic justification and any denominator restriction in the final answer.

For an unfamiliar(3 x+2)(x −4), distributivity gives 3 x²−12 x+2 x −8=3 x²−10 x −8. At x=2, the original 8×(−2)=−16 and expanded 12−20−8=−16 agree. This changes both coefficients and signs rather than repeating a memorised four-product answer.

Rearrange two formulae and verify model quantities

Rearrange two formulae. The subject is the variable isolated on one side. Apply equivalent operations to the whole of each side; do not merely move a symbol across the equality without recording the operation. The same letter can have a different defined role in a different formula, so keep each context separate.

Task	Original formula	Required subject	Rearranged formula	Condition
1	v=u+at	t	t=(v−u)/a	a≠0
2	A=(a+b)h/2	b	b=2 A/h−a	h≠0

For v=u+at, subtract u from both sides: v−u=at. Dividing both sides by a gives $t = \frac{v - u}{a} (a \neq 0)$ . The entire differencev−u is divided by a, not only u. In a stated constant-acceleration model withu 2 m/s, v 14 m/s and a 3 m/s², time is(14−2)/3=4 s. Check the original 2+3×4=14 m/s. These are invented model quantities, not operating guidance for vehicles.

If a=0, the original relation becomesv=u and no unique time follows from it. If v=u, every time in the model domain satisfies it; otherwise none does. The divided formula is not valid by quietly substituting a=0. The nonzero condition describes a real algebraic restriction, not optional decoration.

For A=(a+b)h/2, multiply both sides by 2: 2 A=(a+b)h. Divide by h, then subtract a, giving b=2 A/h−a, for h≠0. In a trapezium-area model withA 30 cm², a 4 cm and perpendicular height h 5 cm, this gives b=60/5−4=8 cm. Check(4+8)×5/2=30 cm². Here a and b are parallel-side lengths, not the acceleration variable from the first task.

For a physical trapezium, height and side lengths must be positive and the chosen numbers must describe a feasible figure. The algebraic rearrangement itself only needs a nonzero divisor. A negative computed base in a proposed geometry model would signal that its givens do not form the intended positive-length trapezium; it would not justify hiding the sign.

Compare inverse steps with substitution into the original formula. The former derives the new subject for all permitted values; the latter checks a particular numerical case and its units. Rearranging an expression cannot be checked solely by looking for matching symbols. Retain both complete formula derivations, the eight manipulation answers, their domains, a first-error correction and the unfamiliar expansion. Record actual later retrieval rather than claiming lasting fluency from one worked example.

Interactive Tasks

Quiz: Test Your Knowledge

Expand(x+3)(x+5). (x²+8 x+15) (!x²+15) (!x²+8 x+8) (!x²+15 x+8)

Collectx²+x². (2 x²) (!x⁴) (!2 x⁴) (!x²)

Simplifya³ timesa⁴. (a⁷) (!a¹²) (!2 a⁷) (!a)

12 x² divided by 3 x, with x nonzero? (4 x) (!4 x²) (!9 x) (!4)

Square the whole 2 x³. (4 x⁶) (!2 x⁶) (!4 x⁹) (!2 x⁹)

Why retainx≠0 after cancelling? (The original denominator is zero atx 0) (!The result 4 x is never defined) (!All polynomials exclude zero) (!Cancellation proves zero is a prime)

Subjectt in v=u+at, anonzero? (The whole differencev−u divided bya) (!v−u divided bya then added tov) (!v+u+a) (!a divided byv−u)

InA 30, a 4, h 5 trapezium model, the other base? (8 cm) (!12 cm) (!56 cm) (!26 cm)

Can one successful substitution prove a polynomial identity? (No; an algebraic derivation is needed) (!Yes; one value proves all values) (!Only if the value is 1) (!Only if a calculator is used)

Expand(3 x+2)(x −4). (3 x²−10 x −8) (!3 x²+14 x −8) (!3 x²−8) (!3 x²−12 x+2)

Memory Game

Like terms	Terms with the same variable powers
Distributivity	Multiplication over every term in a sum
Factorisation	Writing a sum as a product of factors
Identity	Equality valid throughout the common stated domain
Subject	Variable isolated by rearranging a formula
Domain	Permitted values for the original expression

Drag and Drop

x²+8 x+15	Expansion of the source two-bracket product
3 x times 2 x+3	Factor structure of 6 x²+9 x
a⁷	Product ofa³ anda⁴
4 x for nonzero x	Equivalent form of 12 x² divided by 3 x
8 cm	Other base in the stated trapezium model

Crossword Puzzle

DISTRIBUTE	Multiply across every bracket term
FACTORISE	Rewrite a sum as a product
IDENTITY	Equality for all permitted variable values
SUBJECT	Variable isolated in a formula
DOMAIN	Original permitted values
COEFFICIENT	Numerical factor multiplying a variable term

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Expand the source bracket pair: List all four products before collecting.
Distinguish sum and product: Comparex²+x² withx²×x².
Factor a shared term: Show 6 x²+9 x=3 x times 2 x+3 and expand back.
Read a quotient domain: Explain why the original 0/0 cannot be evaluated.

Standard

Complete eight manipulations: Retain algebraic reasoning and stated domains.
Rearrange both formulae: Show inverse operations on both complete sides.
Check model units: Use time 4 s and trapezium base 8 cm in the original formulae.
Repair missing cross-products: Find the first invalid step when only first and last bracket terms are multiplied.

Advanced

Transfer a new expansion: Expand(3 x+2)(x −4) and verify a selected value.
Explain degenerate formula cases: Discussa 0 in v=u+at without dividing by zero.
Compare proof and checking: Give a false identity matching one substitution and a counterexample elsewhere.
Retain actual later work: Keep eight answers, two derivations, domain checks and real delayed retrieval.

Learning Check

Eight manipulation answers: Derive the saved result table with domains.
Two rearranged formulae: Show each inverse operation and divisor restriction.
Numerical and unit checks: Use both formula contexts independently.
First-error and domain analysis: Correct addition-index misuse and invalid cancellation atzero.
Unfamiliar transfer and proof: Expand the new product and distinguish a check from proof.
Portfolio and model limits: Retain actual work and conditions.

Worked checks and assessment guidance

x²+8 x+15; 2 x²+7 x −4; 2 x²+7 x; 3 x(2 x+3); a⁷; 4 xfor x≠0; 4 x⁶; x³ for x≠0. Expansion/factorisation preserve polynomial coefficients; quotients retain original zero exclusions.
v−u=at gives t=(v−u)/a for a≠0.2 A=(a+b)h gives 2 A/h=a+b then b=2 A/h−a for h≠0. Whole sides and whole differencev−u are transformed.
u 2, v 14, a 3 gives t 4 s and 2+3×4=14 m/s. A 30, a 4, h 5 givesb 8 cm and(4+8)×5/2=30 cm². Symbola has a different defined meaning in each context.
x²+x²=2 x², notx⁴; the latter is the product.12 x²/(3 x)=4 x only for x≠0; the simpler expression alone being defined atzero does not extend the original quotient. Squaring 2 x³ also squares 2.
(3 x+2)(x −4)=3 x²−10 x −8; at x=2 both give −16. One substitution cannot prove all values: x²+x equals 2 x at 1 but fails at 2. Distributivity establishes the general identity.
Keep eight transformations, two subject derivations, unit checks, first-error correction and new expansion. Positive-length area models require feasible geometry; actual measurements, vehicle behaviour and delayed mastery are not invented.

Learning evidence

Retain eight algebraic manipulations, two full subject rearrangements, domain restrictions, independent checks and unfamiliar expansion.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, number, ratio, fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

Illustration for the area of a rectangle.svg, Stephan Kulla, CC0. Unchanged original; caption and learning activities are original.

Cuboid.abc.png, Cuboid.png: en:User:Tomruen, CC BY-SA 4.0. Unchanged original; caption and learning activities are original.