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Algebraic expansion factorisation and formulae - Mathematics Year 8

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Introduction

What changes when you expand a bracket, and what must remain equal? Connect distributivity, like terms, common factors and inverse operations rather than memorising isolated letter-moving rules.

You will simplify or expand eight expressions, factorise six expressions and rearrange two formulae. Explain each identity, compare methods, and test with substitution while distinguishing a numerical check from a general algebraic argument.

Expand and simplify eight complete expressions

An expression represents a value but does not by itself ask for a unique solution. An identity states that two expressions agree for every permitted substitution. Distributivity, k(u+v)=ku+kv, multiplies every term inside the bracket by k. Expanding changes the representation, not the value. Factorising uses the same law backwards to write a sum as a product.

Original worked simplification. $2 (3 x - 4) + 5 x = 6 x - 8 + 5 x = 11 x - 8$ . The outside 2 multiplies both 3x and −4. Then 6x and 5x are like terms. The constant −8 is not another x-term. Check x = 2: the original is 2(6−4)+10=14 and the simplified expression is 22−8=14. That check can expose a mistake, while the distributive and like-term steps justify the identity generally.

For −2(4x−3)+x, expansion gives −8x+6+x=−7x+6. The negative multiplier affects every term, so the product with −3 is +6. Expanding only the first term would produce a different expression. At x = −1, both complete forms give 13, which also tests the sign handling.

Eight original expressions, with equivalent collected forms.

Original expression	Equivalent simplified form
2(3x−4)+5x	11x−8
3(x+4)−2x	x+12
−2(4x−3)+x	−7x+6
5a+2a−3	7a−3
4(2y+1)−3(y−2)	5y+10
−(3p−5)+2p	−p+5
2x²+3x−x²+4x	x²+7x
(1/2)(6m−8)+m	4m−4

Cover the second column before attempting the set. Collect only terms with matching variable parts and powers: x² and x remain different types. The fractional multiplier in the last row distributes to both 6m and −8, giving 3m−4+m. A rectangle partition can model positive area sums; the symbolic law, not an invented negative physical side, supports negative coefficients.

Choose between expanding first or substituting first for a particular numerical value. Both can give a useful check. If a task asks for an identity, one successful substitution is insufficient: a wrong expression can agree at one input. Preserve the general symbolic reasoning as well as the numerical check.

Factorise six expressions and reverse-check

Factorising by a common factor divides every term by the chosen factor and places the quotients inside brackets. For the source example, $6 a + 9 = 3 (2 a + 3)$ . Expanding the right side recovers both terms. A common factor must divide each term; removing it from just one term changes the value.

Six original factorisations.

Expression	One valid factored form
6a+9	3(2a+3)
12x−18	6(2x−3)
8y+12	4(2y+3)
15p²−10p	5p(3p−2)
−6z+9	3(−2z+3)
14q²+21q	7q(2q+3)

Take the greatest common numerical factor where useful and include common variables with their lowest shared power. In 15p²−10p, the shared factor is 5p, leaving 3p−2. This is an identity for all p: no step divides the entire equation by p or excludes p = 0. Expanding 5p(3p−2) restores the original even at zero.

Factored forms need not look identical to be equivalent. For −6z+9, −3(2z−3) also works. The chosen table form uses positive factor 3; a negative common factor is valid if every bracket sign changes correctly. Check by expansion instead of marking a different equivalent arrangement wrong just because it differs from the printed answer.

Compare a method with a model. A rectangle with a shared positive side length 3 and adjacent widths 2a and 3 has area 6a+9=3(2a+3) when those widths are positive lengths. The same algebraic identity extends to other permitted a values by distributivity, even where that particular physical interpretation would not use a negative width. State the model's limitation instead of treating a diagram as proof of every sign case.

Use Corbettmaths’ original expanding-brackets explanation to check that every term is multiplied. Its visible 4(x+2) example is separate from our eight-expression and six-factorisation sets. Reverse the distributive step to verify each factored answer. Original: corbettmaths, Expanding Brackets - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Error clinic. Expanding 2(3x−4) as 6x−4 misses the multiplier on the constant. Collecting x²+7x as 8x² combines unlike powers. Factorising 6a+9 as 3(2a+9) fails to divide the second term inside the bracket. Locate and correct the first invalid representation, then expand or substitute a deliberately informative value to check.

Rearrange two formulae with units and conditions

Rearranging a formula changes its subject while preserving the same relationship. Apply inverse operations to both sides, keeping brackets around a whole numerator where needed. A formula can relate several variables; rearranging does not mean choosing a single numerical solution before values are supplied.

Formula one: rectangle perimeter. From P = 2l + 2w, subtract 2w from both sides and divide both sides by 2: $l = \frac{P - 2 w}{2} = \frac{P}{2} - w$ . These are two equivalent rearrangements. Do not write l = P − 2w/2, which divides only the final term. With model perimeter 34 cm and width 5 cm, l = (34−10)/2 = 12 cm. Substitute: 2×12 + 2×5 = 34 cm.

Formula two: cuboid volume. From V = lwh, divide by the whole non-zero product lw to obtain $h = \frac{V}{l w} (l w \neq 0)$ . For positive cuboid lengths this condition is satisfied. With V = 240 cm³, l = 8 cm and w = 5 cm, h = 240/40 = 6 cm. Check 8×5×6 = 240 and note that cm³ divided by cm² leaves cm.

The area of a base is lw, so a second route first names base area B = lw and uses h = V/B. This can reduce repeated calculations, provided B is clearly defined and not silently confused with the source image's uppercase edge label B. Algebraic letters are definitions in a model, not universal names fixed by every picture.

Unfamiliar transfer. Expand −3(2x−5)+4x to get −2x+15 and check x = 0 and x = −2. Factorise 18t²+12t as 6t(3t+2), then expand to confirm. For perimeter 46 cm and width 8 cm, the same rearranged formula gives l = 15 cm; checking 30+16=46 retains the original relationship and units.

Retain eight simplified expressions, six checked factorisations and two complete rearrangements with both numerical contexts. Explain why substitution detects particular mistakes without replacing a general identity proof. A later unfamiliar response after a genuine break is separate evidence; no actual physical measurement or long-term mastery is invented by these ideal models.

Interactive Tasks

Quiz: Test Your Knowledge

What is 2(3x−4)+5x after collecting like terms? (11x−8) (!11x−4) (!6x+1) (!11x+8)

What is −2(4x−3)+x? (−7x+6) (!−7x−6) (!−9x+6) (!−8x−3)

Why can x squared and seven x not be collected into eight x squared? (Their variable powers differ) (!They have the same coefficient) (!Only constants can ever be collected) (!The letter x must be positive)

Which factored description equals 6a+9? (3 times the bracket 2a plus 3) (!3 times the bracket 2a plus 9) (!6 times the bracket a plus 9) (!9 times the bracket 6a plus 1)

What is the common factor used for 15p squared minus 10p in the table? (5p) (!5p squared) (!10p) (!15p)

What checks a factored expression generally? (Expanding it back to the original expression) (!Trying one random value only) (!Counting its printed symbols) (!Removing all brackets without multiplication)

What length follows P=34 cm and w=5 cm in P=2l+2w? (12 cm) (!29 cm) (!24 cm) (!17 cm)

What height follows volume 240 cubic cm with base lengths 8 cm and 5 cm? (6 cm) (!48 cm) (!40 cm) (!192 cm)

What condition is needed when dividing by lw? (The product lw is non-zero) (!The product lw equals zero) (!Only l must be a whole number) (!The letters must have the same name as the image)

What is −3(2x−5)+4x? (−2x+15) (!−2x−15) (!−10x+15) (!−6x−5)

Memory Game

Identity	Equality holding for every permitted substitution
Distributivity	Multiplying each term of a bracket by the outside factor
Expansion	Rewriting a product with a bracket as a sum of products
Factorisation	Rewriting a sum as a product using shared factors
Subject	Variable isolated in the rearranged formula
Coefficient	Numerical factor multiplying a variable term

Drag and Drop

11x−8	Collected form of the first expression
−7x+6	Collected form of the negative-multiplier example
5p	Common factor taken from 15p squared minus 10p
12 cm	Length in the 34-cm perimeter model
6 cm	Height in the 240-cubic-centimetre cuboid model

Crossword Puzzle

IDENTITY	Equality true for every permitted input
EXPANSION	Distributing a product across a bracket
FACTORISATION	Writing a sum as a product
SUBJECT	Isolated variable of a formula
COEFFICIENT	Numerical factor of a variable term
INVERSE	Operation used to undo another operation

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

Complete eight expressions: Expand or simplify every row, preserving variable powers and negative signs.
Check the source expression: Use x 2 in both 2(3 x −4)+5 xand 11 x −8, then state the general distributive reason.
Model a positive area split: Declare positive rectangle lengths and show 3(2 a+3)=6 a+9.
Find the first sign error: Explain why −2 times −3 gives+6 in the negative bracket.

Standard

Factorise all six: Give valid factored forms and expand every one to check.
Compare equivalent arrangements: Explain why 3(−2 z+3)and −3(2 z −3)both equal −6 z+9.
Rearrange perimeter: Derive both l=(P −2 w)/2 andl=P/2−w; use 34 cmand 5 cm.
Rearrange volume: Deriveh=V/(lw), state the non-zero condition and use 240 cm³ with 8 cmand 5 cm.

Advanced

Transfer expansion and factors: Use −3(2 x −5)+4 xand 18 t²+12 twith symbolic checks.
Explain check versus proof: Give a reason one successful substitution cannot prove a universal identity.
Interpret units and variable names: Explaincm³/cm²=cm and distinguish defined base area from an image’s edge label.
Retrieve later: Keep an actual later expansion, factorisation and rearrangement with corrected first errors and checks.

Learning Check

Eight equivalent forms: Give every simplified expression.
Six factored forms: Give and verify all common-factor products.
Perimeter rearrangement: Derive l and interpret the numeric case.
Volume rearrangement: Derive h with its condition and unit check.
Unfamiliar transfer and errors: Expand, factorise and use a new perimeter.
Full evidence and actual later work: Distinguish model, identity and check.

Worked checks and assessment guidance

11 x −8; x+12; −7 x+6; 7 a −3; 5 y+10; −p+5; x²+7 x; 4 m −4. Every outside coefficient multiplies every term; unlike powers remain separate. General distributivity justifies each identity.
3(2 a+3); 6(2 x −3); 4(2 y+3); 5 p(3 p −2); 3(−2 z+3); 7 q(2 q+3). Expansion restores the original expressions. A negative common factor can give another valid arrangement; p 0 remains permitted in the identity.
P −2 w=2 l, so l=(P −2 w)/2=P/2−w. WithP 34 cm, w 5 cm, l 12 cm; 2×12+2×5=34 cm. The entire numerator is divided by 2.
V=lwh givesh=V/(lw)forlwnonzero.240 cm³/(8 cm×5 cm)=6 cm; 8×5×6=240. Defining base area first gives the same result; uppercase picture labels do not set numeric lengths.
−3(2 x −5)+4 x=−2 x+15; 18 t²+12 t=6 t(3 t+2); P 46 cm, w 8 cm→l 15 cm. Correct missed constant multiplication, unlike-power collection and failure to divide every factored term.
Keep eight expansions, six factors, two formulae, unit contexts and independent checks. A positive area picture has domain limits; substitution checks particular values while symbolic laws justify identities. Actual measurements or later retrieval are not invented.

Learning evidence

Retain eight simplified expressions, six factorisations, two full formula rearrangements and numerical unit checks. Explain general identities separately from substitution checks and actual later evidence.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, ratio, algebra, fluency, reasoning and problem solving. Year 8 allocation is editorial within KS3. Tasks, numerical models, examples and feedback are original. The selected Commons sources retain their creators and licences below; videos are embedded from the original named 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.