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Algebraic manipulation expansion and factorisation - Mathematics Year 10

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Introduction

Expansion and factorisation are inverse ways of writing the same expression. Track every product and sign, then verify that the result is equal for all permitted values rather than merely matching one substitution.

Expand six expressions and factorise six monic quadratics. Use a product grid, distributive reasoning and independent re-expansion checks; keep the separately labelled non-monic factorisation extension distinct from the core set.

Expand six expressions with every partial product

The distributive law multiplies every term in one factor by every term in the other. For(x+3)(x+4), the four partial products arex², 4 x, 3 x and 12. Collect the like x terms to obtainx²+7 x+12. The middle coefficient is 3+4, and the constant is 3×4; both relationships will matter when reversing the operation.

A product grid supplies a systematic record. With column componentsx and 3 and row componentsx and 4, the four cells arex², 3 x, 4 x and 12. Add all cells to get the same expanded expression. This prevents a common omission: using only the first and last products and losing both middle cells.

Six complete expansion tasks. In the table, ^ or a superscript indicates a power; adjacent brackets indicate multiplication. Cover the last two columns and account for every partial product before collecting terms.

Task	Original expression	Uncollected or distributed form	Collected expansion
1	(x+3)(x+4)	x²+3 x+4 x+12	x²+7 x+12
2	(x −5)(x+2)	x²+2 x −5 x −10	x²−3 x −10
3	(2 x+3)(x −4)	2 x²−8 x+3 x −12	2 x²−5 x −12
4	3(2 x −5)	6 x −15	6 x −15
5	(x −4)²	x²−4 x −4 x+16	x²−8 x+16
6	(x+6)(x −6)	x²−6 x+6 x −36	x²−36

In row 2, the two middle products 2 x and −5 x collect to −3 x, and the constant −5×2=−10. In row 3, the coefficient 2 multiplies both terms in the other bracket: 2 x×x=2 x² and 2 x×(−4)=−8 x. The 3 also produces 3 x and −12. Combining −8 x+3 x gives −5 x.

Row 4 tests a single bracket: 3 multiplies both 2 x and −5, giving 6 x −15. A learner’s 6 x −5 first fails by leaving the second term undistributed. In row 5, a square means two equal factors: (x −4)(x −4). The two −4 x terms give −8 x, so the result is notx²+16. Row 6 has opposite middle terms which cancel; the negative constant remains −36.

The actual original Corbettmaths grid example expands the different expression(x+5)(x −3) into x²+2 x −15. Follow its four cells, then independently build the grid for our source(x+3)(x+4). The original video’s values are not the answers to all six course tasks. Original: corbettmaths, Expanding Two Brackets using a Grid - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Substitution can expose a mistake quickly. Atx=1, the source product gives 4×5=20, and its expansion 1+7+12=20 agrees. However, one successful substitution is only a spot check, not proof for every x. The distributive-law derivation supplies the general reason for equality.

Factorise six monic quadratics and verify signs

A monic quadratic has leading coefficient 1. Expanding two brackets gives $(x + p) (x + q) = x^{2} + (p + q) x + p q .$ To factorx²+bx+c over integers, seek integersp andq whose sum is b and product isc. Meeting only one condition is insufficient. Some monic quadratics do not have integer linear factors at all.

For the source x²+7 x+12, list positive factor pairs of 12: 1 and 12, 2 and 6, 3 and 4. Their sums are 13, 8 and 7, so choose 3 and 4. The answer(x+3)(x+4) is verified by reproducing all four products. This reverse-check is essential when signs or repeated factors are present.

Six complete monic factorisations. The sixth expression has zero x coefficient; read it asx²+0 x −36. Cover the factor-pair and final columns and check both sum and product.

Task	Quadratic expression	Chosen signed integers	Factorised form
1	x²+7 x+12	3 and 4	(x+3)(x+4)
2	x²−3 x −10	−5 and 2	(x −5)(x+2)
3	x²−9 x+20	−4 and −5	(x −4)(x −5)
4	x²+x −12	4 and −3	(x+4)(x −3)
5	x²−8 x+16	−4 and −4	(x −4)²
6	x²−36	−6 and 6	(x −6)(x+6)

In row 2, a negative constant means opposite signs. The pair −5, 2 has product −10 and sum −3, so it gives(x −5)(x+2). For row 3, the positive 20 requires equal signs, and the negative middle coefficient requires both to be negative: −4−5=−9 and(−4)(−5)=20.

Row 4 uses 4 and −3 because their sum 1 and product −12 match the quadratic. Row 5 repeats −4 twice, so(x −4)² is a compact form of(x −4)(x −4). Row 6 uses −6 and 6; their sum 0 removes the middle term and their product −36 supplies the constant.

Do not force an integer factorisation when no pair works. Forx²+x+1, the only positive integer factor pair of 1 is 1 and 1, whose sum is 2, not 1; the negative pair sums to −2. Thus there is no factorisation into two integer monic linear factors. This is a claim about the chosen factor type, not a proof that all possible number-system factorisations are impossible.

Separate non-monic factorisation extension. For 2 x²+7 x+3, try(2 x+1)(x+3). Re-expansion gives 2 x²+6 x+x+3, which matches. The simple core sum/product rule with constant 3 alone cannot be applied unchanged because the leading coefficient is 2. Keep this extension outside the six monic tasks and justify it through all partial products.

Transfer, interpret and distinguish proof from testing

The expression 3 x²+2 x−x²+5 x collects to 2 x²+7 x. Only terms with the same variable part can combine: the x² terms form one group and the x terms another. Replacing 3 x²+2 x with 5 x³ would combine unlike terms and confuse addition with multiplication. State whether the current operation is collecting, expanding or factorising before choosing a rule.

For unfamiliar expansion, (x+7)(x −2) givesx²−2 x+7 x −14=x²+5 x −14. The integer-factor route reverses it: find product −14 and sum 5, so 7 and −2 reconstruct the two brackets. Atx=3, the product 10×1=10 agrees with 9+15−14=10, but the symbolic expansion is what establishes equality for all real x.

For a new factorisation, x²−2 x −15 has integers −5 and 3: their product −15 and sum −2 give(x −5)(x+3). A common plausible error is choosing 5 and −3. Those have the correct product −15 but the wrong sum 2; their expansion is x²+2 x −15. Always check both conditions rather than accepting a pair after the constant matches.

An original rectangle application has positive side lengthsx+3 and x+4 centimetres. For a literal positive rectangle take x greater than −3. Its area is(x²+7 x+12)cm². Atx=2, the lengths 5 cm and 6 cm give 30 cm², and 4+14+12=30 checks the area expression. The polynomial identity remains valid even for values which do not describe that positive rectangle; the physical interpretation has a narrower domain.

Distinguish an expression from an equation. Factorisingx²+7 x+12 rewrites a value; it does not by itself ask for roots. Only if an equation such as(x+3)(x+4)=0 is supplied can a zero-product argument lead to specific solutions. Do not invent a zero on the other side while manipulating an expression.

For a repeated-factor check, expand(x −4)² as two brackets rather than squaring each displayed term independently. Atx=4, the expression is 0, and the correct expansion 16−32+16=0 agrees; the wrongx²+16 gives 32 and is exposed. Testing can reject a false identity with a single counterexample, but cannot prove an unbounded identity by a few agreeing examples.

Keep six expansions, six monic factorisations, a signed-factor first-error repair and at least one unfamiliar transfer with both symbolic and numerical checks. Reattempt an item later without its answer column and retain the actual working; do not count a copied displayed solution as independent evidence.

Interactive Tasks

Quiz: Test Your Knowledge

Expand the source(x+3)times(x+4). (x²+7 x+12) (!x²+12) (!x²+12 x+7) (!2 x+7)

Expand 3 times the bracket 2 x −5. (6 x −15) (!6 x −5) (!6 x+15) (!5 x −15)

Expand the square of x −4. (x²−8 x+16) (!x²+16) (!x²−16) (!x²−4 x+16)

Expand the product x+6 and x −6. (x²−36) (!x²+36) (!x²+12 x −36) (!2 x −36)

For x²+7 x+12, the signed factor integers are? (3 and 4) (!2 and 6) (!1 and 12) (!-3 and -4)

For x²−9 x+20, the signed factor integers are? (-4 and -5) (!4 and 5) (!-2 and -10) (!2 and 10)

For x²+x −12, the signed factor integers are? (4 and -3) (!3 and -4) (!6 and -2) (!-6 and 2)

Collect 3 x²+2 x−x²+5 x. (2 x²+7 x) (!9 x³) (!2 x²−3 x) (!5 x²+5 x)

New product x+7 and x −2 expands to? (x²+5 x −14) (!x²−5 x −14) (!x²+9 x −14) (!x²+5 x+14)

One matching substitution proves an identity for all x? (No; general algebraic reasoning is still needed) (!Yes; one example is always enough) (!Only if x is 1) (!Only for positive constants)

Memory Game

Expansion	Writing products as sums of partial products
Factorisation	Writing an expression as a product
Monic	Leading polynomial coefficient is 1
Like terms	Terms with the same variable part
Identity	Equality holding for every permitted value
Counterexample	One permitted case disproving a general claim

Drag and Drop

x²+7 x+12	Expansion of the source double bracket
6 x −15	Single-bracket result
x²−8 x+16	Repeated negative-four factor expansion
x²−36	Opposite six-factor expansion
2 x²+7 x	Collected unlike-power groups

Crossword Puzzle

EXPANSION	Distributing factors into partial products
FACTORISATION	Reverse process of writing a product
COEFFICIENT	Numerical multiplier of a variable term
MONIC	Having leading coefficient one
IDENTITY	Equality valid for every permitted value
DISTRIBUTIVE	Law connecting multiplication with sums

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

Make the source grid: Account for all four partial products.
Collect like terms: Separate x² terms from x terms.
Check a negative sign: Explain both middle terms in the square of x −4.
Name the operation: Distinguish an expression rewrite from solving an equation.

Standard

Complete six expansions: Retain all products and collected forms.
Complete six monic factorisations: Check signed pair sum, product and re-expansion.
Use the rectangle context: Find 30 cm² at x 2 and state the positive-length domain.
Repair first errors: Correct a missing distribution or wrong signed factor pair.

Advanced

Transfer both directions: Expand x+7 with x −2 and reverse the result.
Explain a failed integer factor search: State the limited claim for x²+x+1.
Separate non-monic extension: Verify 2 x²+7 x+3 by full re-expansion.
Compare proof and testing: Explain why one matching substitution is not a universal proof.

Learning Check

Six expansions: Give collected forms and retain partial products.
Six monic factorisations: State signed pair and re-expand.
New-value transfer: Check both directions independently.
Contexts and domain: Keep area meaning separate from algebraic validity.
Separate extension and limits: Verify a non-monic product and an integer-factor limitation.
Error and evidence: Retain work rather than copied final answers.

Worked checks and assessment guidance

x²+7 x+12; x²−3 x −10; 2 x²−5 x −12; 6 x −15; x²−8 x+16; x²−36. Every distributed term is included; middle terms combine or cancel according to their signs.
Pairs 3/4, −5/2, −4/−5, 4/−3, −4/−4, −6/6 give the six displayed products. Their sums 7, −3, −9, 1, −8, 0 and products 12, −10, 20, −12, 16, −36 reconstruct each original quadratic.
Product(x+7)(x −2)=x²+5 x −14, with value 10 atx 3. Quadraticx²−2 x −15 factors as(x −5)(x+3); the opposite sign choice would produce middle coefficient+2.
Source rectangle has positive lengths for x greater than −3. Atx 2, 5×6=30 cm² agrees with 4+14+12. The polynomial identity holds beyond the physical rectangle domain. Factorisation alone does not assert an equation equal to zero.
2 x²+7 x+3=(2 x+1)(x+3) by full expansion. No integer monic linear factor pair has product 1 and sum 1 for x²+x+1. This does not claim no factors in every possible number system.
3 distributes to both terms; the square of x −4 includes −8 x. A counterexample can refute an identity, but a few matching substitutions do not prove it. Keep six expansions, six monic factorisations and actual independent transfer with first-error repair.

Learning evidence

Keep six complete expansions, six monic factorisations, signed-factor and distribution repairs, unfamiliar transfer and independent symbolic/numerical checks.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Algebra - Notation vocabulary and manipulation items 1 to 6. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

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.