Recall: what are factors
Splitting 12 as 3 × 4 gives its factors. In the same way (x + 3) and (x + 4) are the factors of x² + 7x + 12.
- Common factor: 4xy + 8xy² = 4xy(1 + 2y)
- Difference of squares: p² − 9q² = (p + 3q)(p − 3q)
Before factorising, always take out the common factor first. Then the work becomes easier. Memory clue: Before factorising always follow "common factor first, then the formula".
Factors of x² + (a + b)x + ab
A rectangle has area x² + 7x + 12. What are its length and width? Find two numbers with product 12 and sum 7: 3 and 4. So the sides are (x + 3) and (x + 4).
This is multiplication run backwards. Multiplying turned brackets into an expression; here we turn the expression back into brackets.
In the figure: x² + 7x + 12 = (x + 3)(x + 4): the factors of 12 are 3 and 4 with sum 7 — the rectangle has four parts x², 4x, 3x, 12.
Formula: x² + (sum)x + (product) = (x + first number)(x + second number)
In words: Find two numbers whose product is the last number and whose sum is the middle number.
Factors of x² + (a + b)x + ab
In the app you can move this figure with your finger.
Open in the app- Easy: x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4) (3 × 4 = 12, 3 + 4 = 7)
- Medium: x² − x − 20 = x² − 5x + 4x − 20 = x(x − 5) + 4(x − 5) = (x − 5)(x + 4) (−5 × 4 = −20, −5 + 4 = −1)
- Exam-type: y² − 11y + 28 = y² − 4y − 7y + 28 = y(y − 4) − 7(y − 4) = (y − 4)(y − 7) (−4 × −7 = 28, −4 − 7 = −11)
Sign rule
| Last number | Middle number | Signs of the two numbers |
|---|---|---|
| + | + | both + |
| + | − | both − |
| − | any | one + , one − (bigger one takes the middle sign) |
Check: multiply the brackets back. If you get the original expression, the answer is right.
3 and 4 work for 12 only when the sum is 7; for x² + 8x + 12 you need 2 and 6.
List all factors of the last number and pick the pair whose sum equals the middle number. Memory clue: x² + (a + b)x + ab: "product is the last number, sum is the middle number".
Spot the difference
| Point | a = 1 (x² + bx + c) | a ≠ 1 (ax² + bx + c) |
|---|---|---|
| Factors of | the last number | first × last |
| Example | x² + 7x + 12 | 2x² + 7x + 3 |
| Pair (sum = middle) | 3 and 4 | 1 and 6 |
| Factors | (x + 3)(x + 4) | (2x + 1)(x + 3) |
Practice questions
- 1x² + 9x + 20
- 2y² − 8y + 15
- 3m² + 2m − 15
- 4a² − 3a − 40
- 5p² + 14p + 33
- 6x² − 12x + 35
- 7x² + 2x − 48
- 8A rectangle has area y² + 10y + 21. Find its sides.
Show answers (try them yourself first)
- 1x² + 9x + 20 = x² + 4x + 5x + 20 = x(x + 4) + 5(x + 4) = (x + 4)(x + 5)
- 2y² − 8y + 15 = y² − 3y − 5y + 15 = y(y − 3) − 5(y − 3) = (y − 3)(y − 5)
- 3m² + 2m − 15 = m² + 5m − 3m − 15 = m(m + 5) − 3(m + 5) = (m + 5)(m − 3)
- 4a² − 3a − 40 = a² − 8a + 5a − 40 = a(a − 8) + 5(a − 8) = (a − 8)(a + 5)
- 5p² + 14p + 33 = p² + 3p + 11p + 33 = p(p + 3) + 11(p + 3) = (p + 3)(p + 11)
- 6x² − 12x + 35 = x² − 5x − 7x + 35 = x(x − 5) − 7(x − 5) = (x − 5)(x − 7)
- 7x² + 2x − 48 = x² + 8x − 6x − 48 = x(x + 8) − 6(x + 8) = (x + 8)(x − 6)
- 8y² + 10y + 21 = y² + 3y + 7y + 21 = y(y + 3) + 7(y + 3) = (y + 3)(y + 7); sides (y + 3) and (y + 7).
Factors of ax² + bx + c (when the first number is not 1)
In 2x² + 7x + 3 the x² has a 2 in front, so we use the product 2 × 3 = 6 and look for numbers 1 and 6 (sum 7).
Multiply the first and last numbers. Find two numbers with that product and with sum equal to the middle number. Split the middle term with them and group in twos.
In the figure: 2x² + 7x + 3 = (2x + 1)(x + 3): the factors of 2 × 3 = 6 are 1 and 6 (sum 7); the rectangle has four parts 2x², x, 6x, 3.
Factors of ax² + bx + c (when the first number is not 1)
In the app you can move this figure with your finger.
Open in the appIn words: Step 1: find (first × last). Step 2: pick two numbers with that product and sum = middle. Step 3: split the middle term. Step 4: group and take out the common factor.
- Easy: 2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) (2 × 3 = 6 ; 1 + 6 = 7)
- Medium: 3m² − 10m + 8 = 3m² − 4m − 6m + 8 = m(3m − 4) − 2(3m − 4) = (3m − 4)(m − 2) (3 × 8 = 24 ; −4 − 6 = −10)
- Exam-type: 6y² + y − 15 = 6y² + 10y − 9y − 15 = 2y(3y + 5) − 3(3y + 5) = (3y + 5)(2y − 3) (6 × −15 = −90 ; 10 − 9 = 1)
Common factor first: 3x² − 12x − 36 = 3(x² − 4x − 12) = 3(x − 6)(x + 2).
After grouping both brackets must be identical. If not, the split numbers were wrong.
When splitting the middle term the product must be (first × last), not just the last number.
Find first × last, then pick factors of it whose sum is the middle number. Memory clue: When a ≠ 1 split using "first × last".
Spot the difference
| Point | Expansion (multiplying) | Factorisation |
|---|---|---|
| Action | removing brackets | making brackets |
| Example | (2x + 1)(x + 3) → 2x² + 7x + 3 | 2x² + 7x + 3 → (2x + 1)(x + 3) |
| Middle term | gets combined | gets split into two terms |
| Check | each undoes the other | each undoes the other |
Practice questions
- 12a² + 9a + 10
- 23x² + 11x + 6
- 32p² − 5p + 2
- 43x² − 7x − 6
- 55m² + 13m + 6
- 64y² − 12y + 5
- 72x² − 2x − 12 (take the common factor first)
- 86a² − 13a + 6
Show answers (try them yourself first)
- 12a² + 9a + 10 = 2a² + 4a + 5a + 10 = 2a(a + 2) + 5(a + 2) = (a + 2)(2a + 5)
- 23x² + 11x + 6 = 3x² + 9x + 2x + 6 = 3x(x + 3) + 2(x + 3) = (x + 3)(3x + 2)
- 32p² − 5p + 2 = 2p² − 4p − p + 2 = 2p(p − 2) − 1(p − 2) = (p − 2)(2p − 1)
- 43x² − 7x − 6 = 3x² − 9x + 2x − 6 = 3x(x − 3) + 2(x − 3) = (x − 3)(3x + 2)
- 55m² + 13m + 6 = 5m² + 10m + 3m + 6 = 5m(m + 2) + 3(m + 2) = (m + 2)(5m + 3)
- 64y² − 12y + 5 = 4y² − 2y − 10y + 5 = 2y(2y − 1) − 5(2y − 1) = (2y − 1)(2y − 5)
- 72x² − 2x − 12 = 2(x² − x − 6) = 2(x − 3)(x + 2)
- 86a² − 13a + 6 = 6a² − 4a − 9a + 6 = 2a(3a − 2) − 3(3a − 2) = (3a − 2)(2a − 3)
Factors of a³ + b³
A 3 cm cube and a 2 cm cube together have volume 27 + 8 = 35. And (3 + 2)(9 − 6 + 4) = 5 × 7 = 35 — a sum of cubes also splits into two factors.
Sum of two cubes: a short bracket (a + b) and a long bracket (a² − ab + b²).
In the figure: A 3 cm cube (27) + a 2 cm cube (8) = 35 = 5 × 7; a³ + b³ = (a + b)(a² − ab + b²): (3 + 2)(9 − 6 + 4).
Formula: a³ + b³ = (a + b)(a² − ab + b²)
In words: Long bracket: square of first, − (product of both), + square of second.
Factors of a³ + b³
In the app you can move this figure with your finger.
Open in the app- Easy: x³ + 216 = x³ + 6³ = (x + 6)(x² − 6x + 36) ( 6² = 36 )
- Medium: 8m³ + 125 = (2m)³ + 5³ = (2m + 5)(4m² − 10m + 25)
- Exam-type: 54y³ + 2 = 2(27y³ + 1) = 2(3y + 1)(9y² − 3y + 1) (take 2 out first)
The first bracket of a³ + b³ is (a + b); in the second bracket the middle sign is − and the last is +.
(a + b)³ and a³ + b³ are different. a³ + b³ ≠ (a + b)³.
Same – Opposite – Positive: the first bracket keeps the sign (+); the middle sign of the long bracket is opposite (−); its last sign is always +. Memory clue: Sum of cubes: (sum of the two)(square − product + square).
Spot the difference
| Point | (a + b)³ | a³ + b³ |
|---|---|---|
| Kind | expansion | factorisation |
| Result | a³ + 3a²b + 3ab² + b³ | (a + b)(a² − ab + b²) |
| For a = 3, b = 2 | 125 | 35 |
| Are they equal | no | no |
Practice questions
- 1x³ + 8
- 2y³ + 1
- 3m³ + 64
- 427a³ + 8
- 58x³ + 27y³
- 6p³ + 125q³
- 716m³ + 2
- 8Write 10³ + 1³ as a product of factors and find its value.
Show answers (try them yourself first)
- 1x³ + 8 = (x + 2)(x² − 2x + 4)
- 2y³ + 1 = (y + 1)(y² − y + 1)
- 3m³ + 64 = (m + 4)(m² − 4m + 16)
- 427a³ + 8 = (3a + 2)(9a² − 6a + 4)
- 58x³ + 27y³ = (2x + 3y)(4x² − 6xy + 9y²)
- 6p³ + 125q³ = (p + 5q)(p² − 5pq + 25q²)
- 716m³ + 2 = 2(8m³ + 1) = 2(2m + 1)(4m² − 2m + 1)
- 810³ + 1³ = 1000 + 1 = 1001 = (10 + 1)(100 − 10 + 1) = 11 × 91
Factors of a³ − b³
Remove a 2 cm cube from a 5 cm cube: 125 − 8 = 117. By the formula (5 − 2)(25 + 10 + 4) = 3 × 39 = 117.
Difference of two cubes: short bracket (a − b), long bracket (a² + ab + b²).
In the figure: A 5 cm cube (125) − a 2 cm cube (8) = 117 = 3 × 39; a³ − b³ = (a − b)(a² + ab + b²): (5 − 2)(25 + 10 + 4).
Formula: a³ − b³ = (a − b)(a² + ab + b²)
In words: Flip the signs of the sum formula: first bracket −, middle term +, last term still +.
Factors of a³ − b³
In the app you can move this figure with your finger.
Open in the app- Easy: y³ − 64 = y³ − 4³ = (y − 4)(y² + 4y + 16)
- Medium: 27m³ − 8 = (3m)³ − 2³ = (3m − 2)(9m² + 6m + 4)
- Exam-type: 250p³ − 16q³ = 2(125p³ − 8q³) = 2(5p − 2q)(25p² + 10pq + 4q²)
The first bracket of a³ − b³ is (a − b); all signs in the second bracket are +.
The long bracket cannot be factorised further; leave it as it is.
Sign order: the first bracket keeps the sign, the middle one flips, the last is always +. Memory clue: Difference of cubes: (difference of the two)(square + product + square).
Spot the difference
| Point | a³ + b³ | a³ − b³ |
|---|---|---|
| Short bracket | (a + b) | (a − b) |
| Middle term | − ab | + ab |
| Last term | + b² | + b² |
Practice questions
- 1x³ − 1
- 2m³ − 125
- 3a³ − 27
- 48y³ − 27
- 564x³ − y³
- 627p³ − 216q³ (take 27 out first)
- 716m³ − 54
- 8Factorise (x + 2)³ − (x − 2)³ with a = x + 2, b = x − 2.
Show answers (try them yourself first)
- 1x³ − 1 = (x − 1)(x² + x + 1)
- 2m³ − 125 = (m − 5)(m² + 5m + 25)
- 3a³ − 27 = (a − 3)(a² + 3a + 9)
- 48y³ − 27 = (2y − 3)(4y² + 6y + 9)
- 564x³ − y³ = (4x − y)(16x² + 4xy + y²)
- 627p³ − 216q³ = 27(p³ − 8q³) = 27(p − 2q)(p² + 2pq + 4q²)
- 716m³ − 54 = 2(8m³ − 27) = 2(2m − 3)(4m² + 6m + 9)
- 8(x + 2)³ − (x − 2)³ = 4 × ((x+2)² + (x+2)(x−2) + (x−2)²) = 4(3x² + 4) = 12x² + 16.