01 · Signs and like terms
Expand and simplify.
2x(3x − 4) − (x − 5)(x + 1)
Hint
Expand the second product before applying its outside minus sign.
Worked solution
= 6x² − 8x − (x² − 4x − 5)
= 6x² − 8x − x² + 4x + 5
= 5x² − 4x + 5
Understand · explore · practise
Expanding and factorising
Move between products and sums, expand two or three brackets, and factorise quadratics and higher powers without losing terms.
Before you startSigned numbers and the laws of indices
01 / Distribute
A multiplier belongs to every term.
Brackets group an expression into one quantity. Multiplying that quantity means multiplying all its terms.
k(a + b) = ka + kb
A minus sign also distributes. For example, −(u − 2v) = −u + 2v. You can think of the outside minus as multiplication by −1.
After expanding, collect like terms: terms with exactly the same variable part. The coefficients of 7x² and −3x² combine; x and x² do not.
7x² − 3x² + 5x = 4x² + 5x
Keep the sign attached to its term. In −3x(2x − 5), multiplying two negative quantities gives a positive term.
Keep the sign with the termWorked example
−3x(2x − 5) + 2x²
There are two terms inside the bracket.
= −6x² + 15x + 2x²
Multiply −3x by both 2x and −5.
= −4x² + 15x
Only the x² terms combine.
At x = 1: −4 + 15 = 11
The original also gives −3(−3) + 2 = 11.
02 / Two brackets
Four products. Then collect.
In (2x + 3)(x + 4), each term in the first bracket multiplies each term in the second. The area model makes it possible to account for all four products.
(2x + 3)(x + 4)
= 2x² + 8x + 3x + 12
= 2x² + 11x + 12
Choose a product below the picture to inspect its region. Both middle regions contain one factor of x, so their areas combine to 11x.
The same method works for more terms and more variables:
(2a − b)(a + 3b − 2)
= 2a² + 6ab − 4a
− ab − 3b² + 2b
= 2a² + 5ab − 3b² − 4a + 2b
The picture uses positive lengths. The expanded identity itself remains true for every real x. For two terms multiplied by three terms, expect six products before collecting.
Every term meets every termArea model
The two x terms combine: 3x + 8x = 11x. The other terms stay separate.
Watch all four areas appear
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Three brackets
Multiply a pair, then use the third.
Do not try to hold every product in your head. Expand two brackets, collect their terms, and multiply the result by the remaining bracket.
You can choose the order. A pair such as (x − c)(x + c) is often useful because its middle terms cancel.
(a + b)(a − b) = a² − b²
For a squared bracket, keep the middle term:
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
Squaring a sum is not the same as squaring its parts. At a = b = 1, (a + b)² is 4, while a² + b² is 2.
Extend the same idea to a fourth power
Square the squared bracket: (a + b)⁴ = (a² + 2ab + b²)². Multiplying all nine term pairs and collecting gives:
a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
The binomial expansion gives a quicker way to organise these coefficients later in Pure 1.
A cubic from three bracketsWorked example
(x − 1)(x + 4)(2x − 3)
Start with the first pair.
= (x² + 3x − 4)(2x − 3)
Collect x terms before continuing.
= 2x³ + 6x² − 8x
− 3x² − 9x + 12
Distribute both 2x and −3 over the whole quadratic.
= 2x³ + 3x² − 17x + 12
Collect equal powers. At x = 0, both forms give 12.
04 / Factorise
Read the multiplication backwards.
Factorising rewrites a sum as a product. First look for a factor shared by every term, including a numerical factor.
14x²y − 21xy² = 7xy(2x − 3y)
For ax² + bx + c, with a ≠ 0, one useful method is to split bx. Find two numbers whose product is ac and whose sum is b, then group the terms.
In the worked example, ac = −72. The numbers 9 and −8 multiply to −72 and add to 1.
For a monic quadratic, x² + bx + c, this reduces to finding two numbers with sum b and product c:
x² − 2x − 15 = (x − 5)(x + 3)
Not every quadratic factorises into brackets with integer coefficients. If none fit, do not invent a pair: other solving methods appear in the quadratics chapter.
Split the middle termWorked example
6x² + x − 12
Product ac = −72; required sum b = 1.
= 6x² + 9x − 8x − 12
Replace x with 9x − 8x.
= 3x(2x + 3) − 4(2x + 3)
Factor each pair. The bracket is now shared.
= (3x − 4)(2x + 3)
Expand the brackets to check all three terms.
05 / Higher powers
Keep going after the first factor.
“Factorise completely” means checking whether any remaining factor can be broken down further in the required number system. In these examples, we use integer coefficients.
4x³ + 2x² − 6x
= 2x(2x² + x − 3)
= 2x(2x + 3)(x − 1)
A cubic with a common x becomes a quadratic after that x is removed. Likewise, an expression containing x⁴, x² and a constant can be treated as a quadratic in u = x².
The difference of squares can also repeat:
a⁴ − b⁴ = (a² − b²)(a² + b²)
= (a − b)(a + b)(a² + b²)
There is no matching real-number rule that splits a² + b² into (a + b)(a − b): those brackets produce a difference.
Using factors to simplify a fraction
(6x² + 9x)/(3x) = 3x(2x + 3)/(3x)
= 2x + 3, x ≠ 0
Cancel a shared factor of the entire numerator and denominator. Do not cancel individual terms across addition. Keep the excluded value x = 0 even after simplification.
A quadratic hidden inside a quarticWorked example
x⁴ − 10x² + 9
Let u = x², so x⁴ = u².
u² − 10u + 9 = (u − 1)(u − 9)
The factor pair is −1 and −9.
= (x² − 1)(x² − 9)
Put x² back in place of u.
= (x − 1)(x + 1)(x − 3)(x + 3)
Each remaining bracket is a difference of squares.
06 / Your turn
Use expansion to check factorisation.
Try these without looking at the solutions. A correct factorisation must expand to the original expression, including its constant and every sign.
02 · Work backwards
Choose p so that (x + p)(x − 6) has no x term. Then write the expanded expression.
Hint
The coefficient of x is p − 6.
Worked solution
(x + p)(x − 6) = x² + (p − 6)x − 6p
p − 6 = 0, so p = 6
(x + 6)(x − 6) = x² − 36
03 · Factor completely
Write as a product of factors with integer coefficients.
10x³ − 7x² − 12x
Hint
Remove x first. For the quadratic, split the middle coefficient using ac = −120.
Worked solution
= x(10x² − 7x − 12)
= x(10x² + 8x − 15x − 12)
= x[2x(5x + 4) − 3(5x + 4)]
= x(2x − 3)(5x + 4)
04 · Match coefficients
Find b and c so this identity is true for every x. Then factorise the cubic completely.
(x − 2)(2x² + bx + c)
= 2x³ + x² − 13x + 6
Hint
Expand the left. Match the x² coefficient and the constant first.
Worked solution
2x³ + (b − 4)x² + (c − 2b)x − 2c
b − 4 = 1 ⇒ b = 5
−2c = 6 ⇒ c = −3
The x coefficient checks: c − 2b = −3 − 10 = −13.
2x² + 5x − 3 = (2x − 1)(x + 3)
So the cubic is (x − 2)(2x − 1)(x + 3).
05 · Repeated structure
Factorise x⁶ − 16x² completely over the integers.
Hint
Take out x². Then apply the difference of squares twice where possible.
Worked solution
x⁶ − 16x² = x²(x⁴ − 16)
= x²(x² − 4)(x² + 4)
= x²(x − 2)(x + 2)(x² + 4)
x² + 4 has no real linear factors, so there is no further integer factorisation.
06 · Interpret a product
An open rectangular tray has internal base dimensions (x + 1) cm and (x + 3) cm and depth (x − 1) cm, where x > 1. Find its capacity as an expanded expression in cm³.
Hint
Pair (x + 1) with (x − 1) before using the third dimension.
Worked solution
V = (x + 1)(x − 1)(x + 3)
= (x² − 1)(x + 3)
= x³ + 3x² − x − 3
The condition x > 1 makes all three lengths positive. The formula is a volume, so the units are cm³.
07 / Recap
Products and sums reveal different things.
- Expand: multiply every term in one factor by every term in the other.
- Collect: only combine terms with identical variable parts.
- Three brackets: expand one pair, collect, then multiply by the third.
- Factorise: begin with a common factor, then look for quadratic factors or a difference of squares.
- Check: expand your factors back out. A spot-check is useful for catching errors, but does not prove an identity.
Section 1 of 7 · Distribute