Expanding and Factorising Brackets
Mathematics (Core) revision notes · Cambridge (CIE) · 0580 · Core
1. Expanding a single bracket
Multiplying through
- Expanding removes a bracket by multiplying the outside term by every term inside it
- Each product keeps the sign that belongs to its term
- 5a(2a − 7) gives 10a² − 35a
- A negative outside the bracket changes the sign of every term inside
- −3y(2y − 4) gives −6y² + 12y
- Writing the negative term in brackets while multiplying keeps the sign rules visible
Worked example
Expanding a single bracket
Expand 4x(3x − 5).
Solution:
- Expanding removes the bracket by multiplying the term outside it by every term inside
- The bracket holds two terms, so there are two products to write
- Take each inside term with the sign in front of it: +3x and −5
- 4x × 3x = 12x², because 4 × 3 = 12 and x × x = x²
- 4x × (−5) = −20x, negative because one factor is negative
- The answer is 12x² − 20x
- The two terms are unlike, one in x² and one in x, so nothing collects and this is the simplest form
- Check with a value: x = 2 gives 8 × 1 = 8 in the question, and 48 − 40 = 8 in the answer
- The minus sign belongs to the 5 and travels with it, which is where most sign errors in this topic start
2. Expanding two brackets
Every term times every term
- Two brackets of two terms need four multiplications
- Work through the first bracket's terms in turn, multiplying each by the whole second bracket
- Collect like terms afterwards, which usually leaves three terms
- (x + 6)(x − 2) gives x² − 2x + 6x − 12, which collects to x² + 4x − 12
- A grid keeps track of the products when the coefficients are awkward
Squaring a bracket
- A bracket squared is that bracket multiplied by itself, so it expands as a pair of brackets like any other
- (x + 3)² is (x + 3)(x + 3), which expands to x² + 6x + 9
- Squaring the two terms separately is the error this guards against: x² + 9 is not the expansion, because the two middle products are missing
- The middle term is always twice the product of the two terms in the bracket
- (2x − 5)² gives 4x² − 20x + 25
Worked example
Expanding a pair of brackets
Expand and simplify (3x − 4)(2x + 5).
Solution:
- Every term in the first bracket must multiply every term in the second, which gives four products from two pairs of brackets
- Take the terms with their signs: the first bracket holds +3x and −4
- Multiply 3x by both terms of the second bracket: 3x × 2x = 6x², and 3x × 5 = 15x
- Multiply −4 by both terms: −4 × 2x = −8x, and −4 × 5 = −20
- Writing all four before collecting is worth a mark on its own: 6x² + 15x − 8x − 20
- Only the two x terms are like terms, so only they combine: 15x − 8x = 7x
- The answer is 6x² + 7x − 20
- Check with a value: x = 1 gives (−1)(7) = −7 in the question, and 6 + 7 − 20 = −7 in the answer
3. Factorising by taking out a common factor
Reversing the expansion
- Factorising writes an expression as a product, which is the reverse of expanding
- Take the largest number dividing every coefficient, together with the highest power of each letter that divides every term
- Dividing each term by that combined factor leaves what belongs inside the bracket
- 20p³ − 15p² has highest common factor 5p², giving 5p²(4p − 3)
- Expanding the answer back is a reliable check, and it costs one line
Exam tip
Take out every common factor
Set in 11 of the 22 papers. Taking out only the number, as in 4(3x² − 2x), or only the letter, as in x(12x − 8), earns one of the two marks; the second needs both, 4x(3x − 2). Expanding back cannot catch this, because a partial answer expands correctly too, so check that the terms left in the bracket share no factor at all.
Worked example
Taking out a common factor
Factorise 12x³ − 18x².
Solution:
- Find the highest common factor of the numbers and of the letters separately
- The highest common factor of 12 and 18 is 6
- Both terms contain x², and that is the highest power common to both
- So the common factor is 6x²
- 12x³ ÷ 6x² = 2x, and −18x² ÷ 6x² = −3
- 12x³ − 18x² = 6x²(2x − 3)
- Check by expanding back