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
Worked example
Two brackets with two letters
Expand and simplify (2x − 3y)(x + 4y).
Solution:
- Treat the second letter exactly like a number: there are still four products, and each takes the signs of both its terms
- 2x × x = 2x², and 2x × 4y = 8xy
- −3y × x = −3xy, and −3y × 4y = −12y²
- 8xy and −3xy are like terms, because yx and xy are the same, so they collect to 5xy
- The answer is 2x² + 5xy − 12y²
- Check with values: x = 2 and y = 1 give 1 × 6 = 6 in the question, and 8 + 10 − 12 = 6 in the answer
3. Expanding three brackets
- Expand any two brackets first and simplify, then multiply that result by the remaining bracket
- The second stage multiplies a two-term bracket by a three-term expression, so a grid is worth using
- (x + 1)(x + 2)(x + 3) expands to x³ + 6x² + 11x + 6
- The highest power in the answer matches the number of brackets, which is a quick check
Worked example
Three brackets, two at a time
Expand and simplify (3x − 2)(x + 1)(x − 5).
Solution:
- Expand two of the brackets first. Any pair works, so take the two with a plain x: (x + 1)(x − 5)
- x² − 5x + x − 5 = x² − 4x − 5
- Multiply that three-term result by the remaining bracket, (3x − 2): each of its two terms times each of the three, six products in all
- 3x × (x² − 4x − 5) = 3x³ − 12x² − 15x
- −2 × (x² − 4x − 5) = −2x² + 8x + 10
- Collect the like terms: −12x² − 2x² = −14x², and −15x + 8x = −7x
- The answer is 3x³ − 14x² − 7x + 10
- Check with a value: x = 2 gives 4 × 3 × (−3) = −36 in the question, and 24 − 56 − 14 + 10 = −36 in the answer
4. 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
When the common factor is a bracket
- The common factor taken out need not be a number or a letter: a whole bracket can be shared
- 3(x + 1) + y(x + 1) has (x + 1) in both terms, giving (x + 1)(3 + y)
- This is the step that finishes a factorisation by grouping, and it is worth spotting on its own
Worked example
Taking out a common factor
Factorise 12x³ − 18x², and factorise 5(x − 2) + y(x − 2).
Solution:
- For the first, 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)
- For the second, the common factor is the whole bracket (x − 2)
- Taking it out leaves 5 and y, so the answer is (x − 2)(5 + y)
- Check either by expanding back
5. Factorising a quadratic with no number in front
- For x² + bx + c, look for a pair of numbers that multiply to give c and add to give b
- Those two numbers go straight into the brackets
- x² + 7x + 12 needs numbers multiplying to 12 and adding to 7, which are 3 and 4, giving (x + 3)(x + 4)
- x² − 3x − 10 needs numbers multiplying to −10 and adding to −3, which are −5 and 2, giving (x − 5)(x + 2)
- A negative constant means the two numbers have opposite signs
Exam tip
A nearly right pair of brackets still scores
Whether you factorise a quadratic on its own, to solve an equation or to simplify a fraction, many schemes give a mark before the brackets are fully right: for brackets that give the right constant or the right x term when expanded, or for the pairs factorised after splitting the middle term, such as x(x − 5) + 2(x − 5). When the pair will not come, still write your best pair of brackets down, then expand them to check.
Worked example
Factorising a simple quadratic
Factorise x² + 2x − 15.
Solution:
- Factorising reverses expanding, so the answer is a pair of brackets
- With no number in front of x², the two brackets each start with x, and the two numbers in them must multiply to −15 and add to +2
- Those two conditions come straight from the expansion: the numbers multiply to give the constant and add to give the x coefficient
- The factor pairs of 15 are 1 × 15 and 3 × 5
- The product needed is negative, so the two numbers have opposite signs
- Test the pairs: 5 and −3 give 5 × (−3) = −15 and 5 + (−3) = 2, which is what is wanted
- So x² + 2x − 15 = (x + 5)(x − 3)
- Check by expanding back: x² − 3x + 5x − 15 = x² + 2x − 15
- The order of the brackets does not matter, but the signs inside them do
6. Factorising a quadratic with a number in front
- For ax² + bx + c, find a pair multiplying to ac and adding to b
- Rewrite bx as the sum of those two terms, which turns three terms into four
- A common bracket appears, and taking it out completes the factorisation
Worked example
Factorising a harder quadratic
Factorise 3x² + 11x + 6.
Solution:
- A number in front of x² means the two numbers no longer simply multiply to the constant, so use the grouping method
- Write the expression as ax² + bx + c, so a = 3, b = 11 and c = 6
- Find two numbers that multiply to ac and add to b
- ac = 3 × 6 = 18, and b = 11
- Factor pairs of 18: 1 × 18, 2 × 9, 3 × 6. The pair 9 and 2 adds to 11
- Split the middle term using those two numbers: 3x² + 9x + 2x + 6
- Factorise the first pair and the second pair separately: 3x(x + 3) + 2(x + 3)
- Both now carry the bracket (x + 3), which confirms the split was right; if the brackets differ, the pair was wrong
- Take the common bracket out: (x + 3)(3x + 2)
- Check by expanding: 3x² + 2x + 9x + 6 = 3x² + 11x + 6
7. The difference of two squares
- Where one square is subtracted from another, the expression factorises into two brackets differing only in sign
- a² − b² factorises to (a + b)(a − b)
- 25x² − 49 is (5x)² − 7², giving (5x + 7)(5x − 7)
- Both terms must be perfect squares and the operation must be a subtraction
- Recognising this pattern is faster than any general method, and it also appears when rationalising a denominator
Worked example
Difference of two squares, factorised fully
Factorise 9x² − 25, and factorise 2x² − 8.
Solution:
- 9x² − 25 is one square subtracted from another: 9x² = (3x)² and 25 = 5²
- The difference of two squares factorises as (first − second)(first + second)
- 9x² − 25 = (3x − 5)(3x + 5)
- 2x² − 8 is not yet a difference of two squares, because 2 and 8 are not squares
- Take out the common factor of 2 first: 2(x² − 4)
- Now x² − 4 is a difference of two squares: x² − 2²
- 2x² − 8 = 2(x − 2)(x + 2)
- Stopping at 2(x² − 4) is not fully factorised, and the syllabus defines "factorise" to mean factorise fully
8. Factorising by grouping
- Four terms with no factor common to all of them may still factorise in pairs
- Factorise the first two terms and the last two separately, aiming to produce the same bracket in both
- That shared bracket is then taken out as a common factor
- ab + 4a + 3b + 12 becomes a(b + 4) + 3(b + 4), which factorises to (b + 4)(a + 3)
- If the two brackets do not match, try pairing the terms differently
Worked example
Factorising by grouping
Factorise xy + 3x + 2y + 6.
Solution:
- Four terms with nothing common to all of them, which is the signal to group them in pairs
- Take the first two together and the last two together, keeping the order
- First pair: xy + 3x, whose common factor is x, giving x(y + 3)
- Second pair: 2y + 6, whose common factor is 2, giving 2(y + 3)
- The two brackets are now identical, and that is the check that the pairing worked: different brackets mean the terms should be paired the other way round
- The expression reads x(y + 3) + 2(y + 3), which is (y + 3) multiplied by x and by 2
- Take the common bracket out: (y + 3)(x + 2)
- Check by expanding: xy + 3x + 2y + 6, which is the expression we started from
Worked example
Grouping when a pair starts with a minus
Factorise 6xy − 9x − 4y + 6.
Solution:
- Nothing is common to all four terms, so factorise them in pairs
- First pair: 6xy − 9x = 3x(2y − 3)
- The second pair starts with a minus, so take the minus out with its factor: −4y + 6 = −2(2y − 3)
- Dividing by −2 flips both signs inside: −4y ÷ (−2) = 2y and 6 ÷ (−2) = −3. Taking out +2 instead gives 2(3 − 2y), a bracket that does not match
- The expression is now 3x(2y − 3) − 2(2y − 3)
- Take the common bracket out: (2y − 3)(3x − 2)
- Check by expanding: 6xy − 4y − 9x + 6, the four terms we started with
9. Choosing a method
| The expression | Method |
|---|---|
| Every term shares a factor | Take out the common factor first |
| Two square terms subtracted | Difference of two squares |
| Three terms, x² coefficient 1 | Find a pair multiplying to c, adding to b |
| Three terms, x² coefficient not 1 | Find a pair multiplying to ac, adding to b, then group |
| Four terms | Factorise in pairs and look for a common bracket |
- Always take out a common factor before anything else, since it makes what remains simpler
- A quadratic factorises with integers only when a pair multiplying to ac and adding to b exists, so a short search settles it
Worked example
Choosing the method
Factorise each expression: (a) 6x² + 9x (b) x² − 36 (c) x² − 9x + 14
Solution:
- (a) Two terms, and both share a factor. 3x divides both, so take it out
- 6x² + 9x = 3x(2x + 3)
- (b) Two terms, one square subtracted from another, and no common factor
- That is the difference of two squares: x² − 6² = (x − 6)(x + 6)
- (c) Three terms with no number in front of x², so look for two numbers multiplying to +14 and adding to −9
- Both must be negative to give a positive product and a negative sum: −2 and −7
- x² − 9x + 14 = (x − 2)(x − 7)
- Always test for a common factor first, because taking one out can turn what is left into one of the standard forms