- Multiply out two brackets to make an expression with an x2x^2x2 term.
- Collect like terms and handle negative signs carefully.
- Write a quadratic back into brackets by finding number pairs.
- Spot special patterns such as repeated brackets and a difference of two squares.
Before the main methods, make sure these words feel familiar.
Useful vocabulary
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A term is a part of an expression separated by plus or minus signs. In 3x2−5x+73x^2 - 5x + 73x2−5x+7, the terms are 3x23x^23x2, −5x-5x−5x and the constant term 7.
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A coefficient is the number multiplying a variable, such as 3 in 3x3x3x.
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Like terms have the same variable and the same power, such as 6x6x6x and −2x-2x−2x.
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A quadratic expression has a squared variable as its highest power, such as x2+4x−12x^2 + 4x - 12x2+4x−12.
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To expand means to remove brackets by multiplying. To factorise means to put an expression back into brackets. A factor is one part being multiplied.
When you expand quadratics, you usually get two middle terms that can be collected.
Collecting like terms

Simplify x2+7x−3x−21x^2 + 7x - 3x - 21x2+7x−3x−21.
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Identify the like terms: 7x7x7x and −3x-3x−3x both contain xxx.
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Collect them: 7x−3x=4x7x - 3x = 4x7x−3x=4x.
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Keep the other terms unchanged: x2+4x−21x^2 + 4x - 21x2+4x−21.
A bracket like x+5x + 5x+5 has two terms. When you multiply two brackets, every term in the first bracket must multiply every term in the second bracket.
Four multiplications

For two brackets, make four products, then collect like terms. This stops you missing the middle terms.
Expand and simplify (x+5)(x−4)

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Multiply the first terms: x×x=x2x \times x = x^2x×x=x2.
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Multiply the outside terms: x×−4=−4xx \times -4 = -4xx×−4=−4x.
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Multiply the inside terms: 5×x=5x5 \times x = 5x5×x=5x.
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Multiply the last terms: 5×−4=−205 \times -4 = -205×−4=−20.
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Collect the like terms: x2−4x+5x−20=x2+x−20x^2 - 4x + 5x - 20 = x^2 + x - 20x2−4x+5x−20=x2+x−20.
Only multiplying the ends
Do not just multiply xxx with xxx and 5 with -4. The two middle multiplications are what create the xxx term.
If a term is like 3x3x3x, multiply the number parts and the letter parts. For example, 3x×2x=6x23x \times 2x = 6x^23x×2x=6x2.
Expand and simplify (4x−1)(2x+3)

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Multiply 4x4x4x by 2x2x2x: 4x×2x=8x24x \times 2x = 8x^24x×2x=8x2.
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Multiply 4x4x4x by 3: 4x×3=12x4x \times 3 = 12x4x×3=12x.
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Multiply -1 by 2x2x2x: −1×2x=−2x-1 \times 2x = -2x−1×2x=−2x.
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Multiply -1 by 3: −1×3=−3-1 \times 3 = -3−1×3=−3.
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Collect the middle terms: 8x2+12x−2x−3=8x2+10x−38x^2 + 12x - 2x - 3 = 8x^2 + 10x - 38x2+12x−2x−3=8x2+10x−3.
Sign check
A negative times a positive is negative. A negative times a negative is positive. Say the signs out loud if you need to.
A squared bracket means the bracket is multiplied by itself. So (a−6)2(a - 6)^2(a−6)2 means (a−6)(a−6)(a - 6)(a - 6)(a−6)(a−6).
Expand and simplify (b−6)2

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Rewrite it as two brackets: (b−6)(b−6)(b - 6)(b - 6)(b−6)(b−6).
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Multiply all four pairs: b2−6b−6b+36b^2 - 6b - 6b + 36b2−6b−6b+36.
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Collect the like terms: −6b−6b=−12b-6b - 6b = -12b−6b−6b=−12b.
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The answer is b2−12b+36b^2 - 12b + 36b2−12b+36.
Missing the middle term
(b−6)2(b - 6)^2(b−6)2 is not b2+36b^2 + 36b2+36. You must expand it as two brackets.
Factorising reverses expanding. For Grade 5 questions, many quadratics begin with x2x^2x2, so the brackets often start as (x )(x )(x \ \ )(x \ \ )(x )(x ).
Product and sum
For x2+bx+cx^2 + bx + cx2+bx+c, find two numbers with product ccc and sum bbb. Product means multiply; sum means add.
Factorise x2+9x+20

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Start with two brackets because x2=x×xx^2 = x \times xx2=x×x: (x )(x )(x \ \ )(x \ \ )(x )(x ).
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Find two numbers that multiply to 20 and add to 9.
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The pair is 4 and 5, because 4×5=204 \times 5 = 204×5=20 and 4+5=94 + 5 = 94+5=9.
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Put them into the brackets: (x+4)(x+5)(x + 4)(x + 5)(x+4)(x+5).
The constant term is the number at the end. Its sign gives you a big clue:
- Positive constant and positive middle term: both signs are positive.
- Positive constant and negative middle term: both signs are negative.
- Negative constant: one sign is positive and one is negative.
Factorise n2−4n−21

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Start with (n )(n )(n \ \ )(n \ \ )(n )(n ).
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Look for two numbers that multiply to -21 and add to -4.
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The numbers are -7 and 3, because −7×3=−21-7 \times 3 = -21−7×3=−21 and −7+3=−4-7 + 3 = -4−7+3=−4.
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Put them into the brackets: (n−7)(n+3)(n - 7)(n + 3)(n−7)(n+3).
A difference of two squares is a subtraction where both parts are square terms.
a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)a2−b2=(a−b)(a+b)
Factorise x2−64

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Recognise both parts as squares: x2x^2x2 is a square and 64 is 828^282.
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Use one minus bracket and one plus bracket.
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The answer is (x−8)(x+8)(x - 8)(x + 8)(x−8)(x+8).
It must be a difference
x2+64x^2 + 64x2+64 is not a difference of two squares because there is no subtraction sign between the square terms.
Sometimes a quadratic comes from squaring one bracket, such as (x+6)2(x + 6)^2(x+6)2 or (x−6)2(x - 6)^2(x−6)2.
Factorise x2−16x+64
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Notice that 64 is 828^282.
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Check the middle term: −8x−8x=−16x-8x - 8x = -16x−8x−8x=−16x.
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So the two brackets are the same: (x−8)(x−8)(x - 8)(x - 8)(x−8)(x−8).
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Write this more neatly as (x−8)2(x - 8)^2(x−8)2.
In the exam
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When expanding, make four multiplications before collecting like terms.
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When factorising, write down the product target and the sum target.
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Check your answer by quickly expanding your brackets again.
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Watch negative signs carefully, especially when the constant term is negative.
Check yourself
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Can you expand (2x−3)(x+5)(2x - 3)(x + 5)(2x−3)(x+5) without missing any terms?
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Can you factorise x2−x−42x^2 - x - 42x2−x−42 by finding a product and a sum?
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Can you explain why x2−81x^2 - 81x2−81 is a difference of two squares?