Expanding and Factorising

Expanding brackets and factorising are reverse processes, and together they sit at the heart of GCSE algebra. Expanding means multiplying out brackets to remove them; factorising means putting the brackets back by spotting common structure. This lesson covers expanding single and double brackets, factorising by taking out a common factor, the difference of two squares, and factorising quadratics of the form $x^{2} + bx + c$x2+bx+c — and, at Higher tier, $ax^{2} + bx + c$ax2+bx+c. The skills here are pure AO1 technique, but choosing the most efficient method and laying out your work so a marker can follow it draws on AO2 as well.

Key Vocabulary

Term	Meaning
Expand	Multiply out brackets to write an expression without them
Factorise	Rewrite an expression as a product, by inserting brackets
Common factor	A number or letter that divides exactly into every term
Quadratic	An expression whose highest power is $2$2, e.g. $x^{2} + 5x + 6$x2+5x+6
Difference of two squares	An expression of the form $a^{2} - b^{2}$a2−b2, which factorises to $(a + b)(a - b)$(a+b)(a−b)
Coefficient	The number multiplying a variable, e.g. the $3$3 in $3x^{2}$3x2
Product	The result of multiplying; factorising writes an expression as a product

Why are these two skills so central? Expanding lets you remove brackets so that an expression can be simplified or an equation solved; factorising lets you rewrite an expression as a product, which is exactly what you need to solve quadratics, to simplify algebraic fractions, and to reveal hidden structure. Because they are inverse processes, you can always use one to check the other: expand your factorised answer and you should be back where you started. Build that checking habit now and it will protect you in every later algebra topic.

Expanding Single Brackets

To expand a single bracket, multiply every term inside by the term outside, keeping track of signs. The underlying rule is the distributive law, $a(b + c) = ab + ac$a(b+c)=ab+ac — the outside factor is "distributed" to each term inside. The single most important thing to watch is the sign of the multiplier: if it is negative, it changes the sign of every term it touches.

Worked Example 1

Expand $4(2x + 3)$4(2x+3).

$4 \times 2x = 8x$4×2x=8x and $4 \times 3 = 12$4×3=12.

Answer: $8x + 12$8x+12

Worked Example 2

Expand $-3(2a - 5)$−3(2a−5).

$-3 \times 2a = -6a$−3×2a=−6a and $-3 \times (-5) = +15$−3×(−5)=+15.

Answer: $-6a + 15$−6a+15

Common error: forgetting that a negative outside the bracket flips the sign of every term, so $-3 \times (-5)$−3×(−5) becomes $+15$+15, not $-15$−15.

Worked Example 3

Expand and simplify $5(x + 2) + 3(2x - 1)$5(x+2)+3(2x−1).

Expand each: $5x + 10$5x+10 and $6x - 3$6x−3. Collect like terms: $5x + 6x = 11x$5x+6x=11x and $10 - 3 = 7$10−3=7.

Answer: $11x + 7$11x+7

Worked Example 3b

Expand and simplify $4(3x - 1) - 2(x - 5)$4(3x−1)−2(x−5).

Expand each bracket, being careful with the negative in front of the second one: $4(3x - 1) = 12x - 4$4(3x−1)=12x−4 and $-2(x - 5) = -2x + 10$−2(x−5)=−2x+10. Now collect like terms: $12x - 2x = 10x$12x−2x=10x and $-4 + 10 = 6$−4+10=6.

Answer: $10x + 6$10x+6

Common error: writing $-2(x - 5) = -2x - 10$−2(x−5)=−2x−10. The negative multiplier turns the $-5$−5 into $+10$+10; getting this sign wrong is one of the most frequent mistakes in the whole topic.

Expanding Double Brackets

Each term in the first bracket multiplies each term in the second — four products in total. A useful memory aid is FOIL (First, Outer, Inner, Last), but the safest habit is simply to be systematic and make sure every term in the first bracket meets every term in the second. Some students prefer a grid (a "multiplication box") with the two terms of one bracket across the top and the two of the other down the side; whichever layout you choose, the key is that nothing is missed and every sign is carried through.

Worked Example 4

Expand $(x + 4)(x + 5)$(x+4)(x+5).

• First: $x \times x = x^{2}$x×x=x2
• Outer: $x \times 5 = 5x$x×5=5x
• Inner: $4 \times x = 4x$4×x=4x
• Last: $4 \times 5 = 20$4×5=20

Collect the middle terms: $5x + 4x = 9x$5x+4x=9x.

Answer: $x^{2} + 9x + 20$x2+9x+20

Worked Example 5

Expand $(2x - 3)(x + 6)$(2x−3)(x+6).

$2x \times x = 2x^{2}$2x×x=2x2; $2x \times 6 = 12x$2x×6=12x; $-3 \times x = -3x$−3×x=−3x; $-3 \times 6 = -18$−3×6=−18. Collect: $12x - 3x = 9x$12x−3x=9x.

Answer: $2x^{2} + 9x - 18$2x2+9x−18

Worked Example 6

Expand $(x - 7)^{2}$(x−7)2.

This means $(x - 7)(x - 7)$(x−7)(x−7). The four products give $x^{2} - 7x - 7x + 49 = x^{2} - 14x + 49$x2−7x−7x+49=x2−14x+49.

Answer: $x^{2} - 14x + 49$x2−14x+49

Common error: writing $(x - 7)^{2} = x^{2} + 49$(x−7)2=x2+49. Squaring a bracket is not squaring each term — the cross terms $-7x - 7x$−7x−7x matter. In general $(x + a)^{2} \ne x^{2} + a^{2}$(x+a)2=x2+a2.

Factorising — Taking Out a Common Factor

Factorising into a single bracket is the reverse of expanding one. Find the highest factor common to every term and write it outside.

Worked Example 7

Factorise $6x + 9$6x+9.

The highest common factor of $6$6 and $9$9 is $3$3: $6x + 9 = 3(2x + 3)$6x+9=3(2x+3).

Answer: $3(2x + 3)$3(2x+3)

Worked Example 8

Factorise fully $12a^{2}b - 8ab^{2}$12a2b−8ab2.

Numbers: HCF of $12$12 and $8$8 is $4$4. Letters common to both terms: one $a$a and one $b$b, i.e. $ab$ab. So take out $4ab$4ab: $12a^{2}b - 8ab^{2} = 4ab(3a - 2b)$12a2b−8ab2=4ab(3a−2b).

Answer: $4ab(3a - 2b)$4ab(3a−2b)

Common error: only partially factorising, e.g. $2(6a^{2}b - 4ab^{2})$2(6a2b−4ab2). The word "fully" requires the largest possible common factor.

The Difference of Two Squares

When you see $a^{2} - b^{2}$a2−b2 (a square subtract a square), it factorises to $(a + b)(a - b)$(a+b)(a−b).

Worked Example 9

Factorise $x^{2} - 25$x2−25.

This is $x^{2} - 5^{2}$x2−52, so it factorises to $(x + 5)(x - 5)$(x+5)(x−5).

Answer: $(x + 5)(x - 5)$(x+5)(x−5)

Worked Example 10

Factorise $9p^{2} - 16$9p2−16.

Here $9p^{2} = (3p)^{2}$9p2=(3p)2 and $16 = 4^{2}$16=42, so $9p^{2} - 16 = (3p + 4)(3p - 4)$9p2−16=(3p+4)(3p−4).

Answer: $(3p + 4)(3p - 4)$(3p+4)(3p−4)

Worked Example 10b

Factorise $50 - 2y^{2}$50−2y2.

First take out the common factor $2$2: $50 - 2y^{2} = 2(25 - y^{2})$50−2y2=2(25−y2). The bracket is now a difference of two squares, $25 - y^{2} = (5 + y)(5 - y)$25−y2=(5+y)(5−y). So the full factorisation is $2(5 + y)(5 - y)$2(5+y)(5−y).

Answer: $2(5 + y)(5 - y)$2(5+y)(5−y)

This shows a key strategy: always look for a common factor first. Taking out the $2$2 revealed a difference of two squares that was hidden before.

Factorising Quadratics $x^{2} + bx + c$x2+bx+c

To factorise $x^{2} + bx + c$x2+bx+c into $(x + p)(x + q)$(x+p)(x+q), find two numbers $p$p and $q$q that multiply to $c$c and add to $b$b. A short table of factor pairs of $c$c often makes the search quick. Pay close attention to signs: if $c$c is positive, both numbers have the same sign (matching the sign of $b$b); if $c$c is negative, the two numbers have opposite signs.

Worked Example 11

Factorise $x^{2} + 7x + 12$x2+7x+12.

Two numbers multiplying to $12$12 and adding to $7$7: that is $3$3 and $4$4.

Answer: $(x + 3)(x + 4)$(x+3)(x+4)

Worked Example 12

Factorise $x^{2} - 2x - 15$x2−2x−15.

Two numbers multiplying to $-15$−15 and adding to $-2$−2: that is $-5$−5 and $+3$+3.

Answer: $(x - 5)(x + 3)$(x−5)(x+3)

Common error: getting the signs the wrong way round. Always check by expanding: $(x - 5)(x + 3) = x^{2} + 3x - 5x - 15 = x^{2} - 2x - 15$(x−5)(x+3)=x2+3x−5x−15=x2−2x−15. ✓

Extended Worked Examples

Worked Example 13 — Expanding three brackets [Higher]

This part is Higher tier. Expand $(x + 1)(x + 2)(x + 3)$(x+1)(x+2)(x+3).

First expand two of them: $(x + 1)(x + 2) = x^{2} + 3x + 2$(x+1)(x+2)=x2+3x+2. Now multiply by $(x + 3)$(x+3):

$(x^{2} + 3x + 2)(x + 3) = x^{3} + 3x^{2} + 3x^{2} + 9x + 2x + 6$(x2+3x+2)(x+3)=x3+3x2+3x2+9x+2x+6.