Factorise x² − 7x + 12.

(x − 3)(x − 4). You need two numbers multiplying to +12 and adding to −7. Both must be negative: −3 and −4. Option C multiplies to 12 but adds to −8.

Factorise 9a² − 30ab + 25b².

(3a − 5b)². The outer terms are (3a)² and (5b)², and 2 × 3a × 5b = 30ab matches the middle term with a minus sign, so it is a perfect square (3a − 5b)². Option D would give no middle term at all.

ax + ay + bx + by factorises to:

(a + b)(x + y). Group as a(x + y) + b(x + y), then take the common bracket out: (a + b)(x + y). Expanding option A back gives the original four terms.

If a − b = 4 and ab = 5, then a² + b² equals:

26. a² + b² = (a − b)² + 2ab = 16 + 10 = 26. Note the plus sign: the −2ab in the expansion of (a − b)² has to be added back.

Which is the complete factorisation of 5x² − 45?

5(x − 3)(x + 3). Take out the 5, then factorise x² − 9 as a difference of two squares. Option A stops one step early, and options B and D still hide a common factor inside a bracket.

The factorisation of x³ + 27 is:

(x + 3)(x² − 3x + 9). Sum of cubes with b = 3: (a + b)(a² − ab + b²). The middle sign inside the quadratic is the opposite of the sign in the first bracket, which rules out option C.

Split the middle term of 2x² + 7x + 3 using:

6 and 1. The product ac = 2 × 3 = 6 and the sum must be 7, so the pair is 6 and 1. That gives 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

The HCF of (x + 1)²(x − 2) and (x + 1)(x − 2)² is:

(x + 1)(x − 2). Each common factor is taken to its lowest power: (x + 1) appears to the power 1 in the second expression and (x − 2) to the power 1 in the first. Option A is the LCM, not the HCF.

Factorization and Algebraic Manipulation — Mathematics Notes, Worked Examples and Quiz

Factorising is expanding, run backwards

Expanding turns a product into a sum: (x + 3)(x + 5) = x² + 8x + 15. Factorising goes the other way, and it is harder for the same reason that unlocking is harder than locking — you have to find the key.

Every factorisation question you meet at this level falls into one of five shapes. Learning to recognise the shape is most of the work; the algebra afterwards is short.

Common factor — 6x² + 9x = 3x(2x + 3). Always look here first.
Grouping (four terms) — ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y).
Difference of two squares — a² − b² = (a − b)(a + b).
Perfect square — a² ± 2ab + b² = (a ± b)².
Trinomial x² + bx + c — find two numbers multiplying to c and adding to b.

Take the common factor out first — always

Given 2x² − 8, students often reach for the difference of two squares and get stuck because 2x² is not a perfect square. Take out the 2 first: 2(x² − 4) = 2(x − 2)(x + 2). A common factor left inside also costs marks under "factorise completely".

Why the middle term is what it is

The trinomial rule — two numbers that multiply to c and add to b — looks like a trick until you see the rectangle it comes from. Draw (x + a)(x + b) as a rectangle of width x + a and height x + b. It splits into four pieces: a square of area x², two strips of area ax and bx, and a corner of area ab.

The two strips are the middle term. That is why the coefficient of x is a + b and the constant is ab — not because someone decided it, but because that is how the area divides up.

Switch to (x+a)(x−a). One strip is added and the other taken away, so they cancel exactly and no middle term survives — which is the whole of the difference-of-two-squares identity, drawn.

The identities to know without thinking

These appear in every paper, usually not as "factorise" but hidden inside a longer question. Recognising one saves several lines of work.

(a + b)² = a² + 2ab + b²(a − b)² = a² − 2ab + b²a² − b² = (a − b)(a + b)(a + b)³ = a³ + 3a²b + 3ab² + b³a³ + b³ = (a + b)(a² − ab + b²)a³ − b³ = (a − b)(a² + ab + b²)the last two are the sum and difference of cubes — note the sign pattern inside the bracket

The two rearrangements the paper actually uses

From (a + b)² = a² + 2ab + b² you get a² + b² = (a + b)² − 2ab, and from the other one a² + b² = (a − b)² + 2ab. Subtracting them gives (a + b)² − (a − b)² = 4ab. Questions that give you a + b and ab and ask for a² + b² are testing exactly these — and there is no need to find a and b themselves.

Factorising a trinomial with a leading coefficient

When the coefficient of x² is not 1, the "two numbers" method needs one extra step. Multiply the first and last coefficients together, split the middle term using that product, then factorise by grouping.

Worked example

Factorise 6x² + 11x − 10.

Multiply the outer coefficients: 6 × (−10) = −60.This is the number the two parts must multiply to; the middle coefficient 11 is what they must add to.
Find two numbers with product −60 and sum +11: they are 15 and −4.A negative product means opposite signs; a positive sum means the larger number is the positive one.
Split the middle term: 6x² + 15x − 4x − 10.The order of the two new terms does not matter — either arrangement groups successfully.
Group in pairs: 3x(2x + 5) − 2(2x + 5).The bracket must come out identical from both pairs. If it does not, check the signs on the second pair.
Take out the common bracket: (3x − 2)(2x + 5).Expand mentally to check: 6x² + 15x − 4x − 10 = 6x² + 11x − 10. ✓

(3x − 2)(2x + 5)

Highest common factor and lowest common multiple

The chapter also asks for the HCF and LCM of algebraic expressions. Factorise everything completely first — after that, the rule is mechanical: HCF takes each common factor to its lowest power, LCM takes every factor that appears to its highest power.

A result worth remembering: for any two expressions, HCF × LCM = the product of the two expressions. It gives you a free check, and sometimes a shortcut when one of the three is missing.

Worked example

Find the HCF and LCM of x² − 9 and x² + 6x + 9.

Factorise both: x² − 9 = (x − 3)(x + 3) and x² + 6x + 9 = (x + 3)².The first is a difference of two squares, the second a perfect square. Never look for common factors before factorising.
The only factor appearing in both is (x + 3), to the power 1 in the first expression.HCF takes the lowest power present in either.
HCF = x + 3.
For the LCM take every distinct factor at its highest power: (x − 3)(x + 3)².(x + 3) appears squared in the second expression, so the LCM must carry the square.

HCF = x + 3; LCM = (x − 3)(x + 3)²

Before you leave this chapter

Common factor first, every single time, before you try anything cleverer.
Four terms → group in pairs. Two squares with a minus → difference of two squares.
For ax² + bx + c, split the middle term using two numbers with product ac and sum b.
a² + b² = (a + b)² − 2ab — the identity behind most "without finding a and b" questions.
HCF takes lowest powers, LCM takes highest powers, and HCF × LCM = the product of the expressions.

Factorization and Algebraic Manipulation