Sec 2 Expansion & Factorisation: Methods & Examples

Secondary 2 G3 Mathematics · Study guide

Connect distribution to factorisation, check every cross-product and use common factors, two-variable quadratics and grouping with confidence.

Concept → worked examples → practice12 min read · practice linked below

Expansion and factorisation describe the same expression

Expansion writes a product as a sum of terms. Factorisation works backwards: it writes a sum as a product. Being able to move between the two forms helps you simplify expressions, solve quadratic equations and work with algebraic fractions.

Before you begin: know the rules for multiplying signed numbers, collecting like terms and multiplying powers. For example, $x\times x=x^2$ , while $x+x=2x$ . Terms such as $x^2$ , $x$ and a constant are unlike terms and cannot be collected together.

The distributive law is the reason the methods work: $a(b+c)=ab+ac$ . With two brackets, each term in one bracket multiplies every term in the other. Keep all four products before collecting the like terms.

For positive

x

, adding the four areas gives

(x+2)(x+3)=x^2+5x+6

. The algebraic identity also holds when

x

is zero or negative; those values are not lengths in this picture.

Expand first, then collect like terms

Worked example 1 — four products and a subtraction

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

Expand the first pair of brackets. Its four products are $2x^2$ , $8x$ , $-3x$ and $-12$ : $(2x-3)(x+4)=2x^2+8x-3x-12=2x^2+5x-12.$
Expand the product being subtracted: $x(x+1)=x^2+x$ . Subtract the whole expression, so both signs change: $\begin{aligned}(2x-3)(x+4)-x(x+1)&=2x^2+5x-12-(x^2+x)\\&=2x^2+5x-12-x^2-x\\&=x^2+4x-12.\end{aligned}$

Check: with $x=2$ , the original expression gives $(1)(6)-2(3)=0$ , and the answer gives $4+8-12=0$ . A substitution can catch a mistake; the expansion above establishes equality for every $x$ .

When several letters occur, collect only terms with the same letter part: $2ab+3ba=5ab$ , because $ab=ba$ , but $2a^2+3ab$ cannot be shortened.

Factorise by checking the four products

Always look for a common factor first. For $x^2+bx+c$ , the two numbers in the brackets must multiply to $c$ and add to $b$ . If the coefficient of $x^2$ is not 1, try its factor pairs too. The middle term is the sum of the two cross-products.

Worked example 2 — common factor, then two brackets

Factorise completely $12x^2+22x+6$ .

Every term is divisible by 2, so begin with $12x^2+22x+6=2(6x^2+11x+3).$ For the remaining quadratic, $3x\times2x=6x^2$ and $1\times3=3$ . The placement of 1 and 3 matters. Trying $(3x+3)(2x+1)$ gives middle terms $3x+6x=9x$ , so that arrangement does not work. Swap the constants:

Multiplication frame for the correct arrangement
Multiply	$2x$	$3$
$3x$	$6x^2$	$9x$
$1$	$2x$	$3$

The cross-products add to $9x+2x=11x$ , as required. Therefore $\boxed{12x^2+22x+6=2(3x+1)(2x+3).}$ Check: expanding the brackets gives $6x^2+11x+3$ , and multiplying every term by 2 returns the original expression.

Worked example 3 — keep the second letter in every factor

Factorise $2a^2-7ab+3b^2$ .

Use $2a$ and $a$ for the squared term. The final product is positive but the middle term is negative, so try two negative terms whose product is $3b^2$ : $-b$ and $-3b$ .

Check all four products, including their signs
Multiply	$a$	$-3b$
$2a$	$2a^2$	$-6ab$
$-b$	$-ab$	$3b^2$

The two middle terms give $-6ab-ab=-7ab$ . Hence $2a^2-7ab+3b^2=(2a-b)(a-3b).$ Both brackets retain $b$ ; replacing $3b$ by 3 would change the original expression. As a numerical check, $a=2,b=1$ gives $8-14+3=-3$ , matching $(4-1)(2-3)=-3$ .

Grouping: create the same bracket twice

For four terms, pair terms that have useful common factors. If the resulting brackets are opposites, take a negative factor from one pair so that the brackets match. The repeated bracket then behaves like one common factor.

Worked example 4 — grouping and a related expression

(a) Factorise $6xy-9x-4y+6$ . (b) Hence factorise $6(2t+1)y-9(2t+1)-4y+6$ .

For (a), group the first two terms and the last two terms. Taking out $-2$ from the second pair gives the same bracket as in the first:

\begin{aligned}6xy-9x-4y+6&=3x(2y-3)-2(2y-3)\\&=(3x-2)(2y-3).\end{aligned}

For (b), recognise the whole expression $2t+1$ in place of $x$ . Substitute it into the factors from (a), keeping brackets:

\begin{aligned}6(2t+1)y-9(2t+1)-4y+6&=[3(2t+1)-2](2y-3)\\&=(6t+1)(2y-3).\end{aligned}

Check: expanding the final answer gives $12ty-18t+2y-3$ . Expanding the original expression gives $12ty+6y-18t-9-4y+6$ , which simplifies to the same result.

Choose a method: common factor first; a quadratic usually needs two brackets; four terms may need grouping. A square or a difference of squares can use a special identity. Learn those patterns in the special algebraic identities guide.

Explore how the middle term is made

A multiplication frame for $(x+a)(x+b)$

Change the two constants. Watch the cross-products combine, and predict the signs of the middle term and constant before reading the result. This is a multiplication frame, so negative entries are signed products, not lengths.

The frame shows $(x+2)(x+3)$ : the middle terms are $2x$ and $3x$ .

Each cell is the product of its row and column labels. Add the two terms containing

x

to find the middle term.

(x + 2)(x + 3) = x² + 5x + 6. The middle coefficient is 2 + 3 = 5; the constant is 2 × 3 = 6.

Common mistakes and useful checks

Missing one of the four products: use a frame or distribute one term at a time.
Subtracting a bracket incompletely: $-(x^2+x)=-x^2-x$ ; both signs change.
Choosing factors from only the first and last terms: the cross-products must also reproduce the middle term.
Stopping too early: in a complete factorisation, take out a common number or letter factor and inspect every remaining bracket.
Confusing an expression with an equation: factorising gives an equivalent expression; it does not give a value of $x$ unless an equation is also provided.

Expansion is the strongest routine check on a factorised answer. A quick substitution is an additional check, but a match at one value alone does not prove two expressions are identical.

Practise choosing the method

Use the Sec 2 expansion and factorisation practice bank for questions and worked solutions. Begin with two-bracket expansion, then try complete factorisation and grouping. For the next application, continue to quadratic equations or algebraic fractions and formulae.