Factorising algebraic expressions

Multiplying out $(x+3)(x+4)$ to get $x^2+7x+12$ is easy — you have four terms and you just add them up. Factorising is that process run in reverse: you are handed the expanded expression $x^2+7x+12$ and asked to rediscover the two brackets that produced it. There is no single button to press; instead you learn to recognise four different shapes an expression can take, each pointing to a different method — pulling out a common factor, grouping four terms in pairs, matching one of the standard identities, or splitting the middle term of a trinomial. This chapter directly reverses the multiplication and identity work you have already done, so every method below is really an old identity read backwards. Getting fast at spotting which shape you are looking at, before you start manipulating symbols, is the real skill — and it is the one this article drills.

The core insight

Factorising reverses multiplication: pick the method (common factor, grouping, identity, or splitting the middle term) that matches the shape of the expression in front of you, then check your answer by multiplying the factors back out.

The core idea

Common factor. Pull out the highest common factor (HCF) of every term — this is always the first thing to check, since a hidden common factor can lurk inside an expression that also matches an identity.

ax+ay=a(x+y)

where: $a$ is the HCF shared by both terms, and $(x+y)$ is what remains once $a$ is divided out of each term.

Grouping. With four terms and no overall common factor, rearrange them into two pairs that each share a factor, pull that factor out of each pair, and the two pairs collapse into a common binomial.

ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b)

Identities read backwards. Every identity you learnt for expanding brackets has a mirror-image use for factorising — spot the right-hand-side pattern and jump straight to the left-hand side.

a^2-b^2=(a-b)(a+b)

a^2+2ab+b^2=(a+b)^2 \qquad a^2-2ab+b^2=(a-b)^2

where: $a$ and $b$ are the two terms whose squares appear; the sign of the middle term ( $+2ab$ or $-2ab$ ) tells you whether the answer is $(a+b)^2$ or $(a-b)^2$ .

Splitting the middle term. For a monic trinomial $x^2+bx+c$ (coefficient of $x^2$ exactly $1$ ), find two numbers $p$ and $q$ whose sum is $b$ and whose product is $c$ , then rewrite $bx$ as $px+qx$ and group.

x^2+bx+c=(x+p)(x+q) \quad\text{where } p+q=b,\ pq=c

where: $b$ is the coefficient of $x$ , $c$ is the constant term, and $p,q$ are the two numbers you search for — this only works when the leading coefficient is exactly $1$ ; a general trinomial $ax^2+bx+c$ with $a\neq1$ needs a different method left for a later class. A reliable order to work through the methods: check for a common factor first (it keeps the numbers small); then look for a matching identity; if it is a four-term expression, try grouping; and if it is a trinomial that is not a perfect square, split the middle term. Applying the common-factor check first matters even when an identity is also present — a common factor can hide inside an expression that also matches $a^2-b^2$ or a perfect-square trinomial, and pulling it out first keeps the remaining numbers smaller and the identity easier to spot. For example, $8x^2-18$ has the common factor $2$ hiding a difference of squares: $8x^2-18=2(4x^2-9)=2(2x-3)(2x+3)$ , which is far tidier than trying to force the identity onto the original coefficients directly. The main skill across all four methods is the same: pattern recognition before manipulation. Look at the expression and ask how many terms it has, whether the terms share an obvious common factor, and whether it resembles the right-hand side of an identity you already know — only after answering these questions should you start moving symbols around.

Figure

An area rectangle split into four smaller cells showing x times x plus 4x plus 3x plus 12 combining into two matched sides x plus 3 and x plus 4

Derivation AFactorising a monic trinomial by splitting the middle term

Show, using only the identity for expanding two binomials, why searching for two numbers $p$ and $q$ with $p+q=b$ and $pq=c$ correctly factorises any monic trinomial $x^2+bx+c$ into $(x+p)(x+q)$ , and why checking both conditions together — not just one — is essential.

Step 1. Start from the known identity for a product of two binomials of this shape, expanded fully:

(x+p)(x+q)=x^2+px+qx+pq=x^2+(p+q)x+pq

Step 2. Compare this expanded form with the trinomial we want to factorise, $x^2+bx+c$ . For the two to be the same expression, term by term, the coefficient of $x$ must match and the constant term must match:

p+q=b \qquad\text{and}\qquad pq=c

Step 3. This shows the search is exact, not a guess: any pair $p,q$ satisfying both conditions at once makes $(x+p)(x+q)$ expand back to exactly $x^2+bx+c$ . Neither condition alone is enough — a pair could sum correctly but multiply to the wrong constant, or vice versa, so both must be checked together before accepting an answer.

Step 4. Once such a pair is found, rewrite the middle term $bx$ as the sum $px+qx$ (legal since $p+q=b$ means $px+qx=bx$ ), which turns the trinomial into a four-term expression that groups cleanly:

x^2+bx+c=x^2+px+qx+c=x(x+p)+q(x+p)=(x+p)(x+q)

Step 5. This confirms the method: splitting the middle term is nothing but running Step 1 in reverse, using grouping as the mechanical tool to get from four terms back to two brackets.

Worked example

Factorise the trinomial $x^2+7x+12$ by splitting the middle term.

Step 1. Identify $b=7$ and $c=12$ , so we need two numbers $p,q$ with $p+q=7$ and $pq=12$ .

Step 2. Test factor pairs of $12$ : $(1,12)$ gives sum $13$ ; $(2,6)$ gives sum $8$ ; $(3,4)$ gives sum $7$ — matches, and $3\times4=12$ also matches. Both conditions hold for $p=3,\,q=4$ .

Step 3. Split the middle term $7x$ as $3x+4x$ :

x^2+7x+12=x^2+3x+4x+12

Step 4. Group into pairs and pull the common factor from each: $x(x+3)+4(x+3)$ .

Step 5. Factor out the shared binomial $(x+3)$ to reach the final factorisation.

x^2+7x+12=(x+3)(x+4)

Check: $(x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12$ ✓, matching the original expression, so the factorisation is $(x+3)(x+4)$ .

Competitive stretch

The same splitting works when $c$ is negative, since $p$ and $q$ just take opposite signs. For $x^2+2x-15$ : we need $p+q=2$ and $pq=-15$ . The pair $5$ and $-3$ works, since $5+(-3)=2$ and $5\times(-3)=-15$ . Splitting gives $x^2+5x-3x-15=x(x+5)-3(x+5)=(x+5)(x-3)$ , checked by expanding back: $x^2-3x+5x-15=x^2+2x-15$ ✓.

Common traps

Stopping after removing only part of the common factor — e.g. writing $6x^2y-9xy^2=3(2x^2y-3xy^2)$ instead of pulling out the full HCF $3xy$ to get $3xy(2x-3y)$ .
In splitting the middle term, picking a pair that satisfies the sum but not the product (or vice versa) — both conditions must hold together, not just one.
Misapplying $a^2-b^2=(a-b)(a+b)$ to a sum of squares, $a^2+b^2$ , which does not factorise this way over ordinary numbers.
Forgetting to check the sign of the middle term when choosing between $(a+b)^2$ and $(a-b)^2$ for a perfect-square trinomial — a middle term of $-2ab$ means the answer is $(a-b)^2$ , never $(a+b)^2$ .
Attempting to split the middle term of a non-monic trinomial (leading coefficient not $1$ ) using the plain $p+q=b,\,pq=c$ shortcut — this simple version of the method only works when the coefficient of $x^2$ is exactly $1$ ; a different technique is needed otherwise, and it is outside this chapter.

Ask the tutor: Not sure which method fits an expression you are stuck on, or want more practice splitting the middle term? Ask the AI tutor to walk through extra factorisation problems with you.

How do I decide which of the four factorisation methods to use?

Work through them in order. First check for a common factor across all terms — pulling it out first keeps the remaining numbers smaller. Then check whether what is left matches a standard identity, such as a difference of squares or a perfect-square trinomial. If it is a four-term expression with no obvious identity, try grouping into pairs. If it is a trinomial that is not a perfect square, split the middle term. Practising this checklist on many expressions is what makes the recognition automatic.

Why does splitting the middle term always require both a sum and a product condition?

Because a monic trinomial $x^2+bx+c$ has two separate pieces of information baked into it — the coefficient $b$ and the constant $c$ — and a valid pair of brackets $(x+p)(x+q)$ must reproduce both at once, since expanding gives $x^2+(p+q)x+pq$ . A pair that only matches the sum but not the product would expand to a different trinomial, so checking both conditions together is not optional caution, it is the actual definition of a correct factor pair.

Can I use splitting the middle term on x squared minus 5x plus 6, where both b and c look different in sign?

Yes — the method does not change with signs, only the search for $p$ and $q$ does. Here $b=-5$ and $c=6$ , so you need $p+q=-5$ and $pq=6$ ; both numbers must be negative for the product to be positive and the sum to be negative, and $-2,-3$ works since $(-2)+(-3)=-5$ and $(-2)\times(-3)=6$ . So $x^2-5x+6=(x-2)(x-3)$ , which you can check by expanding back out.

Factorising algebraic expressions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads