Factorising trinomials

A trinomial like $x^2+7x+12$ looks like a dead end until you notice it is really two brackets that have already been multiplied out. Factorising trinomials is the skill of running that multiplication backwards: split the middle term into two pieces, then pull out common factors in pairs. This is the third and final technique in the Class 8 Factorisation chapter, after common-factor extraction and factorising with identities, and it applies specifically to monic trinomials — expressions of the form $x^2+bx+c$ , where the coefficient of $x^2$ is exactly $1$ . The whole method rests on one identity, $(x+p)(x+q) = x^2+(p+q)x+pq$ , read in reverse: given $b$ and $c$ , hunt for two numbers whose sum is $b$ and whose product is $c$ . Once found, those numbers split the middle term and the expression factorises by grouping. The general case $ax^2+bx+c$ with $a\neq1$ is deliberately left for Class 9 and 10, so nothing here ever asks for a trinomial like $2x^2+7x+3$ .

The core insight

To factorise a monic trinomial $x^2+bx+c$ , find two numbers whose sum is $b$ and whose product is $c$ , use them to split the middle term, and factorise the resulting four terms by grouping.

The core idea

Reversing a familiar expansion. Every trinomial of the form $x^2+bx+c$ that this chapter asks you to factorise came, in principle, from multiplying out two simple brackets $(x+p)$ and $(x+q)$ . Expanding gives:

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

where: $p$ and $q$ are the two unknown numbers being searched for, $(p+q)$ becomes the middle coefficient, and $pq$ becomes the constant term. Factorising is simply the same statement read from right to left.

Turning it into a search. To factorise $x^2+bx+c$ , the job is to find $p$ and $q$ satisfying two conditions at once:

p + q = b, \qquad p \times q = c

where: $b$ is the coefficient of $x$ in the given trinomial and $c$ is its constant term. In practice this means listing the factor pairs of $c$ and checking which pair adds up to $b$ .

Splitting and grouping. Once $p$ and $q$ are known, the middle term $bx$ is written as $px+qx$ , turning three terms into four, which are then grouped in pairs and factored:

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

Reading the sign pattern first. Before hunting for factor pairs, the signs of $b$ and $c$ narrow the search dramatically. If $c$ is positive, $p$ and $q$ must have the same sign — both positive when $b$ is positive, both negative when $b$ is negative. If $c$ is negative, $p$ and $q$ must have opposite signs, and whichever one is larger in size carries the sign of $b$ . Recognising this pattern before listing pairs saves a lot of wasted guessing.

Figure

A number line style diagram showing the split of the middle term seven x into three x plus four x with arrows labelled sum equals seven and product equals twelve pointing to the numbers three and four

Derivation Aof the split-and-group factorisation

The goal is to justify, using nothing beyond the distributive law already known from earlier algebra, why splitting the middle term of a monic trinomial into two pieces and then grouping the four resulting terms in pairs always reproduces the two-bracket factorised form, so that the method is not a trick to memorise but a direct consequence of multiplication being distributive both ways.

Step 1. Start with the trinomial $x^2+bx+c$ and suppose two numbers $p$ and $q$ have been found with $p+q=b$ and $pq=c$ . Since $b=p+q$ , the middle term can be rewritten exactly as $bx = (p+q)x = px+qx$ , with nothing added or removed — this is just distributing $x$ over the sum $p+q$ .

Step 2. Substitute this rewritten middle term back into the trinomial: $x^2+bx+c = x^2+px+qx+c$ . The expression is unchanged in value; it now simply has four terms instead of three, because $bx$ has been replaced by its equal $px+qx$ .

Step 3. Group the four terms into two pairs, the first two and the last two: $(x^2+px) + (qx+c)$ . From the first pair, $x$ is a common factor, giving $x(x+p)$ . From the second pair, since $c=pq$ , the term $c$ can be written as $pq$ , so $qx+c = qx+pq = q(x+p)$ — the same bracket $(x+p)$ appears in both.

Step 4. Because both grouped terms share the factor $(x+p)$ , it can be pulled out once more: $x(x+p)+q(x+p) = (x+p)(x+q)$ . This is exactly the target factorised form, so the split-and-group procedure is guaranteed to work whenever a valid pair $p,q$ with $p+q=b$ and $pq=c$ exists, which follows directly from expanding $(x+p)(x+q)$ in reverse.

Worked example

Factorise $x^2+7x+12$ .

Step 1. Identify $b$ and $c$ . Here $b=7$ and $c=12$ , and since $c$ is positive with $b$ positive, both required numbers must be positive.

Step 2. List factor pairs of $12$ and test their sums: $1\times12$ (sum $13$ ), $2\times6$ (sum $8$ ), $3\times4$ (sum $7$ ). The pair $3$ and $4$ satisfies both conditions, since $3+4=7$ and $3\times4=12$ .

Step 3. Split the middle term using these numbers: $x^2+7x+12 = x^2+3x+4x+12$ .

Step 4. Group and factorise: $x^2+3x+4x+12 = x(x+3)+4(x+3) = (x+3)(x+4)$ .

Step 5. Check by expanding: $(x+3)(x+4) = x^2+4x+3x+12 = x^2+7x+12$ , which matches the original. Final answer: $x^2+7x+12 = (x+3)(x+4)$ .

Competitive stretch

Factorise $x^2-2x-24$ . Here $c=-24$ is negative, so the two numbers have opposite signs, and since $b=-2$ is negative, the larger number in size must be negative. Testing pairs whose product's size is $24$ : $4\times6=24$ works, so take $-6$ and $4$ , since $-6+4=-2$ and $-6\times4=-24$ . Splitting gives $x^2-6x+4x-24 = x(x-6)+4(x-6) = (x-6)(x+4)$ . Checking: $(x-6)(x+4) = x^2+4x-6x-24 = x^2-2x-24$ . Final answer: $x^2-2x-24 = (x-6)(x+4)$ .

Building a mental factor table. Before hunting for the sum that matches $b$ , it helps to jot down all the factor pairs of $c$ first, in order of size. For $12$ : $1\times12$ , $2\times6$ , $3\times4$ . For $24$ : $1\times24$ , $2\times12$ , $3\times8$ , $4\times6$ . Scanning a short list like this for the pair whose sum equals $b$ is far faster and less error-prone than testing numbers at random, especially once $c$ has several factor pairs to choose from.

Common traps

Picking a factor pair with the correct product but the wrong sum — for $x^2-2x-24$ , trying $8$ and $3$ gives product $24$ but no sign choice makes the sum $-2$ , so every factor pair of $|c|$ must be checked against the sum condition, not just the product.
Forgetting that a negative constant term $c$ forces the two numbers to have opposite signs, and searching only among same-sign pairs by mistake.
Making a grouping error after splitting — the pairing $(x^2+px)+(qx+c)$ must yield the same bracket from both halves; if the brackets do not match, the middle term was split with the wrong pair of numbers and a different pair should be tried.
Trying to apply this splitting method to a non-monic trinomial such as $2x^2+7x+3$ . The Class 8 technique only covers trinomials where the coefficient of $x^2$ is exactly $1$ ; the extended method for $a\neq1$ belongs to a later class.

Ask the tutor: Want more trinomials to practise splitting, or a hint when you are stuck between two factor pairs? Ask the AI tutor to generate fresh examples with step-by-step checking.

Why do I need to split the middle term instead of factorising directly?

A trinomial with three terms usually has no common factor shared by all three terms, so a simple common-factor extraction does not work. Splitting the middle term deliberately turns three terms into four so that grouping in pairs becomes possible, because each pair of two terms can share its own common factor. It looks like an extra step, but it is the only way to reach the bracket form without guessing $(x+p)(x+q)$ directly and checking by trial expansion.

What if I cannot find any pair of numbers that works?

First recheck that every factor pair of $c$ has actually been tried, including both sign choices when $c$ is negative — it is easy to stop after two or three pairs before reaching the correct one. If every pair genuinely fails to give the required sum, the trinomial may not factorise neatly over integers at all, which can happen even within the Class 8 syllabus; in that case the question is likely expecting a different trinomial, so it is worth rechecking the original expression was copied correctly.

Does the order of p and q in the final answer matter?

No. Since multiplication is commutative, $(x+p)(x+q)$ and $(x+q)(x+p)$ are exactly the same expression, so either order is an equally correct final answer. What does matter is getting the correct pair of numbers with the correct signs attached to each — swapping which number gets which sign changes the answer entirely.

Factorising trinomials

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads