Splitting the middle term

Some quadratic trinomials factorise instantly by a known identity — $x^2+2xy+y^2$ or $a^2-b^2$ — but most, like $x^2+7x+12$ or $6x^2-7x-3$ , do not fit any identity at all. Splitting the middle term is the general-purpose technique that handles exactly these cases: it breaks the middle term $bx$ into two pieces so the four-term expression can be grouped and factorised by common brackets. This is the third factorisation method in the ICSE Class 9 Algebra chapter, arriving right after expansions and identity-based factorisation. It works for any quadratic trinomial $ax^2+bx+c$ , whether $a=1$ or not, and it is the method examiners reach for whenever the numbers don't scream "identity." What it does not do at Class 9 is solve $ax^2+bx+c=0$ for roots — that step (discriminant, quadratic formula) belongs to Class 10. Here the goal is only the factorised bracket form.

The core insight

Split $bx$ into $mx+nx$ where $m\times n=a\times c$ and $m+n=b$ , then factorise the resulting four terms by grouping.

The core idea

For a trinomial $ax^2+bx+c$ , the whole method rests on finding a pair of numbers $m$ and $n$ satisfying two conditions at once:

m \times n = a \times c \qquad \text{and} \qquad m + n = b

where: $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ , $c$ is the constant term, and $m$ , $n$ are the two numbers you are hunting for. Once found, the middle term $bx$ is rewritten as $mx+nx$ , turning the three-term trinomial into a four-term expression that can be grouped in pairs and factorised.

The special case $a=1$ . When $a=1$ , the rule collapses to something more familiar: find two numbers that multiply to $c$ and add to $b$ . This is exactly the pattern behind $(x+p)(x+q)=x^2+(p+q)x+pq$ from the expansions topic — splitting the middle term is that same idea run in reverse.

Sign-hunting shortcut. If $c$ is positive, $m$ and $n$ share the same sign as $b$ (both positive if $b>0$ , both negative if $b<0$ ). If $c$ is negative, $m$ and $n$ have opposite signs, and whichever one has the larger magnitude takes the sign of $b$ . When the split isn't obvious by inspection, systematically list the factor pairs of $a\times c$ and test each pair's sum against $b$ — this always terminates and never requires guessing blindly.

Why the target is $a\times c$ , not just $c$ . Expanding a general two-bracket product $(px+m')(qx+n')$ gives $pqx^2+(pn'+qm')x+m'n'$ . Matching this to $ax^2+bx+c$ shows the $x^2$ -coefficient comes from multiplying the two "outer" numbers and the constant term from multiplying the two "inner" numbers — while the middle term mixes both pairs together. That is precisely why the split numbers must multiply to the product $a\times c$ , not to $c$ alone, once $a\neq1$ .

Figure

Area diagram showing x squared plus 7x plus 12 split into four regions x squared, 3x, 4x and 12 that regroup into the brackets x plus 3 and x plus 4

Splitting the middle term (main derivation)

Derivation AWhy the four-term split always regroups into two identical brackets

The goal is to show, using nothing beyond the distributive law and the definitions $m\times n=a\times c$ and $m+n=b$ , that once the middle term $bx$ is correctly split into $mx+nx$ , grouping the resulting four terms in pairs always produces two brackets that are identical to each other — which is exactly what allows the final common-bracket factorisation step to go through.

Step 1. Start from the trinomial and substitute the split. Since $m+n=b$ , we have $bx=mx+nx$ , so $ax^2+bx+c=ax^2+mx+nx+c$ . Nothing has changed except how the middle term is written — the expression's value is untouched.

Step 2. Group the first two terms and the last two terms in the order the split produced them: $(ax^2+mx)+(nx+c)$ . This grouping is not arbitrary — it must be adjacent pairs in this order, because a correct split is designed so that only this pairing shares common factors.

Step 3. Factor each bracket separately. From $ax^2+mx$ , a common factor can always be pulled out. From $nx+c$ , a different common factor can be pulled out. Because $m\times n=a\times c$ , the arithmetic works out so that after pulling out these common factors, the leftover bracket in each term is identical.

Step 4. Take out the shared bracket. Once both terms read (common factor) $\times$ (same bracket), the distributive law in reverse gives (same bracket) $\times$ (sum of the two common factors) — the final factorised form. This is why the method never fails to regroup correctly when $m$ and $n$ are chosen right: the identical-bracket outcome is guaranteed by the product-and-sum conditions, not a coincidence of any one example.

Worked example

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

Step 1. Identify $a$ , $b$ , $c$ : here $a=1$ , $b=7$ , $c=12$ , so $a\times c=12$ .

Step 2. Find two numbers multiplying to $12$ and adding to $7$ . Factor pairs of $12$ are $(1,12)$ , $(2,6)$ , $(3,4)$ . The pair $3$ and $4$ sums to $7$ , so $m=3$ , $n=4$ .

Step 3. Split the middle term: $x^2+3x+4x+12$ .

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

Step 5. Take out the common bracket $(x+3)$ : the answer is $\boldsymbol{(x+3)(x+4)}$ .

Check. Expand back: $(x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12$ . Matches, so the factorisation is correct.

Competitive stretch

Factorise $6x^2-7x-3$ , where $a\neq1$ and the signs are trickier.

Step 1. Here $a=6$ , $b=-7$ , $c=-3$ , so $a\times c=-18$ .

Step 2. Since $c$ is negative, $m$ and $n$ have opposite signs. Testing pairs of $-18$ : $2$ and $-9$ give product $-18$ and sum $-7$ . So $m=2$ , $n=-9$ .

Step 3. Split: $6x^2+2x-9x-3$ .

Step 4. Group carefully with signs: $(6x^2+2x)+(-9x-3)=2x(3x+1)-3(3x+1)$ .

Step 5. Take out the common bracket: the answer is $\boldsymbol{(3x+1)(2x-3)}$ .

Check. Expand back: $(3x+1)(2x-3)=6x^2-9x+2x-3=6x^2-7x-3$ . Matches.

Common traps

Hunting for two numbers that add to $b$ and multiply to just $c$ — that shortcut only holds when $a=1$ ; for $a\neq1$ the target product is $a\times c$ .
Sign slips while grouping, such as pulling out $+3$ from $-9x-3$ instead of $-3$ , which breaks the matching bracket.
Stopping after checking only a few factor pairs instead of listing all pairs of $a\times c$ , including negative pairs when $a\times c$ is negative.
Skipping the final expansion check — a single early sign error otherwise slips through unnoticed.

Ask the tutor: Stuck finding the right pair of numbers for a tricky trinomial like $6x^2-7x-3$ ? Ask your AI tutor to walk through the systematic factor-pair search step by step.

Why isn't the target just c when a is not 1?

Because the coefficient of $x^2$ also comes from multiplying two bracket terms. Expanding $(px+m')(qx+n')$ gives $pqx^2+(pn'+qm')x+m'n'$ : the $x^2$ -coefficient is a product too ( $pq$ ), so to match a general $ax^2+bx+c$ the split numbers must combine both the leading coefficient and the constant, giving product $a\times c$ . Only when $a=1$ does $pq=1$ and the target simplifies to plain $c$ .

What if I can't find any pair of numbers that works?

Then list the factor pairs of $a\times c$ systematically and check each one's sum against $b$ before giving up — many "unsplittable" cases are really just an unchecked pair. If you have genuinely tried every factor pair (including negatives) and none sums to $b$ , the trinomial does not factorise into integer brackets at all. Recognising this quickly, rather than forcing a wrong split, saves exam time.

Does the order I write m and n in matter?

No — swapping $m$ and $n$ just swaps which "half" of the middle term you write first, and the grouping still works as long as you keep the first two terms and last two terms together in whichever order you split them. What does matter is getting the signs of $m$ and $n$ right and grouping adjacent pairs, not mixing terms from opposite ends of the expression.