To factorise ax² + bx + c, find two numbers that multiply to give a × c and add to give b. Use these two numbers to split the middle term, then factorise by grouping. This "split the middle term" method works for every quadratic that factorises nicely — including when a is not 1 — and replaces guess-and-check with a reliable, repeatable process.
Many students learn to factorise x² + 5x + 6 by guessing pairs of brackets until something works. This is fine when a = 1 and the numbers are small. It falls apart the moment a is not 1, or the numbers get larger — students either guess wrong repeatedly and run out of time, or give up and lose the marks entirely.
The method below works identically every time, regardless of the size of a, b, or c. Learn it once and every quadratic factorising question becomes the same 4-step process.
Step 1
Multiply a × c
Step 2
Find 2 numbers that multiply to a×c and add to b
Step 3
Split bx into those 2 terms
Step 4
Factorise by grouping
Factorise x² + 7x + 12.
Solution
x² + 3x + 4x + 12x(x + 3) + 4(x + 3)(x + 3)(x + 4)Quick check: expand your answer back out. (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12. Matches the original — correct. This 10-second check catches sign errors before they cost marks.
Factorise 6x² + 11x + 3. This is where guess-and-check usually breaks down — the systematic method handles it exactly the same way.
Solution
6x² + 2x + 9x + 32x(3x + 1) + 3(3x + 1)(3x + 1)(2x + 3)The step students get wrong
In step 4, the number factored out of each group must leave identical brackets in both groups. If your two brackets don't match, you've either paired the split terms in the wrong order, or made an arithmetic slip. Reorder the two middle terms (swap which one comes first) and try grouping again — the order you split them in can affect how cleanly the grouping works.
The same method works when b or c is negative — the sign just changes which factor pair you're looking for.
Example — negative c: factorise x² + 2x − 15
x² + 5x − 3x − 15 = x(x + 5) − 3(x + 5) = (x + 5)(x − 3)Example — negative b and positive c: factorise x² − 9x + 20
x² − 4x − 5x + 20 = x(x − 4) − 5(x − 4) = (x − 4)(x − 5)When a quadratic has no middle term and both remaining terms are perfect squares with a minus sign between them, it factorises instantly without the split-the-middle-term method:
a² − b² = (a + b)(a − b)
Example — factorise 9x² − 25
9x² − 25 = (3x + 5)(3x − 5)Recognise this pattern instantly: "something squared minus something squared" with no middle term is always difference of two squares. This appears constantly in O-Level algebra, including hidden inside algebraic fraction and equation questions — spotting it saves significant time over the full split-the-middle-term method.
Not every quadratic has whole-number factors. If you can't find a factor pair that works after checking all reasonable options, the quadratic likely doesn't factorise over integers — this is a signal to use the quadratic formula or completing the square instead, particularly if the question says "solve" rather than "factorise." Whichever method you reach for, it usually needs the basic moves from simplifying algebraic expressions to keep the working clean.
Don't force it
Spending several minutes hunting for a factor pair that doesn't exist wastes valuable exam time. If a × c doesn't have a factor pair that adds to b within a reasonable number of tries, move to the quadratic formula or completing the square — especially if the question explicitly says "solve" rather than "factorise."
Before spending time hunting for a factor pair, you can use the calculator's Equation mode to find the roots directly. If the roots come out as whole numbers or simple fractions, you now know exactly what your brackets should be — no more guessing whether a factor pair even exists.
On the fx-97SG CW (ClassWiz)
On the fx-97SG X
Turning the roots back into factors
Take worked example 2 from earlier: 6x² + 11x + 3. Entering a = 6, b = 11, c = 3 into Equation mode gives roots x = −1/3 and x = −3/2.
(3x + 1). Root x = −3/2 → factor (2x + 3).(3x + 1)(2x + 3) — matching the answer from the split-the-middle-term method.How to use this in the exam: the calculator does not show your working, so this is a verification tool, not a substitute for knowing the method — most O-Level questions specifically ask you to show factorisation steps. Use Equation mode to instantly confirm whether a quadratic factorises nicely before you commit time to the split-the-middle-term method, and to double-check your final brackets are correct.
Why do I multiply a × c instead of just using c?
When a = 1, a × c is just c, so the shortcut still works. But when a ≠ 1, the two numbers you need must multiply to a × c (not just c) for the grouping step to produce matching brackets. This is what makes the method work consistently regardless of the value of a.
Does it matter which order I split the middle term in?
Usually not, but occasionally swapping the order of the two split terms makes the grouping step work more cleanly, particularly when a is not 1. If your first attempt at grouping doesn't produce matching brackets, try reversing the order of the two terms before assuming you made an error.
What if there's a common factor across all three terms first?
Always check for and remove a common factor before applying the split-the-middle-term method. For example, 2x² + 10x + 12 should first become 2(x² + 5x + 6), then factorise the bracket normally to get 2(x + 2)(x + 3). Skipping this step makes the numbers unnecessarily large and harder to work with. This same factorised form is what you need whenever a quadratic sits on top of or below a fraction bar, as in algebraic fractions.
— Mr Gan Math Tuition
Mr. Gan works with students who want a systematic method that works every time — not trial and error under pressure.
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