Real quadratics rarely arrive tidy. Model the arc of a long throw, the profit of a business, or the strength of a loaded beam and you often get an x^2 term with an awkward number stuck in front — the point where the simple factorising trick stops working and you need a sturdier method.

You already know how to factorise a quadratic when the x^2 coefficient is 1: hunt two numbers that multiply to c and add to b, and they drop straight into the brackets. For x^2 + 7x + 10 the numbers are 2 and 5, and the answer is (x+2)(x+5). Done.

Now try the same trick on 2x^2 + 7x + 3. Two numbers that multiply to 3 and add to 7? There aren't any — and even if there were, they couldn't just drop into brackets, because the brackets now have to start (2x \;\; )(x \;\; ) to produce that leading 2x^2. That 2 scrambles the "inner" and "outer" terms: the numbers in the brackets no longer simply add to give the middle. The old rule needs an upgrade.

The upgrade is the AC method, and the beautiful thing about it is that it turns the new, harder puzzle back into the one you can already solve. A quadratic in the form

is factorised in three moves: find two numbers that multiply to a \cdot c and add to b, use them to split the middle term into two pieces, then factor by grouping. Same hunt as before — just with a \cdot c as the target product instead of plain c.

The two numbers

For 2x^2 + 7x + 3 we have a = 2, b = 7 and c = 3, so a \cdot c = 2 \times 3 = 6. We need two numbers that multiply to 6 and add to 7. Be systematic: list the factor pairs of 6 and check each sum.

The pair is 1 and 6. Listing pairs in order like this beats staring and hoping: for a product like 6 there are only two pairs to try, and even for a \cdot c = 36 there are just five. One of them either adds to b or none does — and either way you know within a minute.

(Notice that when a = 1 this is exactly the rule you already use — multiply to c, add to b — because a \cdot c = c. The AC method isn't a second method to memorise; it's the same method, told in full.)

Splitting and grouping

Now spend the pair. Rewrite the middle term 7x as 1x + 6x — that changes nothing, since 1x + 6x = 7x — but it leaves four terms instead of three. Four terms can be grouped into two pairs, and a common factor pulled out of each pair. Here is the magic the AC pair guarantees: after factoring each pair, the same bracket appears in both. That shared bracket is itself a common factor of the whole expression, so it comes out the front — and the factorisation falls into your lap. Step through it below.

The working, line by line:

A detail worth knowing: the order of the split doesn't matter. Write 7x = 6x + 1x instead and the grouping still works — the pairs are different, but the same bracket still pops out: 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (x+3)(2x+1). Same answer, different route. If your grouping ever fails to produce a shared bracket, you haven't broken mathematics — you've mis-picked the pair. Go back and re-check the multiply-and-add.

Make checking a habit

Factorising has a superpower most topics lack: you can always mark your own work. Factorising and expanding are inverse operations, so multiplying your answer back out must recreate the original — every term, every sign. For our example:

Ten seconds, and the check either passes or catches you. Suppose you'd rushed and written (2x + 3)(x + 1) — a very tempting wrong answer, since it uses all the right numbers. Expanding exposes it instantly:

The middle term is 5x, not 7x — wrong answer, caught before it cost any marks. Do this check every single time. In an exam it converts "I think that's right" into "I know that's right."

When c is negative: 3x^2 - 5x - 2

Negative coefficients change nothing about the method — but they demand care with signs. Here a = 3, b = -5, c = -2, so a \cdot c = 3 \times (-2) = -6. We need two numbers that multiply to -6 and add to -5. A negative product means the pair has opposite signs — one positive, one negative — and a negative sum means the negative one is bigger. List the candidates:

The pair is 1 and -6. Split the middle term -5x as x - 6x and group:

Watch the second group closely: to make the bracket come out as (3x+1) — matching the first — we pull out -2, not 2, because -6x - 2 = -2(3x + 1). Pulling out a negative flips the signs inside the bracket; that flip is exactly what makes the brackets match. And of course, the habit: (3x+1)(x-2) = 3x^2 - 6x + x - 2 = 3x^2 - 5x - 2. ✓

Step zero: hunt a common factor first

Before any of this cleverness, always ask the simplest question first: do all the terms share a common factor? Sometimes that's the whole job. Take 4x^2 + 10x: there's no constant term at all, and every term is divisible by 2x, so

— fully factorised, no AC method required. Reaching for the fancy tool here would be like using a crane to lift a teacup.

Even when the AC method is needed, a common factor taken out first makes everything that follows easier. Look at 2x^2 + 8x + 6: every term is even, so take out the 2 —

Once the 2 is out, the quadratic left inside has a = 1, and you're back on the easy road: two numbers that multiply to 3 and add to 4 — that's 1 and 3 — done. The common factor didn't just save effort; it downgraded the problem to one you mastered ages ago. So the full recipe is: common factor first, then AC, then check by expanding.

See it explained

Another voice, same method: Khan Academy works through factoring a harder quadratic by grouping. Watch for the two moves you now know by name — the multiply-to-ac, add-to-b hunt, and the shared bracket appearing after grouping.