When the leading coefficient is not 1, guessing the two binomials is a search with too many branches. The AC method replaces the guess with arithmetic: one product to compute, one pair of integers to find, and then the grouping of “Factoring by Grouping” finishes the job.
Core ideas
The method is three lines of arithmetic. What matters is why the middle step is allowed, and what it means when the search comes back empty.
mx + nx is (m + n)x, and m + n is b by construction. The expression after the split is equal to the one before it for every value of x — you have written bx as a sum, not altered the polynomial. That equality is the whole licence for the step.
ac is one integer and it has finitely many factor pairs. Listing them and checking each sum is a complete search, so it either produces the pair or establishes that none exists.
An empty search is an answer. If no integer factor pair of ac sums to b, then ax² + bx + c is not a product of two binomials with integer coefficients, and no amount of further staring will make it one.
The material
Why splitting the middle term is legal, and what an empty search proves.
“Factoring Simple Trinomials” factored x² + bx + c by finding two numbers that multiply to c and add to b. That shortcut depends on the leading coefficient being 1. Once a ≠ 1, the constant term of the product is no longer the product of the two numbers you are looking for, because the a distributes into both binomials.
To factor ax² + bx + c with integer coefficients: compute ac; find integers m and n with mn = ac and m + n = b; rewrite bx as mx + nx; then factor the resulting four terms by grouping.
If m + n = b then ax² + mx + nx + c = ax² + (m + n)x + c = ax² + bx + c. The two expressions agree for every x, so one may be replaced by the other anywhere. Nothing is being done to an equation; a single expression is being rewritten.
This is the point worth holding on to. The method looks like a trick because the number ac appears from nowhere, but the only step that could possibly be illegal — replacing bx by mx + nx — is justified by one line of arithmetic that you control.
ac = 6 · 3 = 18. The factor pairs of 18 are 1·18, 2·9 and 3·6, with sums 19, 11 and 9. The pair is 9 and 2. Splitting: 6x² + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3).
The order of m and n does not matter. Splitting as 6x² + 2x + 9x + 3 gives 2x(3x + 1) + 3(3x + 1) = (2x + 3)(3x + 1) — the same two factors in the other order. If your grouping produces two brackets that are not identical, you have made an arithmetic slip, not discovered a second answer.
If ax² + bx + c = (rx + s)(tx + u) with integer coefficients, then a = rt, b = ru + st and c = su. The two middle products m = ru and n = st therefore satisfy mn = ac and m + n = b. Any integer-coefficient binomial factorisation must produce a pair on the ac list.
ac = 8. The positive pairs 1·8 and 2·4 give sums 9 and 6; the negative pairs (-1)(-8) and (-2)(-4) give -9 and -6. None is 3, and the theorem above shows why the completed list rules out every product of integer-coefficient binomials.
Take the common factor out first, always. 12x² + 22x + 6 has ac = 72 and a messier pair list, but 12x² + 22x + 6 = 2(6x² + 11x + 3) = 2(3x + 1)(2x + 3) reduces it to the example above. “Factoring Out the Greatest Common Factor” put the common factor at the top of the decision list for exactly this reason.
Interactive
The ac search, with every pair it tried
Method
Four steps, and the first one is not optional.
Divide out any factor common to all three terms, including a leading -1 if a is negative. Skipping this is what turns a small pair search into a large one.
Multiply the leading coefficient by the constant term, signs included. For 12x² - 17x + 6 this is 72; for 6x² - 7x - 20 it is -120.
List every pair, with its sum. If ac is positive the pair shares the sign of b; if ac is negative the pair has opposite signs. Stop at the pair whose sum is b, or stop when the list is exhausted.
Rewrite bx as mx + nx, group the four terms in pairs, and take out the common binomial. Multiply your answer back out — on this topic checking is faster than the work.
ac = 72 is positive and b = -17 is negative, so both members of the pair are negative. That observation halves the list before it is written.
The second group is -9x + 6, and the factor taken out is -3, not 3. Pulling out 3 would give 3(-3x + 2), whose bracket does not match 3x - 2, and the grouping would appear to fail. When the two brackets differ only by an overall sign, the sign you extracted is the thing to fix.
For a faded ac pass, start 6x² - 7x - 20 from the supplied product ac = -120. Search opposite-sign factor pairs for sum -7, split the middle term, group, and expand. The pair is 8 and -15, giving (2x - 5)(3x + 4); using -8 and 15 preserves the product but reverses the middle coefficient.
You should be able to factor 6x² - 7x - 20 and 12x² - 17x + 6 without hesitation, and to say of 2x² + 3x + 4 that it does not factor into integer-coefficient binomials — naming the completed ac pair search as your reason.
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