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Factoring a trinomial of the form x² + bx + c means finding two binomials (x + p)(x + q) whose product is the original trinomial. Since FOIL gives x² + (p+q)x + pq, we need p + q = b and p × q = c. For x² + 7x + 12, we need two numbers that add to 7 and multiply to 12: 3 and 4, giving (x + 3)(x + 4). When the leading coefficient is not 1 (ax² + bx + c with a > 1), techniques like the AC method or trial-and-error are used. Factoring trinomials is the reverse of FOIL and is the primary method for solving quadratic equations without the quadratic formula.
Start with simple trinomials (leading coefficient of 1) and practice finding the two numbers that satisfy the sum-and-product conditions. Use organized lists of factor pairs. Then introduce trinomials with leading coefficients greater than 1 using the AC method (multiply a × c, find factors that add to b, split the middle term, factor by grouping). Always check by FOILing the answer.
Factoring trinomials is the reverse of the FOIL process you already know. When you multiplied (x + 3)(x + 4) using FOIL, you got x² + 7x + 12. Factoring runs that process backward: given x² + 7x + 12, find the two binomials. Because FOIL produces a coefficient of x equal to the sum of the two constants, and a constant term equal to their product, the problem reduces to: find two numbers p and q such that p + q = 7 and p × q = 12.
The most reliable approach is systematic: list factor pairs of 12 — (1, 12), (2, 6), (3, 4) — and check which pair sums to 7. The pair (3, 4) works: 3 + 4 = 7 and 3 × 4 = 12, so x² + 7x + 12 = (x + 3)(x + 4). Negative signs follow the same logic with signed numbers. For x² − x − 12, you need p + q = −1 and p × q = −12. The pair (3, −4) works: 3 + (−4) = −1 and 3 × (−4) = −12, giving (x + 3)(x − 4). When c is negative, one factor must be positive and one negative; when c is positive and b is negative, both factors are negative.
When the leading coefficient is not 1 — say 2x² + 7x + 3 — the AC method extends the same idea. Multiply the leading coefficient by the constant: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: those are 1 and 6. Use these to split the middle term: 2x² + x + 6x + 3. Now factor by grouping — the GCF work you already know: x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1). The grouping step works precisely because both pairs share the binomial factor (2x + 1).
Always verify by FOILing your answer — this habit catches sign errors immediately. Factoring trinomials builds directly into solving quadratic equations: once you write ax² + bx + c as a product of two binomials, the zero product property lets you find roots by setting each factor to zero. A trinomial that doesn't factor over the integers is called irreducible over ℤ and requires the quadratic formula, which you'll encounter next.