Factoring Trinomials

For x² + bx + c with a leading coefficient of 1, find two numbers that multiply to c and add to b.

x² + 7x + 12: which pair multiplies to 12 and adds to 7? 3 and 4. So it factors as (x + 3)(x + 4).

Signs tell you where to look:

• c positive, b positive → both numbers positive
• c positive, b negative → both negative
• c negative → one of each, and the larger takes b's sign

When the leading coefficient is not 1, use the AC method: multiply a and c, find a pair multiplying to ac and adding to b, split the middle term, then factor by grouping.

Always check by expanding. It is fast and it is definitive.

For monic trinomials, x² + bx + c = (x + p)(x + q) where pq = c and p + q = b. This is the expansion read backwards, and integer solutions exist only when the discriminant b² − 4c is a perfect square — which is why some trinomials are irreducible over the integers even though the quadratic formula still solves them.

For ax² + bx + c with a ≠ 1, the AC method finds p and q with pq = ac and p + q = b, rewrites bx as px + qx, and factors the resulting four terms by grouping. It works because the split reconstructs the two binomial products that FOIL would have generated.

Factoring by grouping is itself the general technique for four-term polynomials: group in pairs, extract each pair's GCF, and if the remaining binomials match, factor that binomial out.

Every factoring problem should end with a check by expansion, and should begin with a GCF extraction — those two habits eliminate most errors in the topic.