Factoring trinomials

Expanding gave us the master pattern:

$(x + a)(x + b) = x^2 + (a + b)x + ab$

Factoring a trinomial is this equation read right-to-left: given x² + 5x + 6, find two numbers whose sum is 5 and whose product is 6.

The hunt

Factor x² + 7x + 12. List the pairs that multiply to 12 and watch the sums:

So x² + 7x + 12 = (x + 3)(x + 4). Multiply back to check: 3 + 4 = 7 ✓, 3 · 4 = 12 ✓.

Signs tell you where to look

The constant's sign narrows the hunt before it starts:

Examples: x² − 8x + 15 needs product +15, sum −8 → (−3, −5) → (x − 3)(x − 5). And x² + 2x − 15 needs product −15, sum +2 → (+5, −3) → (x + 5)(x − 3).

When x² has a coefficient

For 2x² + 7x + 3 the tile picture still works, with one twist: hunt for two numbers with product a·c = 6 and sum 7 → (6, 1). Split the middle term and factor by grouping:

$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$

Not everything factors

x² + x + 1 has no integer pair with product 1 and sum 1 — it's prime over the integers. That's not failure; it's information. The quadratic formula (end of this course) solves even the trinomials that refuse to factor.

Why this matters

Factored trinomials are the key that opens quadratic equations: x² + 5x + 6 = 0 is hard to stare down, but (x + 2)(x + 3) = 0 answers itself. Two lessons from now, that move gets a name — the zero-product rule.