The simple case: leading coefficient 1
Factoring a trinomial x^2 + bx + c is reverse FOIL: you want (x + p)(x + q) where, by multiplying back out, p + q = b and p·q = c. So hunt for a factor pair of c that adds to b. Always list factor pairs of c first, then test their sums.
Factor x^2 + 7x + 12
Need p*q = 12 and p+q = 7
Factor pairs of 12: 1*12 (sum 13)
2*6 (sum 8)
3*4 (sum 7) <- this one
x^2 + 7x + 12 = (x + 3)(x + 4)
Signs: for x^2 - 7x + 12 use -3 and -4 -> (x - 3)(x - 4)
for x^2 - x - 12 need product -12, sum -1 -> +3, -4 -> (x + 3)(x - 4)The AC method for a ≠ 1
When the leading coefficient a is not 1, reverse FOIL by eye gets hard. The [[ac-method|AC method]] makes it mechanical: multiply a·c, find two numbers that multiply to a·c and add to b, split the middle term using them, and finish by grouping.
Factor 6x^2 + 11x + 3 (a=6, b=11, c=3)
a*c = 6*3 = 18
Need two numbers: product 18, sum 11 -> 9 and 2
Split the middle term 11x as 9x + 2x:
6x^2 + 9x + 2x + 3
Group: (6x^2 + 9x) + (2x + 3)
= 3x(2x + 3) + 1(2x + 3)
= (2x + 3)(3x + 1)
Check (FOIL): (2x+3)(3x+1) = 6x^2 + 2x + 9x + 3 = 6x^2 + 11x + 3 ✓