When the leading coefficient is not $1$, as in $6x^2 + 11x + 3$, guessing gets slow. The ac method turns it into something you already know: finding two numbers, followed by grouping.
Por qué funciona
Multiply out $(px + q)(rx + s) = prx^2 + psx + qrx + qs$. The two middle pieces, $psx$ and $qrx$, have coefficients whose product is $ps \cdot qr = pr \cdot qs$, which is $a \cdot c$, and whose sum is $b$. So: find two numbers that multiply to $ac$ and add to $b$, write the middle term as those two pieces, and the expression splits into two groups that share a binomial.
Idea clave
For $ax^2 + bx + c$: find two numbers with product $ac$ and sum $b$, split $bx$ into those two terms, then factor by grouping. Take out any GCF first.
Worked examples
Ejemplo resuelto: The ac method
Factor $6x^2 + 11x + 3$.
- $ac = 6 \cdot 3 = 18$. Two numbers with product $18$ and sum $11$: $2$ and $9$.
- Split: $6x^2 + 2x + 9x + 3$.
- Group: $2x(3x + 1) + 3(3x + 1)$.
- Factor out $(3x + 1)$: $(2x + 3)(3x + 1)$.
Respuesta: $(2x + 3)(3x + 1)$
Ejemplo resuelto: With a negative
Factor $4x^2 - 4x - 15$.
- $ac = -60$. Numbers with product $-60$ and sum $-4$: $-10$ and $6$.
- Split: $4x^2 - 10x + 6x - 15$.
- Group: $2x(2x - 5) + 3(2x - 5) = (2x + 3)(2x - 5)$.
- Check: $4x^2 - 10x + 6x - 15 = 4x^2 - 4x - 15$.
Respuesta: $(2x + 3)(2x - 5)$
Error común
Factoring as if $a$ were $1$, for example writing $6x^2 + 11x + 3 = (x + 2)(x + 9)$ because $2 + 9 = 11$. That multiplies to $x^2 + 11x + 18$, not the original. Always multiply back: the first terms must give $ax^2$ and the last terms $c$.