Factoring Trinomials
Factoring trinomials means expressing a three-term polynomial, usually in the form $ax^2 + bx + c$, as a product of two binomials.
Formula
ax^2 + bx + c = (px + q)(rx + s)
Definition
Factoring a trinomial means finding two binomials that multiply to give it back; for $x^2 + bx + c$ (leading coefficient $1$), find two numbers $p$ and $q$ with product $c$ and sum $b$. For $ax^2 + bx + c$ where $a \neq 1$, use the "ac method": find two numbers with product $ac$ and sum $b$, split the middle term, then factor by grouping. A trinomial factors over $\mathbb{Q}$ if and only if its discriminant $b^2 - 4ac$ is a perfect square; the roots $r_1$ and $r_2$ of the factored form $a(x - r_1)(x - r_2)$ satisfy Vieta's formulas, $r_1 + r_2 = -b/a$ and $r_1r_2 = c/a$, connecting roots directly to coefficients.
Example
Factor $x^2 + 7x + 12$: two numbers with product $12$ and sum $7$ are $3$ and $4$, giving $(x + 3)(x + 4)$. Factor $2x^2 + 7x + 3$: $ac = 6$, needing sum $7$, so $6$ and $1$: split as $2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)$. $6x^2 - 5x - 6$ has discriminant $25 + 144 = 169 = 13^2$, giving roots $(5 \pm 13)/12 = 3/2$ or $-2/3$, so it factors as $(2x-3)(3x+2)$.
Key Insight
The "product and sum" trick, listing factor pairs of $c$ to find the one that adds to $b$, is really a search for integer roots; if no such integers exist, the trinomial does not factor over the integers, and the quadratic formula is needed to find irrational roots instead. Vieta's formulas generalize to any degree-$n$ polynomial: the elementary symmetric polynomials of the roots equal the coefficients (up to sign and leading coefficient), connecting roots to coefficients without ever solving the equation.