How this instrument works
A trinomial ax²+bx+c factors into two binomials whenever it has real roots — and the quadratic formula finds both of those roots directly from the three coefficients, without needing to guess factor pairs by trial and error. Once the two roots, call them r₁ and r₂, are known, the trinomial is exactly a×(x−r₁)×(x−r₂).
For small, tidy integer coefficients, this often matches a quicker mental shortcut: find two numbers that multiply to give a×c and add to give b. But the quadratic formula handles every case uniformly — irrational roots, non-monic leading coefficients, negative coefficients — without needing a special method for each.
This site's companion Reverse FOIL calculator runs the identical relationship in the opposite direction: starting from two chosen numbers p and q, it expands (x+p)(x+q) back out to find the b and c coefficients they'd produce — seeing both directions side by side is exactly what makes factoring by reversing FOIL click for a lot of learners.
- Enter the trinomial's three coefficients into the a, b, and c fields.
- Read Root 1 and Root 2 — the two solutions found via the quadratic formula.
- Write the factored form directly as a(x − root 1)(x − root 2).
- Expand that factored form back out to confirm it reproduces the original trinomial.
Worked example — factoring x²−5x+6
For x²−5x+6 (a=1, b=−5, c=6), the discriminant is (−5)²−4×1×6=25−24=1, so the quadratic formula gives roots of (5+1)⁄2=3 and (5−1)⁄2=2 — meaning x²−5x+6 factors as (x−3)(x−2). Checking with the quicker mental shortcut confirms it: 3 and 2 multiply to 6 and add to 5, exactly matching the coefficients.
A non-monic example, 2x²+7x+3 (a=2, b=7, c=3), has a discriminant of 49−24=25, giving roots of (−7+5)⁄4=−0.5 and (−7−5)⁄4=−3 — so it factors as 2(x+0.5)(x+3), or equivalently (2x+1)(x+3) once the leading 2 is redistributed into the first binomial.
Questions
How do you factor a trinomial using the quadratic formula?
Find its two roots with the quadratic formula, then write the factored form as a(x−root₁)(x−root₂) — every trinomial with real roots factors this way, regardless of whether the coefficients are small tidy integers or not.
What if the discriminant is negative?
The trinomial has no real roots and doesn't factor into real binomials at all — it remains irreducible over the real numbers, though it would still factor using complex roots.
Does this work for non-monic trinomials (where a isn't 1)?
Yes — the quadratic formula already accounts for the leading coefficient directly, so a non-monic trinomial factors exactly the same way, just redistributing that leading coefficient into one of the two binomials afterward if a cleaner integer form is wanted.
How is this different from the Reverse FOIL calculator on this site?
This page starts from a trinomial's coefficients and works backward to find its two roots; the Reverse FOIL page runs the opposite direction, starting from two chosen numbers and expanding them forward into the trinomial they'd produce.
Can every trinomial be factored?
Only over the real numbers if its discriminant is zero or positive — a negative discriminant means the trinomial has no real roots, and no pair of real binomials will reproduce it.