Factor quadratic trinomials ax² + bx + c into binomials instantly — enter the coefficients and get the factored form with full step-by-step working. Free, fast, and mobile-friendly.
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Factoring means rewriting an expression as a product of simpler expressions. For the quadratic trinomial ax² + bx + c, factoring finds two binomials (px + q)(rx + s) that multiply back to the original. Our free Factoring Calculator factors any quadratic with whole-number coefficients instantly, showing the discriminant check, the binomial hunt, and the verification step — or telling you honestly when no integer factorization exists.
Factoring is the gateway skill of algebra. It solves quadratic equations (set each factor to zero), simplifies rational expressions (cancel common factors), finds x-intercepts of parabolas, and reveals the structure hiding inside a messy polynomial. Doing it by hand means guessing factor pairs of a and c until the middle term works out — a search this calculator performs systematically in milliseconds.
The tool accepts positive, negative, and zero coefficients, detects perfect-square trinomials, handles the c = 0 shortcut by pulling out a common x, and explains non-factorable cases instead of failing silently. Whether you are checking homework, studying for the SAT, or just want to see the steps, you get a complete, verified factorization every time.
Factoring: ax² + bx + c
Factoring reverses multiplication. Expanding (x + 2)(x + 3) gives x² + 5x + 6; factoring starts from x² + 5x + 6 and recovers (x + 2)(x + 3). The factored form is more useful than the expanded form because it exposes the roots: by the zero-product property, (x + 2)(x + 3) = 0 exactly when x = −2 or x = −3. A quadratic equation that looks hard becomes two trivial linear equations.
The general target is (px + q)(rx + s) = ax² + bx + c, which forces three conditions: p×r = a, q×s = c, and ps + qr = b. Finding integers satisfying all three is a small search problem: list factor pairs of a, list factor pairs of c, and test combinations until the middle term matches. When a = 1 this simplifies beautifully — you just need two numbers that multiply to c and add to b.
Not every quadratic factors over the integers. x² + x + 1 has discriminant −3, so it has no real roots at all. x² + x + 2 has discriminant −7. And x² + 2x + 2 has discriminant −4. The discriminant D = b² − 4ac predicts all of this: D < 0 means no real factorization, D a non-square means irrational roots, and D a perfect square is the signature of a clean integer factorization.
A special prize case is the perfect-square trinomial: x² + 6x + 9 = (x + 3)², where the discriminant is exactly zero and both binomials coincide. Spotting the pattern a² + 2ab + b² = (a + b)² saves real time, and the calculator flags it whenever it appears.
(px + q)(rx + s) = ax² + bx + c, where p×r = a, q×s = c, and ps + qr = b
When a = 1, this reduces to finding two numbers that multiply to c and add to b.
Here a = 1, so find two numbers with product 6 and sum 5: 2 and 3. The factorization is (x + 2)(x + 3). Verify by FOIL: x² + 3x + 2x + 6 = x² + 5x + 6. The roots are x = −2 and x = −3. Discriminant check: 25 − 24 = 1, a perfect square, as expected.
Now a = 2. Factor pairs of 2: (1, 2). Factor pairs of 3: (1, 3). Try p = 2, r = 1, q = 1, s = 3: ps + qr = 2×3 + 1×1 = 7. Match! The factorization is (2x + 1)(x + 3). Check: 2x² + 6x + x + 3 = 2x² + 7x + 3. Roots: x = −1/2 and x = −3.
Here b = 0 and c = −9. Need numbers with product −9 and sum 0: 3 and −3. The factorization is (x + 3)(x − 3). This is the difference-of-squares pattern a² − b² = (a + b)(a − b), the fastest factorization in algebra.
Discriminant: 1 − 4 = −3, which is negative. No real roots exist, so no factorization over the reals is possible. The calculator reports this directly rather than guessing — and indeed, x² + x + 1 > 0 for every real x.
A ball is thrown upward and its height follows h(t) = −16t² + 32t (feet, seconds). When does it hit the ground? Factor: −16t² + 32t = −16t(t − 2). Setting each factor to zero gives t = 0 (launch) and t = 2 (landing). Factoring turned a physics equation into an instant answer — the flight lasts exactly 2 seconds.
A rectangular garden has area x² + 7x + 12 square meters, where x is a design parameter. Factoring gives (x + 3)(x + 4), revealing the side lengths directly. If the landscaper chooses x = 5, the garden is 8 by 9 meters — the factored form made the dimensions readable at a glance.
Business analysts meet factored quadratics in break-even problems. If profit follows P(x) = −x² + 9x − 20 (thousands), factoring as −(x − 4)(x − 5) shows profit is zero at x = 4 and x = 5 thousand units — the break-even points — and positive between them. The factored form literally maps the profitable zone.
The calculator’s search is a computerized version of the ac method. Multiply a×c, then split that product into two numbers m and n with m + n = b. Rewrite the middle term: ax² + mx + nx + c, then factor by grouping. For 2x² + 7x + 3: ac = 6, and 1 + 6 = 7, so 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1). Same answer the calculator finds, through pure structure.
Why does this always work when an integer factorization exists? Because (px + q)(rx + s) expands to pr·x² + (ps + qr)x + qs, so m = ps and n = qr automatically satisfy m×n = pqrs = (pr)(qs) = ac and m + n = b. The ac method is not a trick — it is the factorization wearing a disguise, reconstructed one step at a time.
For leading coefficients with many factor pairs (like a = 12), the search space grows, but it stays tiny for a computer: even a = 360 has only 24 factor pairs to try against each pair from c. The calculator tries every combination with both sign arrangements, deduplicates swapped binomials, and reports the first clean hit — exhaustive, not lucky.
Pairs of 3: (1, 3). Pairs of 6: (1, 6), (2, 3). Try p = 3, r = 1, q = 2, s = 3: ps + qr = 9 + 2 = 11. Match! Answer: (3x + 2)(x + 3). Check: 3x² + 9x + 2x + 6 = 3x² + 11x + 6. Roots: x = −2/3, x = −3.
Discriminant: 36 − 36 = 0. Numbers with product 9 and sum 6: 3 and 3 — a repeated pair. Answer: (x + 3)(x + 3) = (x + 3)². The double root is x = −3, where the parabola just touches the x-axis.
Pairs of 6: (1, 6), (2, 3). Pairs of 12: (1, 12), (2, 6), (3, 4). With c negative, q and s take opposite signs. Try p = 3, r = 2, q = −4, s = 3: ps + qr = 9 − 8 = 1… wrong sign. Try q = 4, s = −3: ps + qr = −9 + 8 = −1. Match! Answer: (3x + 4)(2x − 3). Check: 6x² − 9x + 8x − 12 = 6x² − x − 12.
Factoring also runs in the opposite direction through the Factor Theorem: (x − r) is a factor of a polynomial exactly when r is a root. That is why the calculator’s roots and factors always agree — finding one is finding the other. For higher-degree polynomials the Rational Root Theorem lists every possible rational root p/q (p dividing the constant term, q dividing the leading coefficient), turning root-hunting into the same kind of finite search the calculator performs here.
Factor x² + 9x + 20. (Hint: two numbers with product 20 and sum 9.) Then verify with the calculator.
Factor 4x² − 9. What special pattern is this? What are the roots?
Show that x² + 3x + 7 cannot be factored over the integers. What does the discriminant tell you?
A startup models monthly profit (in thousands) as P(x) = −2x² + 14x − 24, where x is thousands of units sold. Management asks: at what sales levels does the company break even?
First pull out the common factor −2: P(x) = −2(x² − 7x + 12). The calculator factors the inside instantly: x² − 7x + 12 = (x − 3)(x − 4), since −3 × −4 = 12 and −3 + −4 = −7. So P(x) = −2(x − 3)(x − 4).
Setting each factor to zero: break-even at x = 3 and x = 4 thousand units. Because the leading coefficient is negative, profit is positive between the roots — selling 3,000 to 4,000 units makes money, and anything outside loses it. One factorization answered the strategic question completely: the profitable window is exactly 3k–4k units.
The Factoring Calculator factors quadratic trinomials ax² + bx + c with integer coefficients by exhaustively searching integer binomial pairs (px + q)(rx + s) satisfying pr = a, qs = c, and ps + qr = b. It shows the discriminant, the verification step, the roots, flags perfect-square trinomials, handles the c = 0 shortcut, and explains clearly when no integer factorization exists.
Find two numbers that multiply to 6 and add to 5: 2 and 3. The factorization is (x + 2)(x + 3). Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6.
Multiply a×c, split the product into two numbers m, n with m + n = b, rewrite as ax² + mx + nx + c, and factor by grouping. For 2x² + 7x + 3: ac = 6 = 1 × 6, giving 2x² + x + 6x + 3 = (2x + 1)(x + 3).
List factor pairs of a and of c, then test combinations (p, q, r, s) until ps + qr equals b. For 3x² + 11x + 6, p = 3, r = 1, q = 2, s = 3 works because 3×3 + 2×1 = 11, giving (3x + 2)(x + 3).
If the discriminant b² − 4ac is negative, there are no real roots and no real factorization. If it is positive but not a perfect square, the roots are irrational — use the quadratic formula instead of hunting for integer binomials.
A trinomial that factors as a repeated binomial: x² + 6x + 9 = (x + 3)². The pattern is a² + 2ab + b² = (a + b)², and the discriminant is exactly zero.
By the zero-product property: if (x + 2)(x + 3) = 0, then x + 2 = 0 or x + 3 = 0, so x = −2 or x = −3. Factoring converts one quadratic equation into two trivial linear ones.
Pulling out the greatest common factor first keeps the remaining numbers small and avoids missing a factor. 3x² + 12x + 9 becomes 3(x + 1)(x + 3) — without that first step the search is three times harder.
Factoring is faster and more revealing when integer factors exist, but it fails on non-factorable quadratics. The quadratic formula always works, giving exact roots (possibly irrational or complex) for any coefficients. Use factoring first; fall back to the formula when it does not apply.
Cite this page as: Factoring Calculator. Sigma Calculator. Retrieved October 9, 2026, from https://sigmacalculator.net/math-calculators/factoring-calculator/
Last updated: October 9, 2026