How to use the Factoring Calculator
- Write your expression in descending order: ax² + bx + c.
- Enter the three coefficients, including negative signs.
- Check the discriminant: negative means no real factorisation exists.
- Read the factored form. Non-integer roots produce decimal factors, which are still correct.
- Expand the factors mentally to confirm they reproduce the original trinomial.
How the calculation works
Factoring and root-finding are the same problem viewed from two sides. The factor theorem states that r is a root of a polynomial exactly when (x − r) divides it. So the reliable route to a factorisation is to solve the quadratic first with the quadratic formula, then write the factors from the roots: a(x − r₁)(x − r₂). This always works, whereas guess-and-check trial factoring only works when the roots happen to be simple rationals.
The discriminant b² − 4ac decides what kind of factorisation exists. A positive discriminant gives two distinct real roots and two distinct real factors. Zero gives a repeated root and a perfect square, a(x − r)². A negative discriminant means the parabola never touches the axis, so no real linear factors exist — the expression is irreducible over the reals, though it does factor over the complex numbers.
A perfect-square discriminant is the signal that the classic integer methods will succeed. When b² − 4ac is a perfect square and a divides cleanly, the AC method or simple inspection produces integer factors immediately, which is what most homework expects. When it is not, the decimal factors this calculator reports are the honest answer, and any textbook demanding 'nice' factors has a typo in the coefficients.
ax² + bx + c = a(x − r₁)(x − r₂), where r = (−b ± √(b² − 4ac)) / 2aSource: Factor theorem and the quadratic formula; see OpenStax College Algebra, Sections 1.5 and 2.5.
Worked example
Factor 2x² − 7x + 3 to find where a projectile's height model equals zero.
- Enter a = 2, b = −7, c = 3.
- Discriminant: 49 − 4(2)(3) = 49 − 24 = 25, a perfect square, so integer-friendly factors exist.
- Roots: (7 ± 5)/4 = 3 and 0.5.
- Factored: 2(x − 3)(x − 0.5), which tidies to (x − 3)(2x − 1).
(x − 3)(2x − 1) — expanding returns 2x² − 7x + 3, confirming the factorisation.
Frequently asked questions
Why does my factored form have a fraction in it?+
Because a root is fractional. Multiply that factor by the leading coefficient and divide the other by the same amount: 2(x − 0.5) becomes (2x − 1), which is the tidier equivalent form.
What does 'irreducible over the reals' mean?+
No pair of real linear factors multiplies back to the expression. The parabola sits entirely above or entirely below the x-axis. Over the complex numbers it still factors, into a conjugate pair.
Can I factor a cubic here?+
Not with this tool. Use the polynomial root finder to locate a real root r, divide the cubic by (x − r), then factor the resulting quadratic here.
Is factoring always the best way to solve a quadratic?+
No. Factoring is fastest when the roots are simple integers. Otherwise the quadratic formula is quicker and never fails.
Last reviewed August 31, 2026. We review this page whenever the underlying formula, tax year, published rate or standard changes.