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Factorization Calculator

Use ^ for powers (x^2). Fractions and decimals work as coefficients. Separate several whole numbers with commas for their GCF and LCM.

Try an example

Factored form

x2−9=(x−3)(x+3)x^{2} - 9 = \left(x - 3\right)\left(x + 3\right)

Check: multiplying the factors back out gives the original polynomial.

Roots

x=3x = 3
x=−3x = -3

Steps

  1. Rational root test: any rational root is ±(divisor of 9)/(divisor of 1); candidates 1, -1, 3, -3, 9, -9.
  2. x = 3 is a root, so (x − 3) is a factor.
  3. x = -3 is a root, so (x + 3) is a factor.

Factor a polynomial in x, find the prime factorization and divisors of a whole number, or the greatest common factor and least common multiple of several numbers. Polynomials are factored with the rational root test and standard identities, the steps are shown, and the result is checked by multiplying it back out.

Formula

a2−b2=(a−b)(a+b)\text{a}^{2} - \text{b}^{2} = (\text{a} - \text{b})(\text{a} + \text{b})
a3−b3=(a−b)(a2+ab+b2)\text{a}^{3} - \text{b}^{3} = (\text{a} - \text{b})(\text{a}^{2} + \text{ab} + \text{b}^{2})
rational root=±(factor of constant term)/(factor of leading coefficient)\text{rational root} = \pm (\text{factor of constant term}) / (\text{factor of leading coefficient})

Worked Example

Polynomial6x² + x − 2
Result
Factored form(2x − 1)(3x + 2)
Rootsx = 1/2 and x = −2/3

Any rational root must be ±(a divisor of 2) ÷ (a divisor of 6). Testing those candidates, x = 1/2 and x = −2/3 make the polynomial zero, so (2x − 1) and (3x + 2) are factors, and their product is 6x² + x − 2.

Frequently Asked Questions

How do I factor a quadratic when the x² coefficient isn't 1?

Use the ac method. For 6x² + x − 2, multiply a·c = −12 and find two numbers with that product that add up to b = 1: 4 and −3. Split the middle term, 6x² + 4x − 3x − 2, and group: 2x(3x + 2) − 1(3x + 2) = (2x − 1)(3x + 2).

What if a polynomial has no rational roots?

Then it has no factors with rational coefficients from the rational root test. x² − 2 still splits over the real numbers as (x − √2)(x + √2), and x² + 1 only over the complex numbers, as (x − i)(x + i).

How are the GCF and LCM found from prime factorizations?

Write each number as a product of primes. The GCF takes each shared prime to its lowest power and the LCM takes every prime to its highest: 84 = 2²·3·7 and 126 = 2·3²·7 give a GCF of 2·3·7 = 42 and an LCM of 2²·3²·7 = 252.