Factoring Calculator

About the Factoring Calculator

Factoring rewrites a polynomial as a product of simpler polynomials. It is the reverse of multiplying out brackets, and it is one of the most used skills in algebra: it solves quadratic and cubic equations, simplifies rational expressions, finds the zeros of a graph and prepares integrals for partial fractions.

This calculator factors polynomials in one variable, such as $6x^2 + 11x - 10$ or $x^4 - 16$ , and common two-variable forms such as $x^2 - 9y^2$ or $x^2 - 5xy + 6y^2$ . It always pulls out the greatest common factor first, then looks for special patterns, then uses the AC method or the rational root theorem for higher degrees. Whatever cannot be split further over the rationals is reported as irreducible rather than left unexplained.

What does it mean to factor a polynomial?

To factor a polynomial is to write it as a product whose factors each have lower degree. For example $x^2 - 5x + 6 = (x - 2)(x - 3)$ . Multiplying the factors back together returns the original polynomial, which is the easiest way to check any factorization.

A factorization is complete when no factor can be split again using rational coefficients. The quadratic $x^2 - 2$ is irreducible over the rationals even though it equals $(x - \sqrt2)(x + \sqrt2)$ , because $\sqrt2$ is irrational. School courses and this calculator work over the rationals unless a problem explicitly asks for real or complex factors.

How to use the Factoring Calculator

Type the polynomial using ^ for powers, for example 6x^2 + 11x - 10 or x^3 - 8.
You may enter it in any order or with brackets, such as (x + 1)(x + 2) + 3x; the calculator expands and sorts it first.
Press Factor. The first step writes the polynomial in standard form with descending powers.
Follow the method steps: common factor, special pattern or AC method, and any rational-root tests with synthetic division for cubic or quartic polynomials.
Read the complete factorization at the end and, if you want to verify it, multiply it out with the simplify calculator.

Formulas

Difference of squares

a^2 - b^2 = (a - b)(a + b)

Perfect-square trinomials

a^2 \pm 2ab + b^2 = (a \pm b)^2

Sum and difference of cubes

a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)

AC method

ax^2 + bx + c:\ \text{find } m, n \text{ with } mn = ac,\ m + n = b

Split $bx$ into $mx + nx$ and factor by grouping.

Rational root theorem

p(r/s) = 0 \Rightarrow r \mid a_0,\ s \mid a_n

Candidate roots are factors of the constant term over factors of the leading coefficient.

Worked example: Trinomial 6x² + 11x − 10

The worked example uses the trinomial $6x^2 + 11x - 10$ , a classic case where the leading coefficient is not 1, so simple guess-and-check is slow. The engine multiplies $a \cdot c = 6 \cdot (-10) = -60$ and searches for two integers with product $-60$ and sum $11$ .

Those integers are $15$ and $-4$ . Splitting the middle term as $15x - 4x$ lets the four terms be grouped in pairs, each pair sharing a binomial factor. Multiply the answer back to confirm it.

The solution below is generated by the same engine that powers the calculator above.

Final answer

(3x - 2)(2x + 5)

How we read your problem

Factor a polynomial over the rational numbers

6x^{2} + 11x - 10

Step-by-step solution

Write the polynomial in standard form
$6x^{2} + 11x - 10$
Use the AC method
Multiply a·c = 6·-10 = -60. Find two numbers whose product is -60 and whose sum is b = 11: 15 and -4.
$6x^{2} + 11x - 10 = 6x^2 + 15x - 4x - 10$
Factor by grouping
Group the first two and last two terms, factor each group, then factor out the common binomial. The roots are -5/2 and 2/3.
Write the complete factorization
$(3x - 2)(2x + 5)$

Formulas used

Difference of squares
$a^2 - b^2 = (a - b)(a + b)$
Perfect square trinomial
$a^2 \pm 2ab + b^2 = (a \pm b)^2$
Sum and difference of cubes
$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$
Rational Root Theorem
$a_n x^n + \dots + a_0 = 0,\ \tfrac{p}{q} \text{ root} \Rightarrow p \mid a_0,\ q \mid a_n$

6x^2 + 11x - 10 factors completely over the rationals as (3x - 2)(2x + 5).

Verification

Expanding (3x - 2)(2x + 5) gives back 6x^2 + 11x - 10.

Detailed explanation

The order of methods matters

A reliable factoring routine always starts with the greatest common factor. Taking $2$ out of $2x^2 - 18$ leaves $x^2 - 9$ , which is then recognisable as a difference of squares. Skipping that step hides the pattern and leaves an incomplete answer such as $(2x - 6)(x + 3)$ , which still contains a common factor of 2.

After the common factor, a good routine looks at the shape. Two terms suggest a difference of squares or a sum or difference of cubes; three terms suggest a perfect square or the AC method; four terms often factor by grouping. For degree three and higher the calculator tests rational root candidates and divides out each root it finds with synthetic division, which lowers the degree by one each time and gives the same result grouping would, for example $x^3 + 2x^2 + 3x + 6 = (x + 2)(x^2 + 3)$ .

Why the AC method works

If $ax^2 + bx + c = (px + q)(rx + s)$ , then expanding gives $pr = a$ , $qs = c$ and $ps + qr = b$ . The two numbers $ps$ and $qr$ have product $pq rs = ac$ and sum $b$ . So finding two numbers with product $ac$ and sum $b$ is exactly finding the cross terms, and grouping recovers the binomials.

When no integer pair exists, the quadratic has no rational roots and is irreducible over the rationals. You can confirm this with the discriminant $b^2 - 4ac$ : rational factors exist precisely when it is a perfect square.

Higher-degree polynomials

For a cubic such as $x^3 - 6x^2 + 11x - 6$ , the rational root theorem limits candidates to $\pm1, \pm2, \pm3, \pm6$ . Testing shows $x = 1$ is a root, synthetic division leaves $x^2 - 5x + 6$ , and that quadratic factors as $(x - 2)(x - 3)$ .

Quartics without rational roots can still split into two quadratics, as in $x^4 + 4 = (x^2 - 2x + 2)(x^2 + 2x + 2)$ . The calculator searches integer quadratic pairs for this case before declaring a quartic irreducible.

Common mistakes

Forgetting to take out the greatest common factor first, which leaves a factorization that is not complete.
Treating $a^2 + b^2$ as factorable over the reals. A sum of squares does not factor without complex numbers.
Sign errors in cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ , with a plus sign on the middle term of the quadratic.
Stopping after one step, for instance leaving $x^4 - 16$ as $(x^2 - 4)(x^2 + 4)$ when $x^2 - 4$ factors again.
Dropping a factor of $-1$ when taking out a negative leading coefficient.

Tips

Always check by multiplying the factors back together.
Write the polynomial in descending powers before choosing a method.
For the AC method, list factor pairs of $ac$ systematically from smallest to largest.
Use the discriminant to decide quickly whether a quadratic factors over the rationals.
If you need the roots rather than the factors, set each factor equal to zero.

Frequently asked questions

What is the first step in factoring any polynomial?

Take out the greatest common factor of all the terms, including any common power of the variable. Every other method works better on what remains.

How do I factor a trinomial when the leading coefficient is not 1?

Use the AC method: multiply a by c, find two numbers with that product whose sum is b, split the middle term into those two parts and factor by grouping.

What does irreducible mean?

A polynomial is irreducible over the rationals when it cannot be written as a product of lower-degree polynomials with rational coefficients. For example x squared plus 1 and x squared minus 2 are irreducible over the rationals.

Can the calculator factor cubic and quartic polynomials?

Yes. It tests rational roots with the rational root theorem, divides them out with synthetic division, and for quartics also searches for a split into two quadratics with integer coefficients.

Does it factor expressions with two variables?

It handles common two-variable forms: a shared monomial factor, homogeneous polynomials such as x squared minus 5xy plus 6y squared, differences of squares and grouping by one variable. Expressions with three or more variables are not supported.

How is factoring related to solving equations?

If a product equals zero, one of its factors must be zero. Factoring a polynomial set equal to zero therefore turns one hard equation into several easy ones.

Why is my answer written differently from the textbook?

Factors can appear in any order, and a factor of minus one can be moved between brackets. Multiply both versions out; if they give the same polynomial, both are correct.

Factor your polynomial now

Enter any polynomial above to see its complete factorization and the reasoning behind each step.

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