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Factoring Calculator with Steps

Factor a polynomial completely with the pattern named before it is used — greatest common factor, difference of squares, trinomial, or grouping — and every factor checked for whether it breaks down further.

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How to use it

Type the polynomial and say completely if you want every factor broken down as far as it goes, which is what exam questions normally mean. Powers can be written with a caret, so 16x^4-81 parses. Terms do not need to be in descending order. If you want the roots rather than the factors, ask for the equation set to zero instead, or send it to the algebra solver.

The method behind it

Factoring is multiplication read backwards, so every pattern below is just an expansion you already know, written from right to left. There are only four patterns worth memorising, and the order you check them in matters more than the patterns themselves.

Always take the greatest common factor first. It makes every subsequent step smaller, and skipping it is the single most common reason an otherwise correct answer is marked down as incompletely factored.

With two terms, look for squares or cubes. The difference of squares is the pattern that appears most often, and it is worth recognising even inside larger expressions.

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

Cubes factor too, and the signs follow a fixed pattern: the binomial keeps the original sign, the middle term of the trinomial takes the opposite sign, and the last term is always positive.

a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}), a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

With three terms and a leading coefficient of 1, expanding $(x + p) (x + q)$ shows what to look for: the constants multiply to give the constant term and add to give the middle coefficient.

(x + p) (x + q) = x^{2} + (p + q) x + pq

When the leading coefficient is not 1 that shortcut breaks, because the leading coefficient also multiplies into the middle term. The AC method repairs it: find $p$ and $q$ with $pq = a c$ and $p + q = b$ , split the middle term, and group the resulting four terms.

a x^{2} + b x + c = a x^{2} + p x + q x + c where pq = a c, p + q = b

Four terms with no obvious pattern means grouping: pair them, extract a common factor from each pair, and the same bracket should appear twice. The four patterns and how to tell them apart at a glance are covered in factoring quadratics: the four patterns, and rewriting terms with exponent and radical rules often exposes a pattern that was hiding.

Worked examples

Worked example

6 x^{3} - 27 x^{2} + 30 x

1
All three coefficients are divisible by 3 and every term has at least one x. Pull out 3x before touching anything else.
$3 x (2 x^{2} - 9 x + 10)$
2
The bracket is a trinomial with leading coefficient 2, so use the AC method. Multiply a by c to get the target product.
$a c = 2 \cdot 10 = 20, b = - 9$
3
Find two numbers multiplying to 20 and adding to −9. Both must be negative, and −4 and −5 work.
$(- 4) (- 5) = 20, - 4 + (- 5) = - 9$
4
Split the middle term using those two numbers and group the four terms in pairs.
$3 x (2 x^{2} - 4 x - 5 x + 10) = 3 x [2 x (x - 2) - 5 (x - 2)]$
5
The bracket (x − 2) is common to both pairs, so factor it out. Expanding back confirms the original.
$3 x (2 x - 5) (x - 2)$

Answer

6 x^{3} - 27 x^{2} + 30 x = 3 x (2 x - 5) (x - 2)

Worked example

x^{3} + 3 x^{2} - 4 x - 12

1
Four terms with no common factor across all of them, so try grouping. Pair the first two and the last two.
$(x^{3} + 3 x^{2}) + (- 4 x - 12)$
2
Take the common factor out of each pair. Pulling −4 rather than +4 out of the second pair is what makes the brackets match.
$x^{2} (x + 3) - 4 (x + 3)$
3
The same bracket now appears twice, so treat (x + 3) as a common factor.
$(x + 3) (x^{2} - 4)$
4
Do not stop here. The second factor is a difference of squares, so it breaks down further.
$x^{2} - 4 = (x - 2) (x + 2)$
5
Write all three linear factors. The roots are −3, 2, and −2, which you can confirm by substitution.
$(x + 3) (x - 2) (x + 2)$

Answer

x^{3} + 3 x^{2} - 4 x - 12 = (x + 3) (x - 2) (x + 2)

Which pattern is it?

Counting terms narrows the choice to one or two patterns almost immediately, which is why experienced solvers look at the term count before the coefficients. Work down this tree in order and the pattern identifies itself.

Identify the factoring pattern

Do every one of the terms share a common factor?

Common mistakes

Leaving the common factor behind

Factoring $6 x^{3} - 27 x^{2} + 30 x$ as $(6 x^{2} - 15 x) (x - 2)$ is technically a factorisation, but not a complete one, and it will not earn full marks. Extract $3 x$ first and the rest becomes easier as well as correct.

Factoring a sum of squares

$x^{2} + 9$ does not factor over the real numbers. It is not $(x + 3)^{2}$ , which expands to $x^{2} + 6 x + 9$ , and it is not $(x + 3) (x - 3)$ , which is $x^{2} - 9$ . Only the difference of squares factors.

Stopping one step early

After grouping produced $(x + 3) (x^{2} - 4)$ the work was not finished, because the second factor still factors. Before writing an answer, look at every factor once more and ask whether any of them matches a pattern.

Practice

Answers are checked here — nothing is sent anywhere.

1
$Factor completely: 3 x^{2} + 12 x$
2
$Factor completely: 4 x^{2} - 25$
3
$Factor completely: x^{2} - 7 x + 12$

Frequently asked questions

Does it factor over the integers or further than that?

By default over the integers, which is what completely factored means in most courses. Ask for factorisation over the reals or the complex numbers and you get the irrational or complex linear factors instead.

How do I check a factorisation myself?

Two ways. Expand it back and compare every coefficient, or substitute a convenient number such as x = 2 into both the original and the factored form. If the two values differ, the factorisation is wrong.

What about cubics and quartics?

Those need a root first. The rational root theorem lists the candidates as factors of the constant over factors of the leading coefficient; testing them finds one root, and dividing by that linear factor reduces the degree.

What if the polynomial does not factor?

Then it is prime over the integers and the solution says so rather than forcing something. A quadratic is prime over the integers exactly when its discriminant is not a perfect square.

What does it cost?

The first solution is free and needs no account. A free account gives three solutions a day; Gauth Plus removes the limit at $11.99 a month with a three-day free trial.

Keep going

Algebra solverFactoring inside equations, systems, and rational expressions.Factoring quadratics: the four patternsTell GCF, difference of squares, perfect squares, and grouping apart on sight.Exponent and radical rulesRewriting powers so a hidden square or cube pattern becomes visible.Quadratic formula calculatorWhen the discriminant is not a perfect square and no integer factors exist.