Algebra

How to Factor Polynomials: A Step-by-Step Method Checklist

Q: How can I tell if a quadratic will factor nicely?

Compute b^2 - 4ac. With integer coefficients, the quadratic factors over the integers exactly when this number is a perfect square such as 0, 1, 4, 9 or 16.

By MathSolver Pro Editorial Team · Published October 25, 2026

Factoring is multiplication in reverse. Work through a fixed checklist, starting with the greatest common factor, and almost every polynomial you meet in algebra can be broken into simple factors.

Why factoring matters

To factor a polynomial is to write it as a product of simpler polynomials. Expanding $(x + 3)(x + 4)$ gives $x^2 + 7x + 12$ ; factoring goes the other way. The factored form is often far more useful: it shows the roots of an equation immediately, simplifies rational expressions by cancelling common factors, and reveals where a graph crosses the x-axis.

Every factorization can be checked by multiplying the factors back out. If the expansion does not reproduce the original polynomial exactly, something is wrong, so make that check a habit.

x^2 + 7x + 12 = (x + 3)(x + 4)

The factoring checklist

Teachers and textbooks order the techniques in slightly different ways, but the following checklist works for nearly every polynomial in a first algebra course. Apply it in order and repeat on each factor until nothing more factors.

Take out the greatest common factor (GCF) of all terms.
Count the terms. Two terms: look for a difference of squares or a sum or difference of cubes.
Three terms: look for a perfect square trinomial, then use the product-sum search or the AC method.
Four terms: try grouping.
Degree three or more with no obvious pattern: use the rational root test and polynomial division.
Check each factor again, then multiply out to verify.

Step 1: the greatest common factor

Always begin here. Find the largest number that divides every coefficient and the lowest power of each variable that appears in every term.

Example: $6x^3 + 9x^2 - 15x$

The GCF of 6, 9 and 15 is 3, and every term contains at least $x^1$ , so the GCF is $3x$ .
Divide each term by $3x$ : $6x^3 + 9x^2 - 15x = 3x(2x^2 + 3x - 5)$ .
The bracket is a trinomial, so continue with the checklist: $2x^2 + 3x - 5 = (2x + 5)(x - 1)$ .
Final answer: $3x(2x + 5)(x - 1)$ .

If the leading coefficient is negative, factor out a negative GCF. $-x^2 + 4x - 3 = -(x^2 - 4x + 3) = -(x - 1)(x - 3)$ is much easier than working with a negative $x^2$ term.

Trinomials with leading coefficient 1

For $x^2 + bx + c$ , find two numbers $m$ and $n$ with $mn = c$ and $m + n = b$ . Then $x^2 + bx + c = (x + m)(x + n)$ .

$x^2 + 7x + 12$ : 3 and 4 multiply to 12 and add to 7, so the factors are $(x + 3)(x + 4)$ .
$x^2 - 2x - 15$ : $-5$ and 3 multiply to $-15$ and add to $-2$ , so the factors are $(x - 5)(x + 3)$ .
$x^2 - 9x + 20$ : $-4$ and $-5$ , giving $(x - 4)(x - 5)$ .

Signs give a shortcut. If $c$ is positive, both numbers share the sign of $b$ . If $c$ is negative, the numbers have opposite signs and the larger one in size takes the sign of $b$ .

Trinomials with any leading coefficient: the AC method

When $a \neq 1$ , multiply $a$ by $c$ , find two numbers with product $ac$ and sum $b$ , use them to split the middle term, then factor by grouping.

Example: $3x^2 - 10x + 8$

$ac = 3 \times 8 = 24$ . Two numbers with product 24 and sum $-10$ are $-4$ and $-6$ .
Split the middle term: $3x^2 - 4x - 6x + 8$ .
Group in pairs: $x(3x - 4) - 2(3x - 4)$ .
Factor out the common bracket: $(x - 2)(3x - 4)$ .
Check: $(x - 2)(3x - 4) = 3x^2 - 4x - 6x + 8 = 3x^2 - 10x + 8$ .

A trinomial with integer coefficients factors over the integers exactly when $b^2 - 4ac$ is a perfect square. For $3x^2 - 10x + 8$ the discriminant is $100 - 96 = 4$ , a perfect square, so the search must succeed. If the discriminant is not a perfect square, stop searching and use the quadratic formula instead.

Special products to recognise

Three patterns appear so often that it pays to know them on sight:

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

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

a^3 - b^3 = (a - b)(a^2 + ab + b^2), \qquad a^3 + b^3 = (a + b)(a^2 - ab + b^2)

$9x^2 - 25 = (3x)^2 - 5^2 = (3x - 5)(3x + 5)$ .
$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ : apply the pattern twice.
$4x^2 + 12x + 9 = (2x + 3)^2$ , because $4x^2 = (2x)^2$ , $9 = 3^2$ and $12x = 2(2x)(3)$ .
$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ and $27x^3 + 1 = (3x + 1)(9x^2 - 3x + 1)$ .

A sum of two squares such as $x^2 + 4$ does not factor over the real numbers. Its discriminant is $-16$ , which is negative.

Four terms: factoring by grouping

Split the polynomial into two pairs, take the GCF of each pair, and look for a common bracket.

Example: $x^3 + 3x^2 - 4x - 12$

Group: $(x^3 + 3x^2) + (-4x - 12)$ .
Factor each pair: $x^2(x + 3) - 4(x + 3)$ .
Take out the common bracket: $(x + 3)(x^2 - 4)$ .
The second factor is a difference of squares: $(x + 3)(x - 2)(x + 2)$ .

If the brackets do not match, try rearranging the terms into different pairs before giving up.

Higher degree: the rational root test

For a polynomial with integer coefficients, any rational root $\frac{p}{q}$ in lowest terms has $p$ dividing the constant term and $q$ dividing the leading coefficient. Test candidates until one gives zero, then divide it out.

Example: $2x^3 - 3x^2 - 11x + 6$

Candidates: $\pm 1, \pm 2, \pm 3, \pm 6, \pm\frac{1}{2}, \pm\frac{3}{2}$ .
Test $x = 3$ : $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ , so $(x - 3)$ is a factor.
Divide (synthetic division with 3): the quotient is $2x^2 + 3x - 2$ .
Factor the quotient by the AC method: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ .
Final answer: $(x - 3)(2x - 1)(x + 2)$ , with roots $3$ , $\frac{1}{2}$ and $-2$ .

The polynomial calculator performs the division step by step if you want to compare.

Common factoring mistakes

Forgetting the GCF, which leaves a harder trinomial and an incomplete answer.
Stopping too early: $x^4 - 16 = (x^2 - 4)(x^2 + 4)$ is not finished, because $x^2 - 4$ factors again.
Treating $a^2 + b^2$ like a difference of squares.
Sign slips in grouping: $-4x - 12 = -4(x + 3)$ , not $-4(x - 3)$ .
Not checking by expansion. Multiplying out takes seconds and catches every one of the errors above.

Try your own polynomials in the factoring calculator, which shows which method it used at each step, and use factored forms to simplify rational expressions.

How to tell when a polynomial will not factor

Not every polynomial factors over the integers, and recognising that early saves a lot of time. For a quadratic $ax^2 + bx + c$ with integer coefficients, compute the discriminant $b^2 - 4ac$ . If it is a perfect square, the quadratic factors into linear factors with rational coefficients. If it is positive but not a perfect square, the roots are irrational and the quadratic is called irreducible over the rationals; if it is negative, it does not factor over the real numbers at all.

For example, $x^2 + 3x + 1$ has discriminant $9 - 4 = 5$ , which is not a perfect square, so no pair of integers multiplies to 1 and adds to 3. Its roots, $\frac{-3 \pm \sqrt{5}}{2}$ , come from the quadratic formula. The sum of squares $x^2 + 4$ has discriminant $-16$ and does not factor over the reals, although it does factor as $(x - 2i)(x + 2i)$ over the complex numbers.

For cubics and higher degrees, the rational root test gives a finite list of candidates. If none of them is a root, the polynomial has no linear factor with rational coefficients, though a quartic could still split into two quadratics.

Why factoring matters later

Factoring is not an isolated topic. Solving polynomial equations relies on the zero product property: if $AB = 0$ , then $A = 0$ or $B = 0$ . Simplifying rational expressions depends on cancelling common factors. Finding the x-intercepts and sign behaviour of a polynomial graph uses the factored form directly, and in calculus, limits of the form $\frac{0}{0}$ are often resolved by factoring and cancelling.

As an example, $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ cannot be evaluated by substitution, but factoring the numerator as $(x - 2)(x + 2)$ and cancelling leaves $x + 2$ , so the limit is 4. Partial fractions in integration also start by factoring the denominator completely.

Always check a factorisation by expanding it. Multiplying out takes less time than factoring and catches sign errors immediately.

To check a factorisation of your own, enter it into the factoring calculator or expand your answer with the simplify calculator.

Frequently asked questions

What should I always do first when factoring?

Take out the greatest common factor. It makes every later step easier and is required for a complete factorization.

How can I tell if a quadratic will factor nicely?

Compute $b^2 - 4ac$ . With integer coefficients, the quadratic factors over the integers exactly when this number is a perfect square such as 0, 1, 4, 9 or 16.

Does $x^2 + 9$ factor?

Not over the real numbers. A sum of squares has no real roots. Over the complex numbers it factors as $(x - 3i)(x + 3i)$ .

How do I know when I have factored completely?

Check every factor against the checklist again. When no factor has a common factor, a special-product pattern, or a rational root, the factorization is complete.

What is the first thing to do when factoring?

Always look for a greatest common factor first, including a negative sign if the leading coefficient is negative. Taking it out makes the remaining polynomial smaller and often reveals a familiar pattern such as a difference of squares or a simple trinomial.

Is x squared plus 9 factorable?

Not over the real numbers. A sum of two squares has a negative discriminant, so it has no real roots and no real linear factors. Over the complex numbers it factors as $(x - 3i)(x + 3i)$ .

MathSolver Pro Editorial Team

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Our editorial team writes every guide to explain the reasoning behind each method, not just the answer. Each worked example is checked against the site's own solver engines and automated tests before it is published, and articles are revised whenever a method or an exam format changes.