Polynomial Factorization

Multiplying (x - 2)(x - 3) gives x^2 - 5x + 6. Factoring goes in the opposite direction: starting from x^2 - 5x + 6, you discover that it splits into (x - 2)(x - 3).

Why bother going backwards? Because factored form reveals structure that expanded form hides. The expression x^2 - 5x + 6 does not obviously tell you where it equals zero. The form (x - 2)(x - 3) shouts it: x = 2 and x = 3. Factoring is how you solve polynomial equations, simplify rational expressions, and see the skeleton of a complicated expression. It is the single most useful transformation in school algebra.

This article covers the five standard techniques, in the order you should try them. Each method has a characteristic "shape" that you learn to recognise on sight.

Method 1: Common factor

Always look for this first. If every term of the polynomial shares a factor, pull it out.

Take 6x^3 + 9x^2 - 3x. Every term is divisible by 3x:

The factored form is 3x times a quadratic. You have reduced the problem from a cubic to a quadratic — a simpler object.

The common factor can be a number, a variable, or both. It can also be a whole binomial. For instance, (x + 1)^2(x - 3) + (x + 1)(x - 3)^2 has a common factor of (x + 1)(x - 3):

Always pull out the common factor before trying anything else. It makes whatever remains smaller and easier to handle.

Method 2: Grouping

When there is no common factor across all terms, but you can split the polynomial into groups that each have a common factor, grouping works.

Take x^3 + 3x^2 + 2x + 6. No single factor divides all four terms. But group the first two and the last two:

Factor each group:

Now both groups share the factor (x + 3):

Grouping works when the polynomial was created by multiplying two factors and then expanding — the groups are the ghost of the original structure. The technique is especially common for four-term polynomials, where a 2 + 2 split usually reveals the pattern.

Sometimes you need to rearrange terms before the grouping becomes visible. If the natural left-to-right split does not produce a common binomial, try a different partition.

Method 3: Factoring quadratic trinomials

A quadratic trinomial has the form ax^2 + bx + c. When a = 1, factoring is a search for two numbers that add to b and multiply to c.

Take x^2 + 7x + 12. You need two numbers whose sum is 7 and whose product is 12. Testing: 3 + 4 = 7 and 3 \times 4 = 12. So:

Check by expanding: (x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12. Correct.

When a \neq 1, the method extends. For 6x^2 + 11x - 10, the product a \times c = 6 \times (-10) = -60. You need two numbers that multiply to -60 and add to 11. Those are 15 and -4 (since 15 \times (-4) = -60 and 15 + (-4) = 11). Now split the middle term:

Group and factor:

This "splitting the middle term" approach is the workhorse of quadratic factoring. It reduces every ax^2 + bx + c to grouping.

Method 4: Factoring using identities

Several algebraic identities from Algebraic Identities give instant factorisations. The most commonly used ones are:

• Difference of squares. a^2 - b^2 = (a + b)(a - b)

  Example: x^2 - 49 = (x + 7)(x - 7).

• Perfect square trinomial. a^2 + 2ab + b^2 = (a + b)^2

  Example: x^2 + 10x + 25 = (x + 5)^2, since 10x = 2 \cdot x \cdot 5 and 25 = 5^2.

• Perfect square trinomial (minus). a^2 - 2ab + b^2 = (a - b)^2

  Example: 4x^2 - 12x + 9 = (2x - 3)^2, since 4x^2 = (2x)^2, 9 = 3^2, and 12x = 2 \cdot (2x) \cdot 3.

Recognising these patterns is a matter of practice. The difference of squares is the easiest to spot — two perfect squares separated by a minus sign. Perfect square trinomials are trickier: check that the first and last terms are perfect squares, and that the middle term is exactly 2ab.

Method 5: Factoring cubics — sum and difference of cubes

Two identities handle cubes specifically:

Take 8x^3 - 27. Recognise: 8x^3 = (2x)^3 and 27 = 3^3. This is a difference of cubes with a = 2x and b = 3:

Check: the quadratic factor 4x^2 + 6x + 9 has discriminant 36 - 144 = -108 < 0, so it does not factor further over the reals. The factorisation is complete.

For x^3 + 125 = x^3 + 5^3:

Again, the quadratic has discriminant 25 - 100 = -75 < 0, so it stays as is.

A useful mnemonic for keeping the signs straight: in the linear factor, the sign matches the original (a^3 + b^3 gives a + b; a^3 - b^3 gives a - b). In the quadratic factor, the signs go opposite, positive: in a^3 + b^3, the quadratic is a^2 - ab + b^2 (the ab term has the opposite sign); the last term is always +b^2.

The strategy: which method to use

Faced with a polynomial to factor, apply the methods in this order:

1. Common factor first. Always. Even if the polynomial needs further factoring, pulling out a common factor makes everything smaller.
2. Count the terms. Two terms? Look for difference of squares or sum/difference of cubes. Three terms? Look for a quadratic trinomial or a perfect square pattern. Four or more terms? Try grouping.
3. Check the discriminant. For a quadratic ax^2 + bx + c, compute b^2 - 4ac. If it is a perfect square, the trinomial factors over the integers. If it is negative, the quadratic is irreducible over the reals — you are done.
4. Use the Factor Theorem for higher-degree polynomials. If the polynomial is cubic or higher and none of the pattern-based methods work, try plugging in small integers (0, \pm 1, \pm 2, \pm 3, \dots). If p(a) = 0, then (x - a) is a factor, and you can use long division to find the quotient.

This hierarchy covers every factoring problem in the standard curriculum. The more you practise, the faster you recognise which shape you are looking at.

Drag the red point along the quadratic below to see where p(x) = x^2 - 2x - 8 crosses the x-axis. Each crossing is a root, and the Factor Theorem guarantees a factor (x - a) at each one.

Two worked examples

Common confusions

• "x^2 + 9 = (x + 3)(x - 3)." No — that identity is x^2 - 9 = (x + 3)(x - 3). The difference of squares factors; the sum of squares does not factor over the reals. x^2 + 9 is irreducible.

• "To factor x^2 + 7x + 12, I need two numbers that multiply to 7 and add to 12." The conditions are swapped. You need two numbers that add to 7 (the coefficient of x) and multiply to 12 (the constant term). Mixing them up gives the wrong factorisation.

• "x^3 + 8 = (x + 2)^3." No. (x + 2)^3 = x^3 + 6x^2 + 12x + 8, which is a very different polynomial. The correct factorisation of x^3 + 8 is (x + 2)(x^2 - 2x + 4), using the sum of cubes identity.

• "After pulling out a common factor, the factorisation is done." Often it is not. The expression that remains after extracting the common factor may itself factor further — as in Example 1, where 2x(x^2 - 4x - 5) had a quadratic that split into (x - 5)(x + 1). Always check what remains.

• "If the discriminant of a quadratic is negative, I made a mistake." Not necessarily. A negative discriminant means the quadratic has no real roots and is irreducible over \mathbb{R}. This is a valid endpoint — it means the factorisation is complete, not that something went wrong. The quadratic factors from both cube identities have negative discriminants.

• "Grouping only works with four terms." It works with any number of terms, as long as you can partition them into groups that share a common factor. Six-term polynomials can sometimes be grouped 3 + 3 or 2 + 2 + 2. The four-term case is just the most common.

Going deeper

Irreducibility over the reals vs. over the rationals

When you say a polynomial "cannot be factored further," you need to specify where the coefficients are allowed to live. The quadratic x^2 - 2 does not factor over the rationals (there are no rational numbers whose product is -2 and whose sum is 0). But over the reals, it factors as (x - \sqrt{2})(x + \sqrt{2}).

In school algebra, "factor completely" almost always means "over the rationals" or "over the integers" — whichever gives factors with clean coefficients. The cube-identity quadratics like x^2 + 2x + 4 are irreducible over the reals (negative discriminant), so they are irreducible over the rationals too. But x^2 - 2 is reducible over \mathbb{R} and irreducible over \mathbb{Q}, which is a finer distinction.

The Rational Root Theorem narrows the search

When using the Factor Theorem to find roots of a polynomial with integer coefficients, the Rational Root Theorem limits the candidates. If \dfrac{p}{q} (in lowest terms) is a root of a_n x^n + \cdots + a_1 x + a_0, then p divides a_0 and q divides a_n. So you only need to test a finite list of fractions, not every real number.

For example, to factor 2x^3 + x^2 - 5x + 2, the possible rational roots are \pm \{1, 2, \tfrac{1}{2}\}. Testing: p(1) = 2 + 1 - 5 + 2 = 0. So (x - 1) is a factor. Dividing gives 2x^2 + 3x - 2 = (2x - 1)(x + 2), and the full factorisation is (x - 1)(2x - 1)(x + 2).

Factoring over the complex numbers

Over the complex numbers, every polynomial of degree n factors into exactly n linear factors — this is the Fundamental Theorem of Algebra. The "irreducible" quadratics you met in the cube identities are only irreducible over \mathbb{R}. Over \mathbb{C}, x^2 + 2x + 4 factors as (x - (-1 + i\sqrt{3}))(x - (-1 - i\sqrt{3})), where i = \sqrt{-1}. Every polynomial shatters completely into linear pieces once you allow complex coefficients.

Where this leads next

Factoring is a bridge between raw algebra and the deeper structure of polynomial equations.

• Polynomial Operations — the mechanics of addition, subtraction, multiplication, and division that produce the expressions you factor here.
• Algebraic Identities — the identities (a + b)^2, a^2 - b^2, a^3 \pm b^3 that serve as instant-factoring patterns.
• Quadratic Equations — Introduction — where factoring meets equation-solving: set the factored polynomial to zero and read off the roots.
• Polynomials — Introduction — the definitions and vocabulary that underlie everything in this article.
• Roots and Radicals — square and cube roots appear inside the discriminant and in the factors themselves when the roots are irrational.