How to Factor Polynomials: Six Methods, Step by Step

Factoring polynomials is the bridge between algebra and almost everything that follows — solving equations, simplifying rational expressions, integrating in calculus. This guide goes through the six standard techniques in order, so when you see a polynomial you have a checklist instead of a guess.

The decision tree

For any polynomial, ask in this order:

1. Common factor? Pull it out first.
2. Two terms → difference of squares / cubes.
3. Three terms → perfect square or integer-pair search.
4. Four terms → grouping.
5. High-degree → rational root test, then synthetic division.

Following this order saves time and prevents missed factorisations.

Method 1: Greatest common factor (GCF)

Always pull out the GCF first. It simplifies everything else.

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

• GCF of $6, 9, -15$ is $3$. GCF of $x^3, x^2, x$ is $x$.
• Combined GCF: $3x$.
• $6x^3 + 9x^2 - 15x = 3x(2x^2 + 3x - 5)$.
• Now factor the inner quadratic: find numbers multiplying to $(2)(-5) = -10$ and adding to $3$. Try $5$ and $-2$: ✓.
• Final: $3x(2x + 5)(x - 1)$.

Method 2: Difference of squares

If you see $a^2 - b^2$, immediately apply

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

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

Watch for hidden squares: $4x^2 - 25 = (2x)^2 - 5^2 = (2x - 5)(2x + 5)$.

Method 3: Sum and difference of cubes

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Example: $x^3 - 27 = x^3 - 3^3 = (x - 3)(x^2 + 3x + 9)$.

The middle term in the trinomial factor often confuses students — it has the opposite sign from the original cubes' sign, then a positive last term.

Method 4: Perfect square trinomial

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

Example: $x^2 + 6x + 9 = (x + 3)^2$ — recognise because $9 = 3^2$ and $6 = 2 \cdot 3$.

This pattern shows up everywhere in calculus (completing the square, Gaussian integrals).

Method 5: Integer-pair search for $x^2 + bx + c$

Find two numbers that multiply to $c$ and add to $b$.

Example: Factor $x^2 + 7x + 12$.

• Pairs of $12$: $(1,12), (2,6), (3,4)$. Pair $(3, 4)$ adds to $7$. ✓
• Result: $(x + 3)(x + 4)$.

For $ax^2 + bx + c$ with $a \neq 1$, use the AC method: find pair multiplying to $ac$ and adding to $b$, split the middle term, factor by grouping.

Method 6: Factoring by grouping

Used when you have four terms. Group in pairs, factor each pair, hope for a common binomial.

Example: Factor $x^3 + 2x^2 + 3x + 6$.

• Group: $(x^3 + 2x^2) + (3x + 6) = x^2(x + 2) + 3(x + 2)$.
• Common factor $(x + 2)$: $(x + 2)(x^2 + 3)$.

Grouping also handles trinomials when the AC method requires splitting the middle term.

Method 7 (advanced): Rational root theorem

For higher-degree polynomials with integer coefficients, the rational root theorem says any rational root $p/q$ has $p$ dividing the constant term and $q$ dividing the leading coefficient. Test those candidates with synthetic division — once you find one root $r$, $(x - r)$ is a factor and you can reduce the polynomial's degree.

Example: Factor $x^3 - 2x^2 - x + 2$.

• Possible rational roots: $\pm 1, \pm 2$.
• Test $x = 1$: $1 - 2 - 1 + 2 = 0$. ✓ So $(x - 1)$ is a factor.
• Synthetic division gives $x^2 - x - 2$, which factors as $(x - 2)(x + 1)$.
• Final: $(x - 1)(x - 2)(x + 1)$.

Common mistakes

• Forgetting to pull out the GCF first — leads to ugly factoring and missed simplification.
• Sign errors in difference of squares — $a^2 - b^2 \neq (a - b)^2$. Many students accidentally write the perfect-square form.
• Trying to factor primes. Not every quadratic factors over the integers. $x^2 + 1$ has no real factorisation. Switch to the quadratic formula or accept "irreducible."
• Stopping after one pass. Always check whether each factor can be factored further (especially after pulling out a GCF — the inner expression often factors again).

Practice with our solver

Drop any polynomial into the free Factoring Calculator and we'll show every step, including which method we tried and why. Pair it with the Quadratic Solver when factoring fails for second-degree.

For specific worked examples:

• Factor x² + 7x + 12
• Factor x² - 16
• Solve x² + 5x + 6 = 0 (factoring + zero product property)