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Polynomial factoring

By Daniel Whitman · Last updated April 2026 · 11 min read

Factoring a polynomial is the inverse of expanding it. Where expansion takes $(x + 2)(x + 3)$ and turns it into $x^2 + 5x + 6$, factoring goes the other way — you are given $x^2 + 5x + 6$ and asked to recover the brackets. It is the more useful direction. A factored polynomial tells you, at a glance, where it equals zero, where it changes sign, and how to simplify any algebraic fraction it sits inside.

There are four patterns that cover almost every factoring problem you will see at school: pulling out a common factor, the difference of two squares, sum/product on a quadratic, and grouping for four-term polynomials. The trick is recognising which pattern applies, in what order, and stopping when you are done.

Always check for a common factor first

Before any other technique, look for the greatest common factor (GCF) of every term in the polynomial. If one exists, pull it out:

$$6x^2 + 9x = 3x(2x + 3).$$

Sometimes the GCF is the entire job — the bracket that remains is already irreducible. More often, pulling out the GCF leaves a simpler polynomial that one of the next three patterns can finish off.

Pattern 1 — difference of two squares

Any polynomial of the form $a^2 - b^2$ factors immediately as

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

Examples: $x^2 - 9 = (x + 3)(x - 3)$; $4x^2 - 25 = (2x + 5)(2x - 5)$. The pattern only works when the operation in the middle is subtraction — $a^2 + b^2$ does not factor over the reals.

Pattern 2 — sum/product on a monic quadratic

For $x^2 + bx + c$ (note the leading coefficient is $1$), look for two numbers whose sum is $b$ and whose product is $c$. If you find them — call them $p$ and $q$ — then

$$x^2 + bx + c = (x + p)(x + q).$$

For $x^2 + 5x + 6$, the pair $(2, 3)$ works: $2 + 3 = 5$, $2 \cdot 3 = 6$. So $x^2 + 5x + 6 = (x + 2)(x + 3)$.

For $x^2 - 7x + 12$, the pair $(-3, -4)$ works: their sum is $-7$ and product is $12$. So $x^2 - 7x + 12 = (x - 3)(x - 4)$. The signs of $p$ and $q$ are determined by the signs of $b$ and $c$:

Pattern 3 — non-monic quadratic ($ax^2 + bx + c$, $a \ne 1$)

For something like $2x^2 + 7x + 3$, multiply $a$ and $c$ together: here $2 \cdot 3 = 6$. Now look for two numbers whose product is $6$ and whose sum is $b = 7$. The pair $(1, 6)$ works. Split the middle term using those numbers: $2x^2 + x + 6x + 3$. Now factor in pairs:

$$2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).$$

This is the “ac method.” It always works on quadratics that have rational roots; if no pair of integers exists with product $ac$ and sum $b$, the quadratic does not factor over the rationals and you should either use the quadratic formula or accept that it is irreducible.

Pattern 4 — factoring by grouping

For four-term polynomials, group them in pairs and look for a common bracket:

$$x^3 + 2x^2 + 3x + 6 = x^2(x + 2) + 3(x + 2) = (x^2 + 3)(x + 2).$$

Grouping is the technique you reach for whenever you see a polynomial with four terms, especially after pulling out a GCF. If the two pairs do not produce a matching bracket, try regrouping the terms in a different order — sometimes only one grouping works.

The mistakes I see most often

1. Stopping after one factoring step

$x^4 - 16$ factors as $(x^2 - 4)(x^2 + 4)$, but you are not done — $(x^2 - 4)$ factors further as $(x - 2)(x + 2)$. Always look at each factor and ask whether it itself can be factored.

2. Trying to factor $x^2 + 4$

The sum of two squares does not factor over the real numbers. The pattern $a^2 + b^2 = (a + b)(a - b)$ does not exist. (It does exist over the complex numbers as $(a + bi)(a - bi)$, but that is a different course.)

3. Forgetting the GCF

$2x^2 - 8$ looks like a difference of squares, and it sort of is, but not until you pull out the GCF first: $2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)$. Skipping the GCF step gives an answer that is correct but not fully factored.

4. Sign confusion in sum/product

For $x^2 - 5x + 6$, the pair has product $6$ and sum $-5$. Both factors must be negative, since their product is positive and their sum is negative. The factorisation is $(x - 2)(x - 3)$, not $(x + 2)(x + 3)$.

Where this comes up later

Factoring is the workhorse step in solving polynomial equations, simplifying rational expressions, integrating rational functions in calculus, and finding asymptotes of rational functions. Almost any problem with $x$ in a denominator wants you to factor numerator and denominator first to see what cancels.

References