Every factoring technique
Every way to turn a sum into a product, in the order you should try them.
Work top to bottom. Pull the GCF first — always, no exceptions. Then count terms: two terms means look for a difference pattern, three means a trinomial method (or a perfect-square check), four means grouping. If nothing above cracks it and the degree is 3 or higher, switch tools entirely and go hunting for zeros instead.
Always start here
Factor this out first, every single time — it makes everything after it easier.
Two or three terms — pattern recognition
Four patterns worth recognizing on sight, each running a multiplication rule in reverse:
Difference of squares: a^2 - b^2 = (a + b)(a - b)
Perfect-square trinomial (both signs): the special products from earlier in this chapter, run in reverse: a^2 + 2ab + b^2 = (a+b)^2, \qquad a^2 - 2ab + b^2 = (a-b)^2
Sum of cubes: a^3 + b^3 = (a+b)(a^2 - ab + b^2)
Difference of cubes: a^3 - b^3 = (a-b)(a^2+ab+b^2)
In both cube patterns, the binomial factor keeps the same sign as the original expression; the trinomial factor always keeps the opposite sign in its middle term, and its last term is always positive.
Difference of squares, perfect-square trinomial, sum/difference of cubes.
Three products recur so often they deserve memorizing:
(a + b)^2 = a^2 + 2ab + b^2(a - b)^2 = a^2 - 2ab + b^2(a + b)(a - b) = a^2 - b^2The multiplication rules these patterns run backwards — recognize the shape by knowing the reverse.
General trinomials and four-term expressions
Split the middle term (AC method) or pair up four terms directly.
Degree 3 and higher — when patterns run out
If f(x) = a_n x^n + \cdots + a_1 x + a_0 has integer coefficients and \frac{p}{q} (in lowest terms) is a rational zero, then p divides a_0 and q divides a_n.
Thus every rational zero appears in the finite list \frac{p}{q} : \quad p \mid a_0, \quad q \mid a_n. The theorem does not promise any of the candidates is a zero. It only guarantees that no rational zero hides outside the list.
Generates the candidate roots to test.
Show proof
Substitute x = \frac{p}{q} into f(x) = 0: a_n \left(\frac{p}{q}\right)^n + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + \cdots + a_1\left(\frac{p}{q}\right) + a_0 = 0. Multiply every term by q^n to clear denominators: a_np^n + a_{n-1}p^{n-1}q + \cdots + a_1pq^{n-1} + a_0q^n = 0.
Isolate the constant term: a_0q^n = -\left(a_np^n + a_{n-1}p^{n-1}q + \cdots + a_1pq^{n-1}\right). Every term on the right has a factor of p, so p divides a_0q^n. Because \frac{p}{q} is in lowest terms, p and q share no prime factor, so p shares no prime factor with q^n either — meaning p must divide a_0 on its own.
The same move on the leading term gives a_np^n = -\left(a_{n-1}p^{n-1}q + \cdots + a_0q^n\right), and every term on the right now has a factor of q, so q divides a_np^n. Again q shares no factor with p^n, so q must divide a_n.
x - c is a factor of f(x) if and only if f(c) = 0. Zeros of f and linear factors of f are the same information.
A confirmed root r certifies (x − r) as a factor.
Show proof
By the Remainder Theorem, dividing f(x) by x - c leaves remainder f(c). If f(c) = 0, the remainder is 0, so f(x) = (x-c)q(x) exactly — x - c divides evenly, i.e. it's a factor. Conversely, if x - c is a factor, f(x) = (x-c)q(x) for some q(x), so f(c) = (c-c)q(c) = 0.
Every degree-n polynomial (n \ge 1) factors over \mathbb{C} as f(x) = a_n(x - c_1)(x - c_2)\cdots(x - c_n), so f has exactly n complex zeros, counted with multiplicity.
Every degree-n polynomial factors into exactly n linear factors, counted with multiplicity.
Show proof
Apply the Fundamental Theorem of Algebra to find one zero c_1; by the Factor Theorem, f(x) = (x - c_1)q_1(x) for a degree-(n-1) quotient q_1. Apply the Fundamental Theorem again to q_1 to peel off c_2, and repeat. After n steps the quotient has degree 0 — a constant, which must be the leading coefficient a_n — leaving the full factorization f(x) = a_n(x-c_1)\cdots(x-c_n).
If f has real coefficients and a + bi (b \neq 0) is a zero, then its conjugate a - bi is also a zero. Hence nonreal zeros of real polynomials come in pairs, and every real polynomial of odd degree has at least one real zero.
Nonreal factors always arrive as a matched pair.
Show proof
Conjugation respects addition and multiplication: for z = a+bi and w = c+di, direct expansion gives \overline{z+w} = \overline{z} + \overline{w} and \overline{zw} = \overline{z}\,\overline{w} (each side works out to the same real and imaginary parts). Since any polynomial expression is built from nothing but additions and multiplications, it follows that \overline{f(z)} = a_n\overline{z}^{\,n} + \cdots + a_1\overline{z} + a_0 provided every coefficient a_k is real — because \overline{a_k} = a_k exactly when a_k is real, which is precisely where the hypothesis gets spent.
Now suppose f(a+bi) = 0 for a real-coefficient f. Taking the conjugate of both sides: \overline{f(a+bi)} = f(\overline{a+bi}) = f(a - bi), and \overline{0} = 0, so f(a-bi) = 0 as well — the conjugate is also a zero.
Every nonreal zero therefore drags its conjugate along as a second zero, so nonreal zeros come in pairs. By the Linear Factorization Theorem, a degree-n polynomial has exactly n zeros counted with multiplicity. If n is odd, the nonreal zeros can only ever be removed two at a time — leaving an odd number unaccounted for, and those leftover zeros can't be nonreal (they'd need an unpaired partner), so at least one must be real.
How to actually divide out a confirmed factor.