Methods of factorisation

Obvious roots and Horner’s method

First, we look for an ‘obvious’ root amongst the divisors of the constant term, provided the coefficients are integers (rational root theorem). Example: P(x) = x³ - 6x² + 11x - 6. We test x = 1: P(1) = 1 – 6 + 11 – 6 = 0, so (x – 1) divides P. Horner’s scheme quickly gives the quotient: Q(x) = x² – 5x + 6.

Useful notable identities

Identity	Factored form
a^2 - b^2	(a-b)(a+b)
a^3 - b^3	(a-b)(a² + ab + b²)
a³ + b³	(a + b)(a² - ab + b²)
a² + 2ab + b²	(a + b)²

Complete example

x³ - 6x² + 11x - 6 = (x - 1)(x² - 5x + 6) = (x - 1)(x - 2)(x - 3). The three roots are 1, 2 and 3, each of which is simple.

Irreducible polynomials over R

Over R, every polynomial can be factored into a product of monomials of degree 1 and binomials of degree 2 with a strictly negative discriminant (irreducible over R). Example: x² + 1 is irreducible over R (its roots i and -i are complex) but factors over C into (x - i)(x + i).

Common pitfall

Before concluding that a second-degree factor is irreducible, check its discriminant δ = b² - 4ac. Also, do not forget to test the negative divisors of the constant term, not just the positive ones, when searching for rational roots.