Learn first: Special product formulas
Factorising is the reverse of expanding: a polynomial is written as a product of several simpler expressions. The skill is needed to simplify fractions, solve equations by the “product equals zero” method, apply the interval method to inequalities and speed up mental arithmetic. In factorised form the zeros, the sign and the structure of an expression are visible at once — which is why it is one of the most used transformations in algebra. This lesson covers the main methods in the order they are applied.
This is always the first step. It is the distributive law used backwards: ab + ac = a(b + c). Take the HCF of the coefficients and the lowest power of each variable:
6x²y − 9xy² = 3xy(2x − 3y).
Checking is easy: expanding the brackets must give back the original expression.
The common factor can itself be an expression: x(a − b) + y(a − b) = (a − b)(x + y).
(b − a) = −(a − b) — change the sign to create a common bracket
For example, a(x − 1) − b(1 − x) = a(x − 1) + b(x − 1) = (x − 1)(a + b).
Group the terms, take a common factor out of each group, and then the same bracket appears in every group:
ax + ay − bx − by = a(x + y) − b(x + y) = (a − b)(x + y).
If no common bracket appears at first, group differently — for example, terms with x and without x, or terms with proportional coefficients.
Factorise x³ − x² − 4x + 4.
a² − b² = (a − b)(a + b); a² ± 2ab + b² = (a ± b)²
a³ − b³ = (a − b)(a² + ab + b²); a³ + b³ = (a + b)(a² − ab + b²)
Examples: x² − 9 = (x − 3)(x + 3); 4x² − 12x + 9 = (2x − 3)²; x³ − 8 = (x − 2)(x² + 2x + 4).
To recognise a formula, see the terms as “squares” or “cubes”: 4x² = (2x)², 27 = 3³. In a perfect square the middle term must be exactly 2ab.
Sometimes a perfect square is created by adding and subtracting a term:
x⁴ + 4 = (x² + 2)² − 4x² = (x² − 2x + 2)(x² + 2x + 2)
The formulas also help in mental arithmetic: 101² − 99² = (101 − 99)(101 + 99) = 2·200 = 400.
Factorise a² − b² − 2b − 1.
ax² + bx + c = a(x − x₁)(x − x₂), where x₁, x₂ are the roots of ax² + bx + c = 0
If D = 0, ax² + bx + c = a(x − x₀)²; if D < 0, the quadratic has no real factors.
Quick method via Vieta (a = 1): find numbers with sum −b and product c: x² − 7x + 12 → 3 and 4 → (x − 3)(x − 4).
Factorise 2x² + 5x − 3.
Integer roots of a polynomial with integer coefficients are among the divisors of the constant term. If x₀ is a root, the polynomial is divisible by x − x₀ (the factor theorem).
P(x₀) = 0 ⇔ P(x) = (x − x₀)·Q(x)
For example, x³ − 6x² + 11x − 6 has the root x = 1 (1 − 6 + 11 − 6 = 0); dividing gives (x − 1)(x² − 5x + 6) = (x − 1)(x − 2)(x − 3).
A·B = 0 ⇔ A = 0 or B = 0 (provided both factors are defined)
Solve x³ − 4x = 0.
Never divide both sides by x — the root x = 0 would be lost.
To simplify a fraction, factorise the numerator and denominator and cancel the common factor, but exclude from the domain the values that make it zero.
Simplify (x² − 5x + 6)/(x² − 4).