Factorisation of Polynomials
Factorisation of polynomials is the process of expressing a polynomial as a product of simpler polynomials or irreducible factors. Just as the integer 12 can be written as 2 × 2 × 3, a polynomial like x² − 5x + 6 can be written as (x − 2)(x − 3). Factorisation is one of the most powerful tools in algebra because it allows us to solve polynomial equations, simplify rational expressions, and understand the roots and behaviour of functions. There are several key techniques for factorising polynomials, each suited to different situations. These include taking out a common factor (HCF), recognising special products such as the difference of two squares and perfect square trinomials, factorising quadratic trinomials by inspection or the AC method, and using the Factor Theorem for higher-degree polynomials. Mastering all of these techniques and knowing when to apply each one is central to success in algebra. Factorisation is essentially the reverse of expansion. When you expand (x + 4)(x − 1), you get x² + 3x − 4. Factorisation starts from x² + 3x − 4 and recovers (x + 4)(x − 1). This reversibility makes factorisation indispensable for solving equations: if (x + 4)(x − 1) = 0, then by the zero-product property, x = −4 or x = 1.
- Step 1 – Always Check for a Highest Common Factor (HCF) First
- Step 2 – Recognise Special Products
- Step 3 – Factorise Quadratic Trinomials (ax² + bx + c)
- Step 4 – Use the Factor Theorem for Degree ≥ 3
- Step 5 – Factor by Grouping (for 4-term polynomials)
Step 1 – Always Check for a Highest Common Factor (HCF) First
Before applying any other method, look for a common factor in all terms. For example, 6x³ − 9x² + 3x = 3x(2x² − 3x + 1). Taking out the HCF simplifies the remaining polynomial and often makes further factorisation easier. The HCF may be a number, a variable, or a combination of both.
Step 2 – Recognise Special Products
Three identities appear repeatedly and should be recognised immediately. Difference of two squares: a² − b² = (a + b)(a − b). Perfect square trinomial: a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². Sum/difference of two cubes: a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²). For example, 4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5).
Step 3 – Factorise Quadratic Trinomials (ax² + bx + c)
When a = 1: find two numbers p and q such that p + q = b and p × q = c, then write (x + p)(x + q). For example, x² − 7x + 12: need p + q = −7, pq = 12, so p = −3, q = −4, giving (x − 3)(x − 4). When a ≠ 1, use the AC method: multiply a × c, find two numbers that multiply to ac and add to b, split the middle term, then factor by grouping. For 2x² + 7x + 3: ac = 6, need factors of 6 adding to 7: use 1 and 6. Rewrite as 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).
Step 4 – Use the Factor Theorem for Degree ≥ 3
The Factor Theorem states: (x − a) is a factor of polynomial P(x) if and only if P(a) = 0. To factorise P(x) = x³ − 6x² + 11x − 6, test integer factors of the constant term (±1, ±2, ±3, ±6). P(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor. Perform polynomial long division or synthetic division to find P(x) ÷ (x − 1) = x² − 5x + 6 = (x − 2)(x − 3). Therefore P(x) = (x − 1)(x − 2)(x − 3).
Step 5 – Factor by Grouping (for 4-term polynomials)
Group terms in pairs and factor out the common factor from each pair. For x³ + 2x² − 3x − 6: group as (x³ + 2x²) + (−3x − 6) = x²(x + 2) − 3(x + 2) = (x² − 3)(x + 2). This technique works when each pair shares a common factor and the resulting binomial factors match.
The Key Equation
- P(x) — Polynomial (dimensionless): A sum of terms each consisting of a coefficient multiplied by a non-negative integer power of x.
- a_n — Leading coefficient (dimensionless): The coefficient of the highest-degree term; appears outside the factored product.
- r_i — Root (zero) (dimensionless): A value of x for which P(x) = 0; each root corresponds to a linear factor (x − r_i).
- n — Degree (dimensionless): The highest power of x in the polynomial; equals the number of linear factors over the complex numbers.
Worked Example
Problem. Factorise completely: f(x) = 2x³ − 3x² − 11x + 6
Answer. f(x) = (x + 2)(2x − 1)(x − 3)
Origins · 1 minÉvariste Galois and how this was discovered
Galois developed what is now called Galois Theory, which provides a deep framework for understanding when polynomial equations can be solved by radicals. His work on the symmetry of roots of polynomials — analysed through group theory — revolutionised algebra and answered centuries-old questions about factorisation and solvability of polynomial equations of degree 5 and higher. Galois died in a duel at the age of 20, but the night before he wrote a letter to a friend summarising his mathematical discoveries — work that mathematicians took decades to fully understand after his death.
Related Concepts
- Remainder Theorem — The Remainder Theorem states that the remainder when P(x) is divided by (x − a) equals P(a), which is the foundation of the Factor Theorem used in factorisation.
- Roots and Zeros of Polynomials — Factorisation directly reveals the roots of a polynomial; each linear factor (x − r) corresponds to a root x = r where the polynomial equals zero.
- Quadratic Equations — Factorising quadratic trinomials (degree-2 polynomials) is the most common application of factorisation and is central to solving quadratic equations.
- Polynomial Long Division — After finding one factor using the Factor Theorem, polynomial long division or synthetic division is used to determine the remaining quotient for complete factorisation.
- Expanding Polynomials — Expansion and factorisation are inverse processes; understanding how to expand products of polynomials helps in recognising and reversing these products during factorisation.