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Why Factorisation Works in Secondary Mathematics — Reversing the Distributive Law

SECONDARY MATHEMATICS · ALGEBRA AS STRUCTURE

Factorisation is not a collection of unrelated tricks. At its core, it reverses multiplication over addition and turns a sum into a product without changing the expression’s value.

Start from distribution

The distributive law says a(b + c) = ab + ac. Read from left to right, it expands a product. Read from right to left, it factorises a sum: ab + ac = a(b + c).

For example, 6x + 9 = 3(2x + 3). Expanding the right side returns 6x + 9. Factorisation changes the structure while preserving value.

Common factor first

In 12x² − 18x, both terms contain 6x. Factorising gives 6x(2x − 3). The factor 6x is not guessed; it is a quantity that multiplies each term.

A useful check is to expand: 6x·2x − 6x·3 = 12x² − 18x. If expansion does not return the original expression, the factorisation is incorrect.

Quadratics reveal a hidden product

Consider x² + 7x + 12. We seek two numbers whose product is 12 and whose sum is 7: 3 and 4. Therefore x² + 7x + 12 = (x + 3)(x + 4).

Why? Expanding (x + 3)(x + 4) gives x² + 4x + 3x + 12 = x² + 7x + 12. The familiar “sum and product” search is a compressed way of reversing this expansion pattern.

Difference of two squares

(a − b)(a + b) expands to a² + ab − ab − b² = a² − b². The middle terms cancel. Therefore a² − b² = (a − b)(a + b).

So x² − 25 = (x − 5)(x + 5). The pattern works because 25 = 5² and the expression is a difference. x² + 25 does not factor over the real numbers using this identity.

Perfect-square structure

(x + a)² = x² + 2ax + a². Therefore x² + 10x + 25 = (x + 5)². The middle coefficient 10 is twice 5, and the constant 25 is 5².

Recognising this structure can make later equation solving or algebraic simplification shorter, but recognition should be checked by expansion.

Why factorisation helps solve equations

Suppose x² + 7x + 12 = 0. Factorisation gives (x + 3)(x + 4) = 0. A product is zero when at least one factor is zero, so x = −3 or x = −4.

The zero-product property is available because factorisation exposed a product. The roots were already encoded in the quadratic; the new form makes them visible.

Factorisation and cancellation

In (x² − 4)/(x − 2), factorisation gives (x − 2)(x + 2)/(x − 2). For x ≠ 2, the common nonzero factor cancels and the expression becomes x + 2. The original restriction x ≠ 2 remains.

Factorisation can therefore expose a common factor, but it does not erase domain conditions. This connects to equivalent expressions.

A factorisation decision tree

  1. Is there a common factor in every term? Extract it first.
  2. Is the remaining expression a difference of two squares?
  3. Is it a quadratic that may split into two linear factors?
  4. Does it match a perfect-square pattern?
  5. Expand the proposed factors to verify the result.

Worked practice

A. 8x + 12 = 4(2x + 3).

B. x² + 9x + 20 = (x + 4)(x + 5).

C. 9y² − 16 = (3y − 4)(3y + 4).

D. x² − 12x + 36 = (x − 6)².

E. 3x² + 12x = 3x(x + 4).

The deeper idea

Expansion and factorisation are not competing chapters. They are opposite directions through the same distributive structure. A student who understands that relationship can reconstruct many factorisation patterns instead of depending only on memorised templates.

Continue through Algebraic Structure and the substitution-checking article, then return to the Secondary Mathematics Master Index.