Chapter 12.1 — Introduction to Factorisation

Checking the Uniqueness Claim Against a Second Number

It's worth testing that "always the same primes" claim against a number besides 36, since one example alone isn't proof. Take 60: it can be written as 2×30, 3×20, 4×15, 5×12, or 6×10. Breaking each of those down fully:

Every single row bottoms out at the same four primes — 2, 2, 3, and 5 — just written in a different order depending on which pair was split apart first. This "always the same primes, regardless of starting point" property is what makes prime factorisation useful as a tool at all: it's a single, unambiguous description of a number that doesn't depend on which factor pair happened to be chosen first.

The Same Idea, Applied to Letters Instead of Numbers

An algebraic expression can be broken into factors the same way a number can. 3ab breaks into 3 × a × b, and none of those three pieces can be split any further — they're irreducible. 14x(y + z) breaks into 2 × 7 × x × (y + z); here 2, 7, and x are irreducible on their own, and so is the bracket (y + z) — even though it has two terms inside it, it can't be factored any further as it stands, so it counts as one irreducible piece, not two separate ones. This is an important distinction to get comfortable with early: "irreducible" doesn't mean "a single symbol" — it means "can't be broken down any further," and a bracket like (y+z) qualifies just as much as a lone variable like x does, provided nothing inside it can be pulled out as a common factor of its own.

Reading a Factored Expression Back Off

Factoring and multiplying are exact opposites of each other, which gives a quick way to check any factorisation: multiply the pieces back together and confirm the original expression reappears. 3ab multiplied back out is clearly 3×a×b — nothing to check there. For 14x(y+z) written as 2×7×x×(y+z), multiplying the four irreducible pieces back together should reproduce the original: 2×7×x = 14x, and 14x×(y+z) = 14xy + 14xz — which does indeed match 14x(y+z) once that bracket is expanded out. Being able to multiply a proposed factorisation back out, and comparing it honestly against the original expression, is the single most reliable way to catch a wrong factorisation before it's used for anything further.

Why Bother Factoring an Expression at All

Factorisation earns its place in this course by making division dramatically simpler. Dividing x² + 7x + 12 by x + 3 directly, term by term, isn't obvious at all — but rewrite the numerator as (x+3)(x+4) first, and the division becomes a matter of cancelling a shared bracket:

Without recognising that x² + 7x + 12 factors into (x+3)(x+4) first, this division would need long division of polynomials — factoring turns it into simple cancellation instead, the same shortcut that makes reducing an ordinary numeric fraction like 24/36 straightforward once both numbers are broken into their shared prime factors.

Pulling Out the Biggest Shared Piece

The most direct factoring method looks for the highest common factor shared by every term, exactly the way the H.C.F. of two numbers is found. Take 6x² + 9xy: writing each term as a product of primes and variables, 6x² = 2×3×x×x and 9xy = 3×3×x×y, shows their H.C.F. is 3×x = 3x. Dividing each term by that H.C.F. and writing the result in brackets:

Every factoring problem built on common factors follows this exact three-step routine: break every term into primes and variables, spot what's shared by all of them, then divide it back out into a bracket. The result can always be checked the same way as the numeric examples above — multiplying 3x back into (2x+3y) gives 3x×2x + 3x×3y = 6x²+9xy, exactly the expression that was started with, confirming the H.C.F. step was done correctly.

When No Single Factor Is Shared by Every Term

Some expressions have no factor common to all their terms at once, but still factor cleanly once split into matching pairs. Take x² + xy + x + y: grouping the first two terms and the last two terms separately,

Neither pair shared a factor with the other pair before grouping — but after grouping, both pairs happen to share the exact same bracket, (x + y), which then factors out just like any ordinary common factor would. This method, factorisation by grouping, is really the common-factor method applied twice: once inside each group, and once more across the two groups afterward. Grouping only works, though, if the terms are paired up the right way — pairing x² with x and xy with y instead (rather than x² with xy, and x with y as done above) would give x(x+1) + y(x+1), which happens to reach the same final answer here purely by coincidence, since this particular expression is symmetric enough to tolerate either pairing. In general, finding a grouping that produces a genuinely matching bracket on both sides takes a bit of trial and error, and not every possible pairing of four terms will work.

Two Methods, One Chapter's Worth of Practice Ahead

These two techniques — pulling out a shared H.C.F., and grouping terms into matching pairs — are practised on a wide range of expressions in Exercise 12.1. From there, the chapter moves to a faster route for expressions that match one of the four algebraic identities from the previous chapter, before showing how factoring makes dividing algebraic expressions genuinely manageable — turning what would otherwise be long, awkward division problems into the same kind of bracket-cancelling shortcut used in the x²+7x+12 example above.