Chapter 12.2 — Exercise 12.1 — Common Factors

Scaling Up to Several Variables at Once

The same approach scales to expressions with several variables spread unevenly across the terms:

The 15p/20qr/25rp row is the one that trips people up: even though every one of the three terms has two variables in it, no single letter — not p, not q, not r — appears in all three terms simultaneously, so the H.C.F. collapses down to just the shared number, 5, with nothing else to add to the list beyond 1 and 5.

Why the Factor List Isn't Just "Divisors of the H.C.F."

It's worth being precise about what the "all common factors" column in the table above actually lists, since it's easy to assume it's simply every divisor of the H.C.F. — and for a single-variable H.C.F. like 8, that happens to be true (1, 2, 4, 8 are exactly the divisors of 8). But for 12x²y and 18xy², the H.C.F. is 6xy, which breaks down into four separate irreducible pieces multiplied together: 2, 3, x, and y. Every one of the sixteen listed common factors is some combination of those four pieces, each either included or left out — 1 (none included), 2, 3, x, y (one included), 6, 2x, 2y, 3x, 3y, xy (two included), and so on up to the full 6xy (all four included) — and since each of the four pieces independently is either in or out, there are exactly 2×2×2×2 = 16 possible combinations, matching the sixteen entries shown. Treating "common factors" as "divisors of a single number" only works when the H.C.F. itself is a single plain number with no variables attached.

Turning the H.C.F. Into an Actual Factorisation

Once the H.C.F. is found, factoring an expression is just dividing every term by it and writing the result in a bracket:

The fourth and fifth rows are worth a second look, since their H.C.F.s are the busiest in the table. 36a²b and 60a²bc share 2×2×3×a×a×b = 12a²b — every prime and every variable that both terms have enough of. Dividing it back out of each term (36a²b÷12a²b = 3, and 60a²bc÷12a²b = 5c) gives the bracket (3 − 5c) directly, with no leftover variables stranded outside. The fifth row, 3a²bc + 6ab²c + 9abc², spreads three variables across three terms rather than two, but the process scales up exactly the same way: the H.C.F. 3abc takes exactly one copy of a, b, and c from each term (since every term has at least one of each), leaving a, 2b, and 3c behind in the bracket — one leftover factor per original term, with nothing missing and nothing doubled.

Grouping When No Single Factor Fits All Four Terms

Five four-term expressions in this exercise need grouping rather than a straight H.C.F., since no one factor is shared by all four terms at once. The first splits into two pairs joined by a shared bracket after regrouping the middle terms:

Notice the sign flip in the middle step: 4b(2y − a) and −4b(a − 2y) are the same quantity written two different ways, and making that flip explicit is what lets (a − 2y) emerge as a shared bracket on both sides, rather than two brackets that look almost — but not quite — identical. The remaining four grouping problems follow the identical strategy:

The third row is the trickiest of the four, since its H.C.F.s inside each group are themselves squared terms rather than single variables — a²(a−b²) and −b(a−b²) — but the underlying strategy hasn't changed at all: group two terms at a time, pull the H.C.F. out of each group separately, and confirm both groups leave behind the exact same bracket. The fourth row is worth comparing against it directly, since it looks superficially similar (four terms, two variables, a mix of squared and plain terms) but groups far more simply: p²q and pq share a plain factor of p, while pr² and r² share nothing beyond r² itself, leaving p and −1 as the two group-factors rather than anything squared.

Confirming a Grouping Really Works

A grouping is only valid once both halves genuinely produce the same bracket — if they don't, the terms need to be regrouped a different way rather than forced together. Checking the m² − mn + 4m − 4n example by multiplying the final answer back out is a reliable way to build confidence: (m−n)(m+4) = m² + 4m − mn − 4n = m² − mn + 4m − 4n, matching the original expression term for term, in whatever order those four terms happen to land. The same multiply-back-out check works for every H.C.F. factorisation in this exercise too, not only the grouped ones — it's a habit worth carrying forward into every remaining exercise in this chapter, since a factorisation that can't be verified this way hasn't actually been confirmed correct yet.

From Grouping by Hand to Recognising a Pattern by Sight

Every factorisation in this exercise was found by breaking terms into primes and variables, or by trying different groupings until a shared bracket appeared. Exercise 12.2 introduces a faster route for a specific, very common family of expressions — ones that already match one of the four algebraic identities from the previous chapter — where the factorisation can be read off almost on sight, once the pattern is recognised, instead of being searched for by trial and error the way every grouping problem in this exercise required.