Chapter 12.4 — Exercise 12.3 — Division of Expressions

A Preview of Where This Is Heading

The third example is the clearest preview of where this exercise is heading: once 6a³+9a² is recognised as 3a²(2a+3), dividing by (2a+3) is nothing more than cancelling a shared bracket, exactly the way cancelling a shared numerical factor works in ordinary fractions, with no long division of any kind required.

The Same Cancelling Already Familiar From Plain Fractions

None of the cancelling in this exercise is a genuinely new idea — it's the identical logic already used to reduce a numeric fraction like 24/36 down to 2/3, just applied to expressions with letters mixed in. 24/36 reduces because both numbers share the factor 12: 24 = 12×2 and 36 = 12×3, so the 12 cancels, leaving 2/3. 24a⁵÷18a² works exactly the same way, except the "shared factor" being cancelled includes both a numerical part (6) and a variable part (a²) at once. Keeping this parallel in mind is often the fastest way to sanity-check an algebraic division: if the same cancelling step wouldn't be trusted between two plain numbers, it shouldn't be trusted between two expressions either.

Monomial Divided by Monomial

Breaking both terms into primes and variables, then cancelling what's shared, handles every monomial-by-monomial division in one consistent way:

The second row is worth a second look, since the quotient comes out smaller than 1 in front of the variable rather than as a whole coefficient: 14/42 reduces to 1/3, and x³/x² leaves just x behind, giving x/3 — not 3x, and not 1/(3x), a mix-up that's easy to make when the numerator's exponent is smaller than expected relative to the coefficient. The fifth row is worth checking against its own sign carefully too: −54 divided by 9 gives −6, and that negative sign belongs to the whole quotient, not to any single variable — l²m stays entirely positive, with the one minus sign sitting out in front of the coefficient where it started.

Polynomial Divided by a Single Monomial

Dividing a multi-term polynomial by one monomial means dividing every term by that monomial separately, then recombining the results with whatever signs they land on:

The last row mixes fractions into the division itself: dividing by ½abc is the same as multiplying by 2/(abc), so each term's fraction combines with that 2 before any cancelling of variables happens — (⅔)×2 = 4/3, and (4/3)×2 = 8/3, which then share a factor with each other once bc is factored back out at the end.

Expression Divided by Expression, No Factoring Needed Yet

Six divisions in this exercise already have both sides written as ready-made factors, so the entire task is spotting what to cancel:

The second and third rows both hide a common factor inside one bracket before it can be cancelled: 8x−20 is really 4(2x−5), and 7c−35 is really 7(c−5) — spotting that shared inner factor first is what turns each of these into straightforward cancellation rather than a division that looks stuck. The fifth row hides a similar step inside a trinomial rather than a binomial: 36x⁴ and 9x⁴ cancel down to a plain 4 straightaway, but the remaining bracket, (x²+7x+10), doesn't share any factor with the denominator at all — it simply comes along for the ride, multiplied by that leftover 4, since nothing in it was ever meant to cancel with 9x⁴ in the first place.

Factoring First, Then Dividing

The remaining divisions start from an unfactored polynomial, so factoring has to happen before any cancellation is possible at all:

The last two rows each combine a genuine factorisation with an ordinary monomial cancellation — 15ab(a²−7a+10) needs a²−7a+10 factored into (a−2)(a−5) before the shared (a−2) can cancel against the denominator, and 15lm(2p²−2q²) needs both the common factor 2 and the difference-of-squares pattern p²−q² pulled out before (p+q) is available to cancel. Both rows are a direct callback to the two separate skills built earlier in this chapter — pulling out an H.C.F. from Exercise 12.1, and recognising an identity from Exercise 12.2 — combined here in service of a division rather than a standalone factoring exercise.

Why Cancelling Only Works on Whole Factors

Every division in this lesson cancels an entire bracket or an entire monomial factor — never part of one. In (x²+7x+12)÷(x+3) = (x+3)(x+4)÷(x+3), the (x+3) that cancels is a complete factor of the numerator, put there deliberately by the factorisation step. It would be a serious mistake to look at the unfactored numerator x²+7x+12 and try to "cancel" an x+3 that merely appears to resemble part of it — division only ever cancels shared factors, and a factor only counts once the expression has genuinely been rewritten as a product, not before. A quick way to catch this mistake before it happens: substitute a specific number for the variable and check both sides independently. At x=1, (1+7+12)÷(1+3) = 20÷4 = 5, and the claimed answer x+4 gives 1+4 = 5 too — a match, which is reassuring, though not by itself a full proof that the cancellation was valid for every possible value of x, only that it hasn't been contradicted by this one particular check.

Division's Place in the Chapter

Every factoring technique from Exercise 12.1 and Exercise 12.2 shows up somewhere in this lesson's division problems, since factoring is the step that makes division possible in the first place. Exercise 12.4 closes the chapter by looking at what goes wrong when a step like this — distributing, squaring, or dividing — is done carelessly, turning a correct method into a wrong answer — including, fittingly, a few divisions very similar in shape to the ones practised throughout this exercise.