Chapter 12.3 — Exercise 12.2 — Factorisation by Identities
Factorisation using algebraic identities. This is Lesson 3 of 5 in Chapter 12: Factorisation.
Reading the Four Identities Backwards
Every identity from the algebraic expressions chapter can be read in reverse — instead of expanding a bracket into three terms, spotting that a three-term expression already matches the expanded side lets it collapse straight back into a bracket:
a² + 2ab + b² ≡ (a+b)²
a² − 2ab + b² ≡ (a−b)²
a² − b² ≡ (a+b)(a−b)
x² + (a+b)x + ab ≡ (x+a)(x+b)Recognising which of these four shapes a given expression matches is the entire skill this exercise builds — once the shape is identified, the factorisation follows in one line.
Spotting a Perfect Square Trinomial
The giveaway for a²±2ab+b² is that the first and last terms are both perfect squares, and the middle term is exactly twice the product of their square roots. Checking a² + 10a + 25: the first term is a², the last is 25 = 5², and the middle term 10a is indeed 2×a×5. So it matches a²+2ab+b² with a=a, b=5, giving (a+5)². Eight examples confirm the pattern holds across a wide range of coefficients and variables:
| Expression | a, b identified | Factorised form |
|---|---|---|
| a² + 10a + 25 | a, 5 | (a+5)² |
| l² − 16l + 64 | l, 8 | (l−8)² |
| 36x² + 96xy + 64y² | 6x, 8y | (6x+8y)² |
| 25x² + 9y² − 30xy | 5x, 3y | (5x−3y)² |
| 25m² − 40mn + 16n² | 5m, 4n | (5m−4n)² |
| 81x² − 198xy + 121y² | 9x, 11y | (9x−11y)² |
| (x+y)² − 4xy | x, y (after expanding) | (x−y)² |
| l⁴ + 4l²m² + 4m⁴ | l², 2m² | (l²+2m²)² |
The seventh row needs an extra step before the pattern is visible: (x+y)² − 4xy has to be expanded first — x²+2xy+y²−4xy = x²−2xy+y² — before it reveals itself as a perfect square of (x−y) rather than of (x+y). Skipping that expansion and trying to match the pattern directly against the unexpanded form would miss it entirely.
Spotting a Difference of Two Squares
The a²−b² pattern is easier to spot: two terms only, connected by a minus sign, both of them perfect squares. x² − 36 becomes x² − 6², matching a=x, b=6, giving (x+6)(x−6). Across a run of related examples:
| Expression | a, b identified | Factorised form |
|---|---|---|
| x² − 36 | x, 6 | (x+6)(x−6) |
| 49x² − 25y² | 7x, 5y | (7x+5y)(7x−5y) |
| m² − 121 | m, 11 | (m+11)(m−11) |
| 81 − 64x² | 9, 8x | (9+8x)(9−8x) |
| x²y² − 64 | xy, 8 | (xy+8)(xy−8) |
| 6x² − 54 | x, 3 (after removing factor 6) | 6(x+3)(x−3) |
| x² − 81 | x, 9 | (x+9)(x−9) |
| 2x − 32x⁵ | 1, 4x² (after removing factor 2x) | 2x(1+4x²)(1+2x)(1−2x) |
| 81x⁴ − 121x² | 9x, 11 (after removing factor x²) | x²(9x+11)(9x−11) |
The sixth and eighth rows both need a common factor pulled out before the difference-of-squares pattern becomes visible: 6x² − 54 = 6(x²−9) only reveals x²−3² once the 6 is factored out first, and 2x − 32x⁵ = 2x(1−16x⁴) needs the 2x removed before 1−16x⁴ shows itself as 1²−(4x²)² — which then factors again into (1+4x²)(1−4x²), and again into (1+4x²)(1+2x)(1−2x), since 1−4x² is itself a second difference of squares hiding inside the first.
Squares Hiding Behind a Bracket
Three more difference-of-squares problems treat an entire bracketed expression as if it were a single variable:
p² − 2pq + q² − r² = (p−q)² − r² = (p−q+r)(p−q−r)
(x+y)² − (x−y)² = [(x+y)+(x−y)][(x+y)−(x−y)] = (2x)(2y) = 4xyThe first treats (p−q) as a single unit — recognising that p²−2pq+q² is itself a perfect square before subtracting r² — while the second applies a²−b² with a=(x+y) and b=(x−y) directly, and the two brackets on the right collapse into plain 2x and 2y once the pluses and minuses inside them are combined.
Common Factor First, Identity Second
Three expressions need a plain H.C.F. step before any identity applies at all — no different from Exercise 12.1, just as a first move rather than the whole solution:
lx² + mx = x(lx + m)
7y² + 35z² = 7(y² + 5z²)
3x⁴ + 6x³y + 9x²z = 3x²(x² + 2xy + 3z)None of these three actually needs the identities at all — they're included here as a reminder that not every factoring problem in an "identities" exercise is really about an identity; sometimes the simplest common-factor check comes first and the job is already done.
Grouping Into an (x+a)(x+b) Shape
A handful of four-term expressions factor by grouping into two matching binomials, arriving at the (x+a)(x+b) shape indirectly:
| Expression | Factorised form |
|---|---|
| x² − ax − bx + ab | (x−a)(x−b) |
| 3ax − 6ay − 8by + 4bx | (x−2y)(3a+4b) |
| mn + m + n + 1 | (m+1)(n+1) |
| 6ab − b² + 12ac − 2bc | (6a−b)(b+2c) |
| p²q − pr² − pq + r² | (p−1)(pq−r²) |
| xy + xz − 5y − 5z | (y+z)(x−5) |
These six all resemble the grouping problems from Exercise 12.1 more than the identity-matching problems earlier in this lesson — a reminder that the boundary between "common factor," "grouping," and "identity" isn't always sharp, and a given expression sometimes yields to more than one of these three strategies.
A Fourth Power Splits Into Two Squares First
Some expressions need the a²−b² pattern applied twice in a row, once to strip off an outer layer, and again on what's left inside. x⁴ − y⁴ is really (x²)² − (y²)², so the first pass gives (x²+y²)(x²−y²) — and the second factor, x²−y², is itself a difference of squares, splitting further into (x+y)(x−y):
x⁴ − y⁴ = (x²+y²)(x²−y²) = (x²+y²)(x+y)(x−y)The same two-pass idea applies even when the "b" side of the very first split is itself a bracket rather than a single letter. a⁴ − (b+c)⁴ splits first into (a²+(b+c)²)(a²−(b+c)²), and the second factor splits again since it's still a difference of two squares:
a⁴ − (b+c)⁴ = (a²+(b+c)²)(a+(b+c))(a−(b+c)) = (a²+b²+c²+2bc)(a+b+c)(a−b−c)l² − (m+n)² factors the same way but only needs one pass, since l² and (m+n)² are already both plain squares with nothing left to split further: (l+m+n)(l−m−n). And 49x² − 16/25 factors exactly like any numeric-fraction version of the pattern, with a=7x and b=4/5: (7x+4/5)(7x−4/5).
Splitting a Square Into Two Squares, Twice
x⁴ − 2x²y² + y⁴ combines both identities in sequence: it's a perfect square first (of x²−y²), and then that result is itself a difference of two squares:
x⁴ − 2x²y² + y⁴ = (x²−y²)² = [(x+y)(x−y)]² = (x+y)²(x−y)²A related problem, 4(a+b)² − 9(a−b)², treats each bracket as a single unit inside a difference-of-squares pattern: with A = 2(a+b) and B = 3(a−b), it becomes (A+B)(A−B), which simplifies down to (5a−b)(5b−a) once the brackets inside A and B are expanded and their like terms combined.
Turning x² + (Sum)x + (Product) Into Two Binomials
The fourth identity works in reverse by splitting the middle coefficient into two numbers whose product matches the constant term. For a² + 10a + 24, the constant 24 needs two factors that add to 10 — checking 4×6=24 and 4+6=10 confirms the pair:
a² + 10a + 24 = a² + (4+6)a + (4)(6) = (a+4)(a+6)Three more examples apply the identical search-for-a-factor-pair strategy, including two where the correct pair is negative:
| Expression | Sum needed | Product needed | Pair found | Factorised form |
|---|---|---|---|---|
| x² + 9x + 18 | 9 | 18 | 3, 6 | (x+3)(x+6) |
| p² − 10p + 21 | −10 | 21 | −7, −3 | (p−7)(p−3) |
| x² − 4x − 32 | −4 | −32 | −8, 4 | (x−8)(x+4) |
The last row is the one worth pausing on: when the constant term is negative, the two factors making it up must have opposite signs, and it's the larger-magnitude one that carries the sign of the middle coefficient — here −4x is negative, so −8 (not 4) takes the negative sign, since −8+4=−4 while 8+(−4) would give the wrong total entirely.
From Recognising Patterns to Reversing Division
Every factorisation in this lesson turned a recognisable expanded pattern back into a compact bracket. Exercise 12.3 puts that skill to direct use in division — factoring both the numerator and denominator of an algebraic fraction first is what usually turns an intimidating division problem into a simple cancellation.