Exploring Algebraic Identities — Detailed Notes
An algebraic identity is an equation that is true for all values of its variables (e.g. (a+b)² = a²+2ab+b²), unlike an ordinary equation which holds only for particular values (e.g. x²−1 = 24 is true only for x = ±5). Identities make squaring, multiplying and factoring much faster, and can be ‘seen’ using squares, rectangles, cubes and algebra tiles.
4.2–4.4 The square identities
(a + b)² = a² + 2ab + b² • (a − b)² = a² − 2ab + b²
(a + b)(a − b) = a² − b² • (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
These hold for all numbers (positive, negative, rational). Uses: expand binomials — (5x+2y)² = 25x²+20xy+4y²; square numbers fast — 43² = (40+3)² = 1600+240+9 = 1849, and 29² = (30−1)² = 841; and factorise by recognising the pattern — x²+4x+4 = (x+2)², 36x²+12x+1 = (6x+1)².
Note (a+b)² ≠ a²+b²; the difference is the 2ab term. Śhrīdharāchārya’s trick a² = (a+b)(a−b)+b² gives 55² = 60×50+25 = 3025.
4.5–4.6 Factorising quadratics (splitting the middle term)
(x + a)(x + b) = x² + (a + b)x + ab. To factor x² + (sum)x + (product): find two numbers whose sum is the coefficient of x and whose product is the constant term, then split the middle term. E.g. x²+7x+12 → need sum 7, product 12 → 3 and 4 → (x+3)(x+4). For x²−5x+6 → −2 and −3 → (x−2)(x−3).
For ax²+bx+c, (px+a)(qx+b) = pq·x² + (pb+qa)x + ab — algebra tiles make this visual. Example: x²−4x−96 = (x−12)(x+8); set each factor to 0 to solve, keeping only values that make sense (a length can’t be negative).
4.7 The cube and sum/difference identities
(a + b)³ = a³ + 3a²b + 3ab² + b³ • (a − b)³ = a³ − 3a²b + 3ab² − b³
a³ + b³ = (a + b)(a² − ab + b²) • a³ − b³ = (a − b)(a² + ab + b²)
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
The cube (a+b)³ can be seen as a cube of edge (a+b) split into 2 cubes (a³, b³) and 6 cuboids (3a²b + 3ab²). A useful fact: (a−b) is a factor of aⁿ − bⁿ for every n.
4.8 Simplifying rational algebraic expressions
Factor the numerator and denominator, then cancel the common factor (allowed only when it is non-zero). Example: (x²−7x+12)/(5x²+5x−100) = [(x−3)(x−4)]/[5(x−4)(x+5)] = (x−3)/[5(x+5)].
Key Terms
identity, equation, binomial, expansion, factorisation, splitting the middle term, algebra tiles, difference of squares, sum/difference of cubes, (a+b+c)² identity, rational algebraic expression.
End-of-Chapter Exercises — Full Solutions
Complete worked solutions to the end-of-chapter problems (starred * = HOTS).
1Use suitable identities to find the products.
(i) (−3x+4)² = 9x² − 24x + 16. (ii) (2s+7)(2s−7) = 4s² − 49. (iii) (p²+½)(p²−½) = p⁴ − 1/4. (iv) (2n+7)(2n−7) = 4n² − 49. (v) (s−2t)(s²+2st+4t²) = s³ − 8t³. (vi) (½r − 4)² = ¼r² − 4r + 16. (vii) (−3m+4k−l)² = 9m² + 16k² + l² − 24km − 8kl + 6lm. (viii) (x − ⅓y)³ = x³ − x²y + ⅓xy² − 1/27·y³. (ix) (7/2·k − 2/3·m)³ = 343/8·k³ − 49/2·k²m + 14/3·km² − 8/27·m³.
2Find the values using suitable identities.
(i) 17×21 = (19−2)(19+2) = 361 − 4 = 357. (ii) 104×96 = (100+4)(100−4) = 10000 − 16 = 9984. (iii) 24×16 = (20+4)(20−4) = 400 − 16 = 384. (iv) 147³ = (150−3)³ = 3,176,523. (v) 199³ = (200−1)³ = 7,880,599. (vi) 127³ = (130−3)³ = 2,048,383. (vii) (−107)³ = −(100+7)³ = −1,225,043. (viii) (−299)³ = −(300−1)³ = −26,730,899.
3Factor the algebraic expressions (using the standard identities).
(i) 4y² + 1 + 1/(16y²) = (2y + 1/(4y))². (ii) 9m² − n²/25 = (3m − n/5)(3m + n/5). (iii) 27b³ − 1/(64b³) = (3b − 1/(4b))(9b² + 3/4 + 1/(16b²)). (iv) x² + (5/6)x + 1/6 = (x + 1/2)(x + 1/3). (v) 27u³ − (27/5)u² + (9/25)u − 1/125 = (3u − 1/5)³. (vi) 64y³ + z³/125 = (4y + z/5)(16y² − 4yz/5 + z²/25). (vii) p³ + q³ + 27r³ − 9pqr = (p + q + 3r)(p² + q² + 9r² − pq − 3qr − 3rp). (viii) 9m² − 12m + 4 = (3m − 2)². (x) 4x² + 9y² + 36z² + 12xy + 36yz + 24zx = (2x + 3y + 6z)². (The remaining fraction-heavy parts (ix), (xi) factor the same way, using a³+b³+c³−3abc and (a−b)³ respectively.)
4Simplify the rational expressions (denominators ≠ 0).
(i) (4x² + 4x + 1)/(4x² − 1) = (2x+1)² / [(2x−1)(2x+1)] = (2x+1)/(2x−1). (ii) (3a³ − 24b³)/(9a² − 36b²) = [3(a−2b)(a²+2ab+4b²)] / [9(a−2b)(a+2b)] = (a²+2ab+4b²)/[3(a+2b)]. (iii) (s³ + 125t³)/(s² − 2st − 35t²) = [(s+5t)(s²−5st+25t²)] / [(s+5t)(s−7t)] = (s²−5st+25t²)/(s−7t).
5Find possible length and breadth of the rectangles with the given areas.
(i) 25a² − 30ab + 9b² = (5a − 3b)² → length = breadth = (5a − 3b) (it is a square). (ii) 36s² − 49t² = (6s − 7t)(6s + 7t) → length (6s + 7t), breadth (6s − 7t).
6Find possible dimensions of the cuboids with the given volumes.
(i) 6a² − 24b² = 6(a − 2b)(a + 2b) → dimensions 6, (a − 2b), (a + 2b). (ii) 3ps² − 15ps + 12p = 3p(s² − 5s + 4) = 3p(s − 1)(s − 4) → dimensions 3p, (s − 1), (s − 4).
7A square playground of side 40 m has a path of width s m around it. Find the area of the path in terms of s.
Outer square side = (40 + 2s). Area of path = (40 + 2s)² − 40² = 1600 + 160s + 4s² − 1600 = 4s² + 160s = 4s(s + 40) square metres.
8A number plus its reciprocal equals 10/3. Find the number.
x + 1/x = 10/3 ⇒ 3x² + 3 = 10x ⇒ 3x² − 10x + 3 = 0 ⇒ (3x − 1)(x − 3) = 0 ⇒ x = 1/3 or x = 3. So the number is 3 (or its reciprocal 1/3).
9A rectangular pool has area 2x² + 7x + 3 (sq. hastas) and width 2x + 1 hastas. Find its length.
Factor the area: 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Dividing by the width (2x + 1), the length = (x + 3) hastas.
10(*) If both (x − 2) and (x − 1/2) are factors of px² + 5x + r, show that p = r.
(x − 2) a factor ⇒ p(2)² + 5(2) + r = 0 ⇒ 4p + r = −10. (x − 1/2) a factor ⇒ p(1/4) + 5(1/2) + r = 0 ⇒ (multiply by 4) p + 4r = −10. Subtracting the two equations: (4p + r) − (p + 4r) = 0 ⇒ 3p − 3r = 0 ⇒ p = r.
11(*) If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.
First, a²+b²+c² = (a+b+c)² − 2(ab+bc+ca) = 25 − 20 = 5. Now a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab − bc − ca) = 5 × (5 − 10) = 5 × (−5) = −25.
12(*) Show that n³ − n is divisible by 6 for every natural number n.
Factor: n³ − n = n(n² − 1) = (n − 1)·n·(n + 1) — the product of three consecutive integers. Among any three consecutive integers at least one is a multiple of 2 and exactly one is a multiple of 3, so the product is divisible by 2 × 3 = 6. Hence n³ − n is always divisible by 6.
13(*) Find the value of (i) x³ + y³ − 12xy + 64 when x + y = −4; (ii) x³ − 8y³ − 36xy − 216 when x = 2y + 6.
(i) Write as x³ + y³ + 4³ − 3·x·y·4 (since 3·4 = 12 and 4³ = 64) = (x + y + 4)(x²+y²+16−xy−4y−4x). As x + y = −4, (x + y + 4) = 0, so the whole value is 0. (ii) Write as x³ + (−2y)³ + (−6)³ − 3·x·(−2y)·(−6) (the last term = 36xy) = (x − 2y − 6)(…). As x = 2y + 6, (x − 2y − 6) = 0, so the value is 0.