Chapter 4 exercise answers: Exploring Algebraic Identities

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(i) (7x + 4y)²

1. Match with (a + b)²: a = 7x, b = 4y.
2. a² = 49x², b² = 16y², 2ab = 2 × 7x × 4y = 56xy.
3. Add the three pieces.

Answer49x² + 56xy + 16y²

(ii) (7/5 x + 3/2 y)²

1. Match with (a + b)²: a = 7x/5, b = 3y/2.
2. a² = 49x²/25, b² = 9y²/4, 2ab = 2 × (7x/5) × (3y/2) = 21xy/5.
3. Add the three pieces.

Answer49x²/25 + 21xy/5 + 9y²/4

(iii) (2.5p + 1.5q)²

1. Write 2.5p as 5p/2 and 1.5q as 3q/2 — match with (a + b)²: a = 5p/2, b = 3q/2.
2. a² = 25p²/4, b² = 9q²/4, 2ab = 2 × (5p/2) × (3q/2) = 15pq/2.
3. Add the three pieces.

Answer25p²/4 + 15pq/2 + 9q²/4

(iv) (3/4 s + 8t)²

1. Match with (a + b)²: a = 3s/4, b = 8t.
2. a² = 9s²/16, b² = 64t², 2ab = 2 × (3s/4) × 8t = 12st.
3. Add the three pieces.

Answer9s²/16 + 12st + 64t²

(v) (x + 1/(2y))²

1. Match with (a + b)²: a = x, b = 1/(2y).
2. a² = x², b² = 1/(4y²), 2ab = 2 × x × 1/(2y) = x/y.
3. Add the three pieces.

Answerx² + x/y + 1/(4y²)

(vi) (1/x + 1/y)²

1. Match with (a + b)²: a = 1/x, b = 1/y.
2. a² = 1/x², b² = 1/y², 2ab = 2/(xy).
3. Add the three pieces.

Answer1/x² + 2/(xy) + 1/y²

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(i) (64)²

1. Write 64 = 60 + 4.
2. (60 + 4)² = 60² + 2×60×4 + 4² = 3600 + 480 + 16.

Answer4096

(ii) (105)²

1. Write 105 = 100 + 5.
2. (100 + 5)² = 100² + 2×100×5 + 5² = 10000 + 1000 + 25.

Answer11025

(iii) (205)²

1. Write 205 = 200 + 5.
2. (200 + 5)² = 200² + 2×200×5 + 5² = 40000 + 2000 + 25.

Answer42025

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(i) 9x² + 24xy + 16y²

1. The ends are 9x² = (3x)² and 16y² = (4y)².
2. Test the middle term: 2 × 3x × 4y = 24xy — this matches.
3. So it is a perfect square.

Answer(3x + 4y)²

(ii) 4s² + 20st + 25t²

1. The ends are 4s² = (2s)² and 25t² = (5t)².
2. Test the middle term: 2 × 2s × 5t = 20st — this matches.
3. So it is a perfect square.

Answer(2s + 5t)²

(iii) 49x² + 28xy + 4y²

1. The ends are 49x² = (7x)² and 4y² = (2y)².
2. Test the middle term: 2 × 7x × 2y = 28xy — this matches.
3. So it is a perfect square.

Answer(7x + 2y)²

(iv) 64p² + 32/3 pq + 4/9 q²

1. The ends are 64p² = (8p)² and 4/9 q² = (2q/3)².
2. Test the middle term: 2 × 8p × (2q/3) = 32pq/3 — this matches.
3. So it is a perfect square. (Taking out 4/9 first gives the same thing in another form: (4/9)(12p + q)².)

Answer(8p + 2q/3)²

(v) 3a² + 4ab + 4/3 b²

1. Take out 1/3 as a common factor: 3a² + 4ab + 4/3 b² = (1/3)(9a² + 12ab + 4b²).
2. Inside, the ends are (3a)² and (2b)², and the middle term 2 × 3a × 2b = 12ab matches.
3. So the bracket is (3a + 2b)².

Answer(1/3)(3a + 2b)²

(vi) 9/5 s² + 6sv + 5v²

1. Take out 1/5 as a common factor: 9/5 s² + 6sv + 5v² = (1/5)(9s² + 30sv + 25v²).
2. Inside, the ends are (3s)² and (5v)², and the middle term 2 × 3s × 5v = 30sv matches.
3. So the bracket is (3s + 5v)².

Answer(1/5)(3s + 5v)²

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(i) (79)²

1. Write 79 = 80 − 1.
2. (80 − 1)² = 80² − 2×80×1 + 1² = 6400 − 160 + 1.

Answer6241

(ii) (193)²

1. Write 193 = 200 − 7.
2. (200 − 7)² = 200² − 2×200×7 + 7² = 40000 − 2800 + 49.

Answer37249

(iii) (299)²

1. Write 299 = 300 − 1.
2. (300 − 1)² = 300² − 2×300×1 + 1² = 90000 − 600 + 1.

Answer89401

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(i) 117²

1. Write 117 = 100 + 17, and use (a + b)² = a² + 2ab + b².
2. 100² + 2×100×17 + 17² = 10000 + 3400 + 289.

Answer13689

(ii) 78²

1. Write 78 = 80 − 2, and use (a − b)² = a² − 2ab + b².
2. 80² − 2×80×2 + 2² = 6400 − 320 + 4.

Answer6084

(iii) 198²

1. Write 198 = 200 − 2, and use (a − b)² = a² − 2ab + b².
2. 200² − 2×200×2 + 2² = 40000 − 800 + 4.

Answer39204

(iv) 214²

1. Write 214 = 200 + 14, and use (a + b)² = a² + 2ab + b².
2. 200² + 2×200×14 + 14² = 40000 + 5600 + 196.

Answer45796

(v) 1104²

1. Write 1104 = 1100 + 4, and use (a + b)² = a² + 2ab + b².
2. 1100² + 2×1100×4 + 4² = 1210000 + 8800 + 16.

Answer1218816

(vi) 1120²

1. Write 1120 = 1100 + 20, and use (a + b)² = a² + 2ab + b².
2. 1100² + 2×1100×20 + 20² = 1210000 + 44000 + 400.

Answer1254400

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(i) 16y² − 24y + 9

1. The ends are 16y² = (4y)² and 9 = (3)².
2. The middle term is negative, so test with −3: 2 × 4y × (−3) = −24y — this matches.

Answer(4y − 3)²

(ii) 9/4 s² + 6st + 4t²

1. The ends are 9/4 s² = (3s/2)² and 4t² = (2t)².
2. Test the middle term: 2 × (3s/2) × 2t = 6st — this matches.

Answer(3s/2 + 2t)²

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

1. Group by the square terms: m²/9 = (m/3)², k²/4 = (k/2)², 9n² = (3n)².
2. Test the three cross terms of (m/3 + k/2 + 3n)²: 2×(m/3)(k/2) = mk/3, 2×(m/3)(3n) = 2mn, 2×(k/2)(3n) = 3kn — all three match the given terms.

Answer(m/3 + k/2 + 3n)²

(iv) p²/16 − 2 + 16/p²

1. The ends are p²/16 = (p/4)² and 16/p² = (4/p)².
2. The middle term is negative, so test with a minus: 2 × (p/4) × (4/p) = 2 — this matches the −2.

Answer(p/4 − 4/p)²

(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc

1. The three square terms are 9a² = (3a)², 4b² = (2b)², c² = (c)².
2. Test the three cross terms of (3a − 2b + c)²: 2×3a×(−2b) = −12ab, 2×3a×c = 6ac, 2×(−2b)×c = −4bc — all three match the given terms.

Answer(3a − 2b + c)²

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(i) (p + 3q + 7r)²

1. Match with a + b + c: a = p, b = 3q, c = 7r.
2. Squares: a² = p², b² = 9q², c² = 49r².
3. Cross terms: 2ab = 6pq, 2bc = 42qr, 2ca = 14pr.
4. Add all six pieces.

Answerp² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)²

1. Match with a + b + c: a = 3x, b = −2y, c = 4z.
2. Squares: a² = 9x², b² = 4y², c² = 16z².
3. Cross terms: 2ab = −12xy, 2bc = −16yz, 2ca = 24xz.
4. Add all six pieces.

Answer9x² + 4y² + 16z² − 12xy − 16yz + 24xz

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1. Expand each bracket with (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
2. (a + b − c)² = a² + b² + c² + 2ab − 2bc − 2ca.
3. (a − b + c)² = a² + b² + c² − 2ab − 2bc + 2ca.
4. (a − b − c)² = a² + b² + c² − 2ab + 2bc − 2ca.
5. Add the three. The squares give 3a² + 3b² + 3c². The ab terms give 2ab − 2ab − 2ab = −2ab. The bc terms give −2bc − 2bc + 2bc = −2bc. The ca terms give −2ca + 2ca − 2ca = −2ca.
6. So the left side is 3a² + 3b² + 3c² − 2ab − 2bc − 2ca, which is not the same as 2a² + 2b² + 2c².
7. One example settles it: at a = 1, b = 0, c = 0 the left side is 1 + 1 + 1 = 3, but the right side is 2.

AnswerNo, it is not an identity. For example, at a = 1, b = 0, c = 0 the left side is 3 and the right side is 2.

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(i) s² − 11s + 24 = (____)(____)

1. We need two numbers whose product is 24 and whose sum is −11.
2. The product is positive and the sum is negative, so both numbers are negative: −8 and −3 (−8 × −3 = 24, −8 + (−3) = −11).
3. So s² − 11s + 24 = (s − 8)(s − 3).

Answer(s − 8)(s − 3)

(ii) (____)(x + 1) = 3x² − 4x − 7

1. 3x² can only come from 3x × x, so the missing bracket starts with 3x: (3x + c)(x + 1).
2. The constant term is c × 1 = −7, so c = −7.
3. Check the middle term: 3x × 1 + (−7) × x = 3x − 7x = −4x — this matches.

Answer(3x − 7)(x + 1) = 3x² − 4x − 7, so the blank is 3x − 7

(iii) 10x² − 11x − 6 = (2x − ___)(___ + 2)

1. Multiply the first and last coefficients: 10 × (−6) = −60. Find two numbers with product −60 and sum −11: −15 and 4.
2. Split the middle term: 10x² − 15x + 4x − 6 = 5x(2x − 3) + 2(2x − 3) = (2x − 3)(5x + 2).
3. Compare with (2x − ___)(___ + 2): the first blank is 3 and the second blank is 5x.

Answer10x² − 11x − 6 = (2x − 3)(5x + 2), so the blanks are 3 and 5x

(iv) 6x² + 7x + 2 = (____)(____)

1. Multiply the first and last coefficients: 6 × 2 = 12. Find two numbers with product 12 and sum 7: 3 and 4.
2. Split the middle term: 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (2x + 1)(3x + 2).
3. Check: (2x + 1)(3x + 2) = 6x² + 4x + 3x + 2 = 6x² + 7x + 2 — this matches.

Answer(2x + 1)(3x + 2)

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(i) (41)²

1. 41 = 40 + 1, so use (a + b)² = a² + 2ab + b².
2. (40 + 1)² = 1600 + 80 + 1.

Answer1681

(ii) (27)²

1. 27 = 30 − 3, so use (a − b)² = a² − 2ab + b².
2. (30 − 3)² = 900 − 180 + 9.

Answer729

(iii) (23 × 17)

1. 23 = 20 + 3 and 17 = 20 − 3, so use (a + b)(a − b) = a² − b².
2. 23 × 17 = 20² − 3² = 400 − 9.

Answer391

(iv) (135)²

1. 135 = 130 + 5, so use (a + b)² = a² + 2ab + b².
2. (130 + 5)² = 16900 + 1300 + 25.

Answer18225

(v) (97)²

1. 97 = 100 − 3, so use (a − b)² = a² − 2ab + b².
2. (100 − 3)² = 10000 − 600 + 9.

Answer9409

(vi) (18 × 29)

1. 18 = 20 − 2 and 29 = 20 + 9, so use (x + a)(x + b) = x² + (a + b)x + ab with x = 20, a = −2, b = 9.
2. 18 × 29 = 20² + (−2 + 9) × 20 + (−2) × 9 = 400 + 140 − 18.

Answer522

(vii) (34 × 43)

1. 34 = 40 − 6 and 43 = 40 + 3, so use (x + a)(x + b) = x² + (a + b)x + ab with x = 40, a = −6, b = 3.
2. 34 × 43 = 40² + (−6 + 3) × 40 + (−6) × 3 = 1600 − 120 − 18.

Answer1462

(viii) (205)²

1. 205 = 200 + 5, so use (a + b)² = a² + 2ab + b².
2. (200 + 5)² = 40000 + 2000 + 25.

Answer42025

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(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc

1. The squares are 9a² = (3a)², b² = (b)², 4c² = (2c)².
2. Test the cross terms of (3a − b + 2c)²: 2×3a×(−b) = −6ab, 2×3a×2c = 12ac, 2×(−b)×2c = −4bc — all match.

Answer(3a − b + 2c)²

(ii) 16s² + 25t² − 40st

1. The ends are 16s² = (4s)² and 25t² = (5t)².
2. Middle term is negative, test with −5t: 2×4s×(−5t) = −40st — matches.

Answer(4s − 5t)²

(iii) r² − r − 42

1. Need two numbers with sum −1 and product −42: −7 and 6 (−7+6 = −1, −7×6 = −42).
2. So r² − r − 42 = (r − 7)(r + 6).

Answer(r − 7)(r + 6)

(iv) 49g² + 14gh + h²

1. The ends are 49g² = (7g)² and h² = (h)².
2. Test the middle term: 2×7g×h = 14gh — matches.

Answer(7g + h)²

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw

1. The squares are 64u² = (8u)², 121v² = (11v)², 4w² = (2w)².
2. Test the cross terms of (8u − 11v − 2w)²: 2×8u×(−11v) = −176uv, 2×8u×(−2w) = −32uw, 2×(−11v)×(−2w) = 44vw — all three match.

Answer(8u − 11v − 2w)²

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(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)

1. Factor the top: 3p² − 3pq − 18q² = 3(p − 3q)(p + 2q).
2. Factor the bottom: p² + 3pq − 10q² = (p − 2q)(p + 5q).
3. The top and bottom have no bracket in common, so nothing cancels — this is already the simplest form.

Answer3(p − 3q)(p + 2q) / [(p − 2q)(p + 5q)]

(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²)

1. The top is a perfect cube: n³ − 3n²m + 3nm² − m³ = (n − m)³.
2. The bottom is 5 times a perfect square: 5m² − 10mn + 5n² = 5(m − n)² = 5(n − m)².
3. Cancel two factors of (n − m): (n − m)³ / [5(n − m)²] = (n − m)/5.

Answer(n − m)/5

(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)

1. The top matches a³ + b³ + c³ − 3abc with a = w, b = −v, c = x, which factors as (w − v + x)(w² + v² + x² + wv + vx − wx).
2. The bottom is (w − v + x)², since its three cross terms −2wv, +2wx, −2vx match 2×w×(−v), 2×w×x, 2×(−v)×x.
3. Cancel one (w − v + x): the answer is (w² + v² + x² + wv + vx − wx) / (w − v + x).

Answer(w² + v² + x² + wv + vx − wx) / (w − v + x)

(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)

1. The top is a perfect square: 4y² − 20yz + 25z² = (5z − 2y)².
2. The bottom is a difference of squares: 25z² − 4y² = (5z − 2y)(5z + 2y).
3. Cancel one (5z − 2y): the answer is (5z − 2y)/(5z + 2y).

Answer(5z − 2y)/(5z + 2y)

(v) (x² + x − 6)(x² − 7x + 12)/((x² − 6x + 8)(x² − 9))

1. Factor every quadratic: x²+x−6 = (x+3)(x−2), x²−7x+12 = (x−3)(x−4), x²−6x+8 = (x−2)(x−4), x²−9 = (x−3)(x+3).
2. Top = (x+3)(x−2)(x−3)(x−4). Bottom = (x−2)(x−4)(x−3)(x+3).
3. Top and bottom have exactly the same four brackets, so they cancel completely.

Answer1

(vi) (p⁴ − 16)/(p² − 4p + 4)

1. The top is a difference of squares twice over: p⁴ − 16 = (p² − 4)(p² + 4) = (p − 2)(p + 2)(p² + 4).
2. The bottom is a perfect square: p² − 4p + 4 = (p − 2)².
3. Cancel one (p − 2): the answer is (p + 2)(p² + 4)/(p − 2).

Answer(p + 2)(p² + 4)/(p − 2)

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(i) (−3x + 4)²

1. Use (a + b)² with a = −3x, b = 4.
2. a² = 9x², 2ab = −24x, b² = 16.

Answer9x² − 24x + 16

(ii) (2s + 7)(2s − 7)

1. Use (a + b)(a − b) = a² − b² with a = 2s, b = 7.
2. = 4s² − 49.

Answer4s² − 49

(iii) (p² + 1/2)(p² − 1/2)

1. Use (a + b)(a − b) = a² − b² with a = p², b = 1/2.
2. = p⁴ − 1/4.

Answerp⁴ − 1/4

(iv) (2n + 7)(2n − 7)

1. Use (a + b)(a − b) = a² − b² with a = 2n, b = 7.
2. = 4n² − 49.

Answer4n² − 49

(v) (s − 2t)(s² + 2st + 4t²)

1. This matches a³ − b³ = (a − b)(a² + ab + b²) with a = s, b = 2t.
2. = s³ − 8t³.

Answers³ − 8t³

(vi) (1/(2r) − 4r)²

1. Use (a − b)² with a = 1/(2r), b = 4r.
2. a² = 1/(4r²), 2ab = 4, b² = 16r².

Answer16r² − 4 + 1/(4r²)

(vii) (−3m + 4k − l)²

1. Use (a + b + c)² with a = −3m, b = 4k, c = −l.
2. Squares: 9m², 16k², l². Cross terms: 2ab = −24km, 2bc = −8kl, 2ca = 6lm.
3. Add all six pieces.

Answer16k² + l² + 9m² − 8kl − 24km + 6lm

(viii) (x − (1/3)y)³

1. Use (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = x, b = y/3.
2. a³ = x³, 3a²b = x²y, 3ab² = xy²/3, b³ = y³/27.

Answerx³ − x²y + xy²/3 − y³/27

(ix) (7/2 k − 2/3 m)³

1. Use (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 7k/2, b = 2m/3.
2. a³ = 343k³/8, 3a²b = 49k²m/2, 3ab² = 14km²/3, b³ = 8m³/27.

Answer343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27

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(i) 17 × 21

1. 17 = 19 − 2, 21 = 19 + 2, so 17 × 21 = 19² − 2² = 361 − 4.

Answer357

(ii) 104 × 96

1. 104 = 100 + 4, 96 = 100 − 4, so 104 × 96 = 100² − 4² = 10000 − 16.

Answer9984

(iii) 24 × 16

1. 24 = 20 + 4, 16 = 20 − 4, so 24 × 16 = 20² − 4² = 400 − 16.

Answer384

(iv) 147³

1. 147 = 150 − 3, use (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 150, b = 3.
2. 150³ − 3×150²×3 + 3×150×3² − 3³ = 3375000 − 202500 + 4050 − 27.

Answer3176523

(v) 199³

1. 199 = 200 − 1, use (a − b)³ with a = 200, b = 1.
2. 200³ − 3×200²×1 + 3×200×1² − 1³ = 8000000 − 120000 + 600 − 1.

Answer7880599

(vi) 127³

1. 127 = 130 − 3, use (a − b)³ with a = 130, b = 3.
2. 130³ − 3×130²×3 + 3×130×3² − 3³ = 2197000 − 152100 + 3510 − 27.

Answer2048383

(vii) (−107)³

1. (−107)³ = −(107)³. 107 = 100 + 7, use (a + b)³ with a = 100, b = 7.
2. 100³ + 3×100²×7 + 3×100×7² + 7³ = 1000000 + 210000 + 14700 + 343 = 1225043, so (−107)³ = −1225043.

Answer−1225043

(viii) (−299)³

1. (−299)³ = −(299)³. 299 = 300 − 1, use (a − b)³ with a = 300, b = 1.
2. 300³ − 3×300²×1 + 3×300×1² − 1³ = 27000000 − 270000 + 900 − 1 = 26730899, so (−299)³ = −26730899.

Answer−26730899

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(i) 4y² + 1 + 1/(16y²)

1. 4y² = (2y)² and 1/(16y²) = (1/(4y))².
2. Middle term: 2 × 2y × 1/(4y) = 1, which matches.
3. So the expression is a² + 2ab + b² = (a + b)² with a = 2y, b = 1/(4y).

Answer(2y + 1/(4y))²

(ii) 9m² − 1/(25n²)

1. 9m² = (3m)² and 1/(25n²) = (1/(5n))².
2. Use a² − b² = (a − b)(a + b).

Answer(3m − 1/(5n))(3m + 1/(5n))

(iii) 27b³ − 1/(64b³)

1. 27b³ = (3b)³ and 1/(64b³) = (1/(4b))³.
2. Use a³ − b³ = (a − b)(a² + ab + b²) with a = 3b, b = 1/(4b).
3. a² = 9b², ab = 3b × 1/(4b) = 3/4, b² = 1/(16b²).

Answer(3b − 1/(4b))(9b² + 3/4 + 1/(16b²))

(iv) x² + 5x/6 + 1/6

1. We need two numbers whose sum is 5/6 and whose product is 1/6.
2. 1/2 + 1/3 = 5/6 and 1/2 × 1/3 = 1/6.
3. So x² + 5x/6 + 1/6 = (x + 1/2)(x + 1/3).

Answer(x + 1/2)(x + 1/3)

(v) 27u³ − 1/125 − 27u²/5 + 9u/25

1. Rearrange: 27u³ − 27u²/5 + 9u/25 − 1/125.
2. Compare with (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 3u, b = 1/5.
3. 3a²b = 3 × 9u² × 1/5 = 27u²/5 and 3ab² = 3 × 3u × 1/25 = 9u/25. Both match.

Answer(3u − 1/5)³

(vi) 64y³ + z³/125

1. 64y³ = (4y)³ and z³/125 = (z/5)³.
2. Use a³ + b³ = (a + b)(a² − ab + b²) with a = 4y, b = z/5.

Answer(4y + z/5)(16y² − 4yz/5 + z²/25)

(vii) p³ + 27q³ + r³ − 9pqr

1. Take a = p, b = 3q, c = r. Then a³ + b³ + c³ = p³ + 27q³ + r³ and 3abc = 9pqr.
2. Use a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Answer(p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − rp)

(viii) 9m² − 12m + 4

1. 9m² − 12m + 4 = (3m)² − 2 × 3m × 2 + 2².
2. This is a² − 2ab + b² = (a − b)² with a = 3m, b = 2.

Answer(3m − 2)²

(ix) 9x³ − (8/3)y³ + z³/3 + 6xyz

1. Take out 1/3: 9x³ − (8/3)y³ + z³/3 + 6xyz = (1/3)(27x³ − 8y³ + z³ + 18xyz).
2. Take a = 3x, b = −2y, c = z. Then a³ + b³ + c³ = 27x³ − 8y³ + z³ and −3abc = −3 × 3x × (−2y) × z = 18xyz.
3. Use a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
4. a² + b² + c² = 9x² + 4y² + z²; −ab = 6xy, −bc = 2yz, −ca = −3zx.

Answer(1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3zx)

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy

1. As printed: 4x² + 9y² + 36z² + 12xz + 36yz + 24xy.
2. The squares point to a = 2x, b = 3y, c = 6z, and (2x + 3y + 6z)² = 4x² + 9y² + 36z² + 12xy + 36yz + 24xz.
3. The printed line has 24xy and 12xz, the other way round, so it is not a perfect square. It cannot be factorised at all: a computer check confirms it has no factors. This is a misprint.
4. The book does not say which correction is meant, and two single swaps each give a perfect square.
5. Swap 12 and 24 (12xy + 24xz): 4x² + 9y² + 36z² + 12xy + 36yz + 24xz = (2x + 3y + 6z)².
6. Swap 9 and 36 (36y² + 9z²): 4x² + 36y² + 9z² + 12xz + 36yz + 24xy = (2x + 6y + 3z)², since 2 × 2x × 6y = 24xy, 2 × 6y × 3z = 36yz and 2 × 3z × 2x = 12xz.

AnswerAs printed: it cannot be factorised (a misprint). Read with 12 and 24 swapped: (2x + 3y + 6z)². Read with 9 and 36 swapped: (2x + 6y + 3z)².

(xi) 27u³ − 1/216 − 9u²/2 + u/4

1. Rearrange: 27u³ − 9u²/2 + u/4 − 1/216.
2. Compare with (a − b)³ with a = 3u, b = 1/6: a³ = 27u³, b³ = 1/216.
3. 3a²b = 3 × 9u² × 1/6 = 9u²/2 and 3ab² = 3 × 3u × 1/36 = u/4. Both match.

Answer(3u − 1/6)³

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Open NCERT p. 89Checked by computer

(i) (4x² + 4x + 1)/(4x² − 1)

1. Top: 4x² + 4x + 1 = (2x + 1)².
2. Bottom: 4x² − 1 = (2x − 1)(2x + 1).
3. Cancel one (2x + 1): (2x + 1)/(2x − 1).

Answer(2x + 1)/(2x − 1)

(ii) 9(3a³ − 24b³)/(9a² − 36b²)

1. Top: 9(3a³ − 24b³) = 27(a³ − 8b³) = 27(a − 2b)(a² + 2ab + 4b²).
2. Bottom: 9a² − 36b² = 9(a − 2b)(a + 2b).
3. Cancel 9 and one (a − 2b): 3(a² + 2ab + 4b²)/(a + 2b).

Answer3(a² + 2ab + 4b²)/(a + 2b)

(iii) (s³ + 125t³)/(s² − 2st − 35t²)

1. Top: s³ + 125t³ = (s + 5t)(s² − 5st + 25t²).
2. Bottom: s² − 2st − 35t² = (s − 7t)(s + 5t).
3. Cancel one (s + 5t): (s² − 5st + 25t²)/(s − 7t).

Answer(s² − 5st + 25t²)/(s − 7t)

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Open NCERT p. 89Checked by computerAnswers can differ: one example

(i) 25a² − 30ab + 9b²

1. Area = length × breadth, so we factorise the area.
2. 25a² − 30ab + 9b²: the ends are (5a)² and (3b)², and 2 × 5a × 3b = 30ab, with a minus sign — so it is a perfect square, (5a − 3b)².
3. So one possible answer is length = 5a − 3b and breadth = 5a − 3b: the rectangle is then a square (this needs 5a > 3b so the side is positive).
4. The question asks for possible expressions, so other splits of the same area also work, for example length = 5(5a − 3b) and breadth = (5a − 3b)/5.

Answerlength = 5a − 3b, breadth = 5a − 3b

(ii) 36s² − 49t²

1. 36s² − 49t² = (6s)² − (7t)², a difference of squares.
2. So 36s² − 49t² = (6s + 7t)(6s − 7t).
3. One possible answer is length = 6s + 7t and breadth = 6s − 7t (this needs 6s > 7t). Other splits, such as 2(6s + 7t) and (6s − 7t)/2, also give the same area.

Answerlength = 6s + 7t, breadth = 6s − 7t

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Open NCERT p. 89Checked by computerAnswers can differ: one example

(i) 6a² − 24b²

1. Volume = length × breadth × height, so we split the volume into three factors.
2. Take out the common factor 6: 6a² − 24b² = 6(a² − 4b²).
3. a² − 4b² = a² − (2b)² = (a − 2b)(a + 2b), so 6a² − 24b² = 6(a − 2b)(a + 2b).
4. One possible answer is length = 6, breadth = a − 2b, height = a + 2b (this needs a > 2b). The question asks for possible expressions, so other splits also work, for example 2, 3(a − 2b) and a + 2b.

Answerlength = 6, breadth = a − 2b, height = a + 2b

(ii) 3ps² − 15ps + 12p

1. Take out the common factor 3p: 3ps² − 15ps + 12p = 3p(s² − 5s + 4).
2. For s² − 5s + 4, find two numbers with product 4 and sum −5: −4 and −1. So s² − 5s + 4 = (s − 4)(s − 1).
3. So 3ps² − 15ps + 12p = 3p(s − 4)(s − 1).
4. One possible answer is length = 3p, breadth = s − 4, height = s − 1 (this needs s > 4). Other splits also work, for example 3, p(s − 4) and s − 1.

Answerlength = 3p, breadth = s − 4, height = s − 1

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Open NCERT p. 90Checked by computer

1. With the path of width s all around, the outer square has side (40 + 2s) metres.
2. Area of the path = area of outer square − area of playground = (40 + 2s)² − 40².
3. This is a difference of squares: (40 + 2s − 40)(40 + 2s + 40) = (2s)(80 + 2s).
4. = 160s + 4s².

Answer4s² + 160s square metres

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Open NCERT p. 90Checked by computer

1. Let the number be x. Then x + 1/x = 10/3.
2. Multiply both sides by 3x: 3x² + 3 = 10x, so 3x² − 10x + 3 = 0.
3. Split the middle term: need two numbers with sum −10 and product 9 (3×3): −9 and −1.
4. 3x² − 9x − x + 3 = 0 → 3x(x − 3) − 1(x − 3) = 0 → (3x − 1)(x − 3) = 0.
5. So x = 3 or x = 1/3 — and indeed each is the reciprocal of the other.

AnswerThe number is 3 or 1/3.

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Open NCERT p. 90Checked by computer

1. Area = length × width, so length = (2x² + 7x + 3)/(2x + 1).
2. Factor the area by splitting the middle term: need two numbers with sum 7 and product 6 (2×3): 1 and 6.
3. 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
4. So length = (2x + 1)(x + 3) / (2x + 1) = x + 3.

Answerx + 3 hastas

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Open NCERT p. 90One way to think about it

1. Since x − 2 is a factor, px² + 5x + r = (x − 2) × (another factor). At x = 2 the right side is 0 × (…) = 0, so the left side is 0 too: p(2)² + 5(2) + r = 0, that is 4p + r = −10.
2. In the same way x − 1/2 is a factor, so putting x = 1/2 gives p(1/2)² + 5(1/2) + r = 0, that is p/4 + 5/2 + r = 0. Multiply by 4: p + 4r = −10.
3. Subtract the second equation from the first: (4p + r) − (p + 4r) = −10 − (−10) = 0, so 3p − 3r = 0.
4. Divide by 3: p = r, as required.
5. (In fact 4p + p = −10 gives p = r = −2, and −2x² + 5x − 2 = −(x − 2)(2x − 1) does have both factors, since 2x − 1 = 2(x − 1/2).)

In shortPutting x = 2 and x = 1/2 gives 4p + r = −10 and p + 4r = −10; subtracting gives 3p − 3r = 0, so p = r (in fact p = r = −2).

Open NCERT p. 90One way to think about it

1. Use the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
2. Also, a² + b² + c² = (a + b + c)² − 2(ab + bc + ca).
3. So a² + b² + c² − ab − bc − ca = (a + b + c)² − 3(ab + bc + ca).
4. Substitute a + b + c = 5 and ab + bc + ca = 10: (5)² − 3(10) = 25 − 30 = −5.
5. So a³ + b³ + c³ − 3abc = (a + b + c) × (−5) = 5 × (−5) = −25, as required.

In shorta³ + b³ + c³ − 3abc = 5 × (5² − 3×10) = 5 × (−5) = −25.

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Open NCERT p. 90One way to think about it

1. Factor: n³ − n = n(n² − 1) = n(n − 1)(n + 1), using a² − b² = (a−b)(a+b) on n²−1.
2. So n³ − n is the product of three consecutive whole numbers: (n − 1), n, and (n + 1).
3. Among any three consecutive whole numbers, at least one is a multiple of 3 (every third number is), so the product is divisible by 3.
4. Among any three consecutive whole numbers, at least one is even (numbers alternate odd/even), so the product is divisible by 2.
5. A number divisible by both 2 and 3 (which share no common factor) is divisible by 2 × 3 = 6.
6. So n³ − n = (n − 1)n(n + 1) is always divisible by 6.

In shortn³ − n = (n − 1)n(n + 1), a product of three consecutive integers, which always contains a multiple of 2 and a multiple of 3 — so it is always divisible by 6.

Open NCERT p. 90Checked by computer

(i) x³ + y³ − 12xy + 64

1. Write 64 = 4³, so the expression is x³ + y³ + 4³ − 3(x)(y)(4), which matches a³ + b³ + c³ − 3abc with a = x, b = y, c = 4.
2. This factors as (x + y + 4)(x² + y² + 16 − xy − 4y − 4x).
3. Since x + y = −4, the first bracket (x + y + 4) is 0, so the whole expression is 0.

Answer0

(ii) x³ − 8y³ − 36xy − 216

1. Write −8y³ = (−2y)³ and −216 = (−6)³, so the expression is x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6), which matches a³ + b³ + c³ − 3abc with a = x, b = −2y, c = −6.
2. This factors as (x − 2y − 6)(x² + 4y² + 36 + 2xy + 6x − 12y).
3. Since x = 2y + 6, the first bracket (x − 2y − 6) is 0, so the whole expression is 0.

Answer0

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