5. Expansion Formulae
Solutions to Practice Sets 5.1, 5.2, 5.3 and 5.4, with every step of the expansion shown. Maharashtra State Board (Balbharati), Standard Eight Mathematics.
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Practice Set 5.1
Formula: (x + a)(x + b) = x2 + (a + b)x + ab
(1) (a + 2)(a − 1)
(a + 2)(a − 1)
= a2 + (2 − 1)a + (2)(−1)
Answer: a2 + a − 2
(2) (m − 4)(m + 6)
(m − 4)(m + 6)
= m2 + (−4 + 6)m + (−4)(6)
Answer: m2 + 2m − 24
(3) (p + 8)(p − 3)
(p + 8)(p − 3)
= p2 + (8 − 3)p + (8)(−3)
Answer: p2 + 5p − 24
(4) (13 + x)(13 − x)
(13 + x)(13 − x)
= 132 + (x − x)13 + (x)(−x)
= 169 + 0 − x2
Here the common term is 13. This is also (a + b)(a − b) = a² − b².
Answer: 169 − x2
(5) (3x + 4y)(3x + 5y)
(3x + 4y)(3x + 5y)
= (3x)2 + (4y + 5y)(3x) + (4y)(5y)
= 9x2 + 9y × 3x + 20y2
Here the common term is 3x.
Answer: 9x2 + 27xy + 20y2
(6) (9x − 5t)(9x + 3t)
(9x − 5t)(9x + 3t)
= (9x)2 + (−5t + 3t)(9x) + (−5t)(3t)
= 81x2 + (−2t)(9x) − 15t2
Answer: 81x2 − 18xt − 15t2
(7) (m + 23)(m − 73)
(m + 23)(m − 73)
= m2 + (23 − 73)m + (23)(−73)
= m2 + (−53)m − 149
Answer: m2 − 53m − 149
(8) (x + 1x)(x − 1x)
(x + 1x)(x − 1x)
= x2 + (1x − 1x)x + (1x)(−1x)
= x2 + 0 − 1x2
Answer: x2 − 1x2
(9) (1y + 4)(1y − 9)
(1y + 4)(1y − 9)
= (1y)2 + (4 − 9)(1y) + (4)(−9)
= 1y2 + (−5)(1y) − 36
Here the common term is 1/y.
Answer: 1y2 − 5y − 36
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Practice Set 5.2
Formula: (a + b)3 = a3 + 3a2b + 3ab2 + b3
(1) (k + 4)3
(k + 4)3
= (k)3 + 3 × k2 × 4 + 3 × k × (4)2 + (4)3
= k3 + 12k2 + 3 × k × 16 + 64
Answer: k3 + 12k2 + 48k + 64
(2) (7x + 8y)3
(7x + 8y)3
= (7x)3 + 3(7x)2(8y) + 3(7x)(8y)2 + (8y)3
= 343x3 + 3 × 49x2 × 8y + 3 × 7x × 64y2 + 512y3
Answer: 343x3 + 1176x2y + 1344xy2 + 512y3
(3) (7 + m)3
(7 + m)3
= (7)3 + 3(7)2(m) + 3(7)(m)2 + (m)3
= 343 + 3 × 49 × m + 21m2 + m3
Answer: 343 + 147m + 21m2 + m3
(4) (52)3
(52)3
= (50 + 2)3
= (50)3 + 3 × (50)2 × 2 + 3 × 50 × (2)2 + (2)3
= 125000 + 15000 + 600 + 8
Answer: 140608
(5) (101)3
(101)3
= (100 + 1)3
= (100)3 + 3 × (100)2 × 1 + 3 × 100 × (1)2 + (1)3
= 1000000 + 30000 + 300 + 1
Answer: 1030301
(6) (x + 1x)3
(x + 1x)3
= x3 + 3 × x2 × 1x + 3 × x × (1x)2 + (1x)3
= x3 + 3x + 3x × 1x2 + 1x3
Answer: x3 + 3x + 3x + 1x3
(7) (2m + 15)3
(2m + 15)3
= (2m)3 + 3(2m)2(15) + 3(2m)(15)2 + (15)3
= 8m3 + 3 × 4m2 × 15 + 3 × 2m × 125 + 1125
Answer: 8m3 + 12m25 + 6m25 + 1125
(8) (5xy + y5x)3
(5xy + y5x)3
= (5xy)3 + 3(5xy)2(y5x) + 3(5xy)(y5x)2 + (y5x)3
= 125x3y3 + 3 × 25x2y2 × y5x + 3 × 5xy × y225x2 + y3125x3
= 125x3y3 + 3 × 5xy + 3 × y5x + y3125x3
Answer: 125x3y3 + 15xy + 3y5x + y3125x3
Practice Set 5.3
Formula: (a − b)3 = a3 − 3a2b + 3ab2 − b3
(1) (2m − 5)3
(2m − 5)3
= (2m)3 − 3(2m)2(5) + 3(2m)(5)2 − (5)3
= 8m3 − 3 × 4m2 × 5 + 3 × 2m × 25 − 125
Answer: 8m3 − 60m2 + 150m − 125
(2) (4 − p)3
(4 − p)3
= (4)3 − 3(4)2(p) + 3(4)(p)2 − (p)3
= 64 − 3 × 16 × p + 12p2 − p3
Answer: 64 − 48p + 12p2 − p3
(3) (7x − 9y)3
(7x − 9y)3
= (7x)3 − 3(7x)2(9y) + 3(7x)(9y)2 − (9y)3
= 343x3 − 3 × 49x2 × 9y + 3 × 7x × 81y2 − 729y3
Answer: 343x3 − 1323x2y + 1701xy2 − 729y3
(4) (58)3
(58)3
= (60 − 2)3
= (60)3 − 3 × (60)2 × 2 + 3 × 60 × (2)2 − (2)3
= 216000 − 21600 + 720 − 8
Answer: 1,95,112
(5) (198)3
(198)3
= (200 − 2)3
= (200)3 − 3 × (200)2 × 2 + 3 × 200 × (2)2 − (2)3
= 8000000 − 240000 + 2400 − 8
Answer: 77,62,392
(6) (2p − 12p)3
(2p − 12p)3
= (2p)3 − 3(2p)2(12p) + 3(2p)(12p)2 − (12p)3
= 8p3 − 3 × 4p2 × 12p + 3 × 2p × 14p2 − 18p3
Answer: 8p3 − 6p + 32p − 18p3
(7) (1 − 1a)3
(1 − 1a)3
= (1)3 − 3(1)2(1a) + 3(1)(1a)2 − (1a)3
Answer: 1 − 3a + 3a2 − 1a3
(8) (x3 − 3x)3
(x3 − 3x)3
= (x3)3 − 3(x3)2(3x) + 3(x3)(3x)2 − (3x)3
= x327 − 3 × x29 × 3x + 3 × x3 × 9x2 − 27x3
Answer: x327 − x + 9x − 27x3
Short cut used below. Write the two expansions one under the other:
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a − b)3 = a3 − 3a2b + 3ab2 − b3
Adding: (a + b)3 + (a − b)3 = 2a3 + 6ab2
Subtracting: (a + b)3 − (a − b)3 = 6a2b + 2b3
The terms with opposite signs cancel, so only half the work is needed.
(1) (2a + b)3 − (2a − b)3
(2a + b)3 − (2a − b)3
Here the first term is 2a and the second is b. Use the subtracting form.
= 6(2a)2(b) + 2(b)3
= 6 × 4a2 × b + 2b3
Answer: 24a2b + 2b3
(2) (3r − 2k)3 + (3r + 2k)3
(3r − 2k)3 + (3r + 2k)3
Here the first term is 3r and the second is 2k. Use the adding form.
= 2(3r)3 + 6(3r)(2k)2
= 2 × 27r3 + 6 × 3r × 4k2
Answer: 54r3 + 72rk2
(3) (4a − 3)3 − (4a + 3)3
(4a − 3)3 − (4a + 3)3
= −[(4a + 3)3 − (4a − 3)3]
= −[6(4a)2(3) + 2(3)3]
= −[288a2 + 54]
The order is (a − b)³ − (a + b)³, which is the negative of the subtracting form.
Answer: −288a2 − 54
(4) (5x − 7y)3 + (5x + 7y)3
(5x − 7y)3 + (5x + 7y)3
= 2(5x)3 + 6(5x)(7y)2
= 2 × 125x3 + 6 × 5x × 49y2
Answer: 250x3 + 1470xy2
Practice Set 5.4
Formula: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(1) (2p + q + 5)2
(2p + q + 5)2
= (2p)2 + (q)2 + (5)2 + 2(2p)(q) + 2(q)(5) + 2(2p)(5)
Answer: 4p2 + q2 + 25 + 4pq + 10q + 20p
(2) (m + 2n + 3r)2
(m + 2n + 3r)2
= (m)2 + (2n)2 + (3r)2 + 2(m)(2n) + 2(2n)(3r) + 2(m)(3r)
Answer: m2 + 4n2 + 9r2 + 4mn + 12nr + 6mr
(3) (3x + 4y − 5p)2
(3x + 4y − 5p)2
= (3x)2 + (4y)2 + (−5p)2 + 2(3x)(4y) + 2(4y)(−5p) + 2(3x)(−5p)
Answer: 9x2 + 16y2 + 25p2 + 24xy − 40py − 30px
(4) (7m − 3n − 4k)2
(7m − 3n − 4k)2
= (7m)2 + (−3n)2 + (−4k)2 + 2(7m)(−3n) + 2(−3n)(−4k) + 2(7m)(−4k)
Answer: 49m2 + 9n2 + 16k2 − 42mn + 24nk − 56km
Take each term with its sign. The square of a negative term is positive; a product of two negative terms is positive.
(1) (x − 2y + 3)2 + (x + 2y − 3)2
(x − 2y + 3)2 + (x + 2y − 3)2
= [x2 + 4y2 + 9 − 4xy − 12y + 6x] + [x2 + 4y2 + 9 + 4xy − 12y − 6x]
The terms 4xy and 6x cancel.
= 2x2 + 8y2 + 18 − 24y
Answer: 2x2 + 8y2 + 18 − 24y
(2) (3k − 4r − 2m)2 − (3k + 4r − 2m)2
(3k − 4r − 2m)2 − (3k + 4r − 2m)2
= [9k2 + 16r2 + 4m2 − 24kr + 16rm − 12km] − [9k2 + 16r2 + 4m2 + 24kr − 16rm − 12km]
= 9k2 + 16r2 + 4m2 − 24kr + 16rm − 12km − 9k2 − 16r2 − 4m2 − 24kr + 16rm + 12km
Answer: 32rm − 48kr
(3) (7a − 6b + 5c)2 + (7a + 6b − 5c)2
(7a − 6b + 5c)2 + (7a + 6b − 5c)2
= [49a2 + 36b2 + 25c2 − 84ab − 60bc + 70ac] + [49a2 + 36b2 + 25c2 + 84ab − 60bc − 70ac]
The terms 84ab and 70ac cancel.
= 98a2 + 72b2 + 50c2 − 120bc
Answer: 98a2 + 72b2 + 50c2 − 120bc