NCERT Solutions for Class 8 Maths - Algebraic Expressions and Identities - 2 (Exercise 8.3 and 8.4)

Exercise 8.3

Q1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r 
Ans: (4p) × (q  +  r) = (4p × q) + (4p × r) =  4pq + 4pr

(ii) ab, a – b
Ans: (ab) × (a - b) = (ab × a) + [ab×(-b)] = a²b - ab² 

(iii) a + b, 7a²b² 
Ans: (a  +  b) × (7a² b²) = (a × 7a²b²) + (b × 7a²b²) =  7a³b² +  7a²b³ 

(iv) a² – 9, 4a 
Ans: (a² - 9) × (4a) = (a² × 4a) + ( -  9) × (4a) =  4a³ -  36a

(v) pq + qr + rp, 0
Ans: (pq  +  qr  +  rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) =  0

Q2. Complete the table.

Ans: The table can be completed as follows. 

Q3. Find the product.
(i) (a²) × (2a²²) × (4a²⁶)
(ii)
(iii)
(iv) x × x² × x³ × x⁴
Ans: 

Q4. (a) Simplify 3x(4x – 5) + 3 and find its values for
(i) x = 3 and 
(ii) x = 1/2 
(b) Simplify a(a² + a + 1) + 5 and find its value for 
(i) a = 0, 
(ii) a = 1
(iii) a = –1. 
Ans: 
(a) 3x (4x − 5) + 3 = 12x² − 15x + 3
(i) For x = 3,
=12x² - 15x + 3
=12(3)² - 15(3) + 3
= 108 - 45 + 3 
= 66

(ii) For x = 1/2
=12x² - 15x + 3

(b) a(a² + a + 1) + 5 = a³ + a² + a + 5
(i) For a = 0, a³ + a² + a + 5 = 0 + 0 + 0 + 5 = 5
(ii) For a = 1, a³ + a² + a + 5 =(1)³ +(1)² + 1 + 5
= 1 + 1 + 1 + 5 = 8
(iii) For a = - 1, a³ + a² + a + 5 =(-1)³ +(-1)² + (-1) + 5
=- 1 + 1 - 1 + 5 = 4

Q5: Solve the following 
(a) Add: p (p - q), q (q - r) and r (r - p)
Ans: First expression = p (p - q) = p² - pq
Second expression = q (q - r) = q² - qr
Third expression = r (r - p) = r² - pr
Adding the three expressions, we obtain

Therefore, the sum is p² - pq + q² - qr  + r² - pr

(b) Add: 2x(z – x – y) and 2y(z – y – x)
Ans: First expression  =  2x (z  -  x  -  y) =  2xz - 2x² - 2xy
Second expression = 2y (z - y - x) = 2yz - 2y² - 2yx
Adding the two expressions, we obtain

Therefore, the sum is 2xz - 2x² - 4xy  + 2yz  - 2y²

(c) Subtract: 3l(l – 4m + 5n) from 41(10n + 3m + 2l)
Ans: 3l (l - 4m + 5n) = 3l² - 12lm + 15ln
= 4l (10n - 3m + 2l) = 40ln - 12lm + 8l²
Subtracting these expressions, we obtain
8l² - 12lm + 40ln
3l² - 12lm + 15ln
(-) (+) (-)

Therefore, the result is 5l² + 25ln

(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(–a + b + c)
Ans: Expand both expressionsPerform subtractionTherefore, the result is

Exercise 8.4

Q1. Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q² ) and (3pq – 2q²)
(vi)
Ans: 
(i) (2x + 5) and (4x – 3)
= 2x × 4x – 2x × 3 + 5 × 4x – 5 × 3
= 8x² – 6x + 20x -15
= 8x² + 14x -15

(ii) ( y – 8)and (3y – 4)
= y × 3y – 4y – 8 × 3y + 32
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32

(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
= 2.5l × 2.5 l + 2.5l × 0.5m – 0.5m × 2.5l – 0.5m × 0.5m
= 6.25l² + 1.25 lm – 1.25 lm – 0.25 m²
= 6.25l²– 0.25 m²

(iv) (a + 3b) and (x + 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) and (3pq – 2q²)
= 2pq × 3pq – 2pq × 2q² + 3q² × 3pq – 3q² × 2q²
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 5pq³ – 6q⁴
(vi)

= 3a⁴ – 2a² b² + 12 a²  b² – 8b4
= 3a⁴ + 10a²  b² – 8b4

Q2. Find the product.
(i) (5 – 2x) (3 + x)
= 5 (3 + x) – 2x (3 + x)
=15 + 5x – 6x – 2x²
= 15 – x -2 x ²

(ii) (x + 7y) (7x – y)
= x(7x-y) + 7y ( 7x-y)
=7x² – xy + 49xy – 7y²
= 7x² – 7y² + 48xy

(iii) (a²+ b) (a + b²)
= a²  (a + b²) + b(a + b²)
= a³ + a²b² + ab + b³
= a³ + b³ + a²b² + ab

(iv) (p²– q²) (2p + q)
= p² (2p + q) – q² (2p + q)
=2p³ + p²q – 2pq² – q³
= 2p³ – q³ + p²q – 2pq²

Q3. Simplify.
(i) (x²– 5) (x + 5) + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x
(ii) (a²+ 5) (b³+ 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 5b³ + 3a² + 20

(iii) (t + s²)(t² – s)
= t (t² – s) + s²(t² – s)
= t³ – st + s²t² – s³
= t³ – s³ – st + s²t²

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
=(ac – ad + bc – bd) + (ac + ad – bc – bd) + (2ac + 2bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²
= 3x² + 4xy – y²

(vi) (x + y)(x²– xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x² + 6xy + 4.5x – 6xy – 16y² – 12y – 4.5x + 12y 
= 2.25x² – 16y²

(viii) (a + b + c)(a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + b² – c² + 2ab