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 
(ii) ab, a – b 
(iii) a + b, 7a²b² 
(iv) a² – 9, 4a 
(v) pq + qr + rp, 0
Ans:

Q2. Complete the table.

Ans: We have:

Q3. Find the product.
(i) (a²) × (2a²²) × (4a²⁶)
(ii)
(iii)
(iv) x × x² × x³ × x⁴
Ans: 
(i) (a²) × (2a²²) × (4a²⁶) = 2 × 4 × a² × a²² × a²⁶ = 8a⁵⁰


(iv) x × x² × x³ × x⁴ = x¹⁰

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 and 
(iii) a = –1. 
Ans: 
(a) 3x (4x – 5) + 3
= 3x ( 4x) – 3x( 5) +3
= 12x² – 15x + 3
(i) Putting x=3 in the equation we gets 12x² – 15x + 3 =12(3²) – 15 (3) +3
= 108 – 45 + 3
= 66
(ii) Putting x=1/2 in the equation we get
12x² – 15x + 3 = 12 (1/2)² – 15 (1/2) + 3
= 12 (1/4) – 15/2 +3
= 3 – 15/2 + 3
= 6- 15/2
= (12- 15 ) /2
= -3/2
(b) a(a² + a + 1) + 5
= a x a² + a x a + a x 1 + 5 =a³+a²+a+ 5
(i) putting a=0 in the equation we get 0³ + 0² + 0 + 5 = 5
(ii) putting a=1 in the equation we get 1³ + 1² + 1+5 = 1 + 1 + 1 + 5 = 8
(iii) Putting a = -1 in the equation we get (-1)³+ (-1)² + (-1) + 5 = -1 + 1 – 1 + 5 = 4

Q5: (a) Add: p(p – q), q(q – r) and r(r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 41(10n + 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(–a + b + c)
Ans:
(a) p ( p – q) + q ( q – r) + r ( r – p)
= (p² – pq) + (q² – qr) + (r² – pr)
= p² + q² + r² – pq – qr – pr
(b) 2x (z – x – y) + 2y (z – y – x)
= (2xz – 2x² – 2xy) + (2yz – 2y² – 2xy)
= 2xz – 4xy + 2yz – 2x² – 2y²
(c) 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)
= (40ln – 12lm + 8l²) – (3l² – 12lm + 15ln)
= 40ln – 12lm + 8l² – 3l² +12lm -15 ln
= 25 ln + 5l²
(d) 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))
= (-4ac + 4bc + 4c²) – (3a² + 3ab + 3ac – ( 2ab – 2b² + 2bc ))
=-4ac + 4bc + 4c² – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)
= -4ac + 4bc + 4c² – 3a² – 3ab – 3ac +2ab – 2b² + 2bc
= -7ac + 6bc + 4c² – 3a² – ab – 2b²

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)(4x – 3)
= 2x × 4x – 2x × 3 + 5 × 4x – 5 × 3
= 8x² – 6x + 20x -15
= 8x² + 14x -15
(ii) ( y – 8)(3y – 4)
= y × 3y – 4y – 8 × 3y + 32
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32
(iii) (2.5l – 0.5m)(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) (x + 5)
= ax + 5a + 3bx + 15b
(v) (2pq + 3q²) (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)
(ii) (x + 7y) (7x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²) (2p + q)
Ans: 
(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
(ii) (a²+ 5) (b³+ 3) + 5
(iii) (t + s²) (t² – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y) (x²– xy + y²)
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c) (a + b – c)
Ans: 
(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