NCERT Solution Class 8th Maths Chapter – 8 Algebraic Expressions and Identities Exercise 8.4

1. Multiply the binomials.

(i) (2× + 5) and (4× – 3)

Solution:
(i) (2× + 5)(4× – 3)
= 2x × 4x – 2x × 3 + 5 × 4x – 5 × 3
= 8x² – 6x + 20x -15
= 8x² + 14x -15

(ii) (y – 8) and (3y – 4)

Solution:
(ii) ( y – 8)(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)

Solution:
(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 + 1.25lm – 1.25lm – 0.25m²
= 6.25l² – 0.25m²

(iv) (a + 3b) and (x + 5)

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

(v) (2pq + 3q² ) and (3pq – 2q²)

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

(vi) (3/4a² + 3b²) and 4(a² – 2/3b²)

Solution:
(vi) (3/4a² + 3b²)4(a² – 2/3b²)
= (3/4a² + 3b²)) × 4(a² – 2/3b²)
= (3/4a² + 3b²) × (4a² – 8/3b²)
= 3/4a² × (4a² – 8/3b²) + 3b² × (4a² – 8/3b²)
= 3/4a² × 4a² – 3/4a² × 8/3b² + 3b² × 4a² – 3b² × 8/3b²
= 3a⁴ – 2a²b² + 12a²b² – 8b⁴
= 3a⁴ + 10a²b² – 8b⁴

2. Find the product.

(i) (5 – 2x) (3 + x)

Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= 15 + 5x – 6x – 2x²
= 15 – x – 2x²

(ii) (x + 7y) (7x – y)

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

(iii) (a² + b) (a + b²)

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

(iv) (p² – q²) (2p + q)

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

3. Simplify.

(i) (x² –  5) (x + 5) + 25

Solution:
(i) (x² – 5) (x + 5) + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x

(ii) (a² + 5) (b³ + 3) + 5

Solution:
(ii) (a² + 5) (b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 5b³ + 3a² + 20

(iii) (t + s²)(t² – s)

Solution:
(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)

Solution:
(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)

Solution:
(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)

Solution:
(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

Solution:
(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)

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