MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4

Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5 m) and (2.5l +0.5 m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q²) and (3pq – 2q²)

Solution:
(i) (2x + 5) × (4x – 3) = 2x (4x – 3) + 5(4x – 3) = 8x² – 6x + 20x – 15 = 8x² + 14x – 15
(ii) (y – 8) × (3y – 4) = y(3y – 4) – 8(3y – 4)
= 3 y² – 4y – 24y + 32 = 3y² – 28y + 32
(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) – 0.5 m (2.5l + 0.5m)
= 6.25l² + 1.25lm – 1.25lm – 0.25m² = 6.25l² – 0.25m²
(iv) (a + 3b) × (x + 5) = a(x + 5) + 3b(x + 5)
= ax + 5a + 3 bx + 15 b
(v) (2pq + 3q²) × (3pq – 2q²)
= 2pq (3pq – 2q²) + 3q² (3pq – 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6 p²q² + 5pq³ – 6q⁴
MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 2

Question 2.
Find the product
(i) (5 – 2x)(3 + x)
(ii) (x + 7y)(7x – y)
(iii) (a² + b)(a + b²)
(iv) (p² – q²)(2p + q)
Solution:
(i) (5 – 2x) × (3 + x) = 5(3 + x) – 2x(3 + x)
= 15 + 5x – 6x – 2x² – 15 – x – 2x²
(ii) (x + 7y)(7x – y) = x(7x – y) + 7y(7x – y)
=7x² – xy + 49xy – 7y² = 7x² + 48xy – 7y²
(iii) (a² + b) (a + b²) = a² (a + b²) + b (a + b²)
= a³ + a²b²) + ab + b³
(iv) (p² – q²) (2p + q) = p² (2p + q) – q² (2p + q)
= 2p³ + p²q – 2q²p – q³

Question 3.
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)
Solution:
(i) (x² – 5) (x + 5) + 25
= x²(x +5) – 5(x + 5) +25
=x³+ 5x² – 5x – 25 + 25 = r³+ 5r² – 5x

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

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

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

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

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

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x (1.5x + 4y + 3) – 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)
= a (a+ b – c) + b (a+ b – c) + c (a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + 2 ab + b² – c²