NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

NCERT Solutions Class 8 Mathematics 
Chapter – 9 (Algebraic Expressions and Identities) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 9 Algebraic Expressions and Identities Exercise 9.4 has been provided here to help the students in solving the questions from this exercise. 

Chapter 9: Algebraic Expressions and Identities

• NCERT Solution Class 8 Maths Ex – 9.1
• NCERT Solution Class 8 Maths Ex – 9.2
• NCERT Solution Class 8 Maths Ex – 9.3
• NCERT Solution Class 8 Maths Ex – 9.5

Exercise – 9.4 

1. 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) and

Solution –

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

(ii) (y – 8) and (3y – 4)
= ( 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)
= (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) and (x + 5)
= (a + 3b) (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) and
= ×

=  ×

= a² × (4a² – b² ) + 3b² × (4a² – b² )

= a² × 4a² – a² × b² + 3b² × 4a² – 3b² × b²

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

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 -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

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
= x³ + 5x² – 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(t ² – s) + s²(t ² – s)
= t³ – st + s²t ² – 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) + 2(ac + bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= (ac + ac + 2ac) + (ad – ad) + (bc – bc) + (2bd – bd – bd)
= 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
= (2x² + x²) + (y² – 2y²) + (xy + 2xy – xy + 2xy)
= 3x² – y² + 4xy

(vi) (x + y)(x²– xy + y²)
= x(x² – xy + y²) + y(x² – xy + y² )
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³ + (xy² – xy²) + (x²y – x²y)
= 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
= 2.25 x² + 6xy + 4.5x – 6xy – 16y² -12y – 4.5x + 12y
= 2.25x² + (6xy – 6xy) + (4.5x – 4.5x) – 16y² + (12y -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 + ca + bc – c²
= a² + b² – c² + (ab + ab) + (bc – bc) + (ca – ca)
= a² + b² – c² + 2ab