 # NCERT Solutions for class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

NCERT Solutions have been structured in a logical and easy language for quick revisions. Well-illustrated solutions for CBSE Class 8 Maths Exercise 9.4 are extremely accurate and solved using step by step problem solving approach. This exercise help students to learn how to perform multiplication of two polynomials. Download free NCERT Solutions for Maths Chapter 9 prepared by BYJU’S subject experts and practice offline.

### Download PDF of NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4   ## Exercise 9.4 Page No: 148

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) (3/4 a2 + 3b2) and 4( a2 – 2/3 b2 )

Solution :

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

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

(iii) (2.5l – 0.5m)(2.5l + 0.5m)

= 2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 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 x 3pq – 2pq x 2q2 + 3q2 x 3pq – 3q2 x 2q2
= 6p2q2 – 4pq3 + 9pq3 – 6q4

= 6p2q2 + 5pq3 – 6q4

(vi) (3/4 a² + 3b² ) and 4( a² – 2/3 b²)

= (3/4 a² + 3b² ) x 4( a² – 2/3 b² )

=(3/4 a² + 3b² ) x (4a² – 8/3 b² )

=3/4 a² x (4a² – 8/3 b² ) + 3b² x (4a² – 8/3 b² )

=3/4 a² x 4a² -3/4 a² x 8/3 b² + 3b² x 4a² – 3b² x 8/3 b²

=3 a4 – 2a²b² + 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 + a2 b2 + ab + b3
= a3 + b3 + a2 b2 + 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

= 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

### Access Other Exercise Solutions of Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Exercise 9.1 Solutions : 4 Questions (Short answers)

Exercise 9.2 Solutions : 5 Questions (Short answers)

Exercise 9.3 Solutions : 5 Questions (Short answers)

Exercise 9.5 Solutions : 8 Questions (Short answers)

### NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Algebraic Expressions and Identities Exercise 9.4 explains about multiplying a binomial by a binomial, multiplying a binomial by a trinomial and simplification of the polynomials. For more practice, download the NCERT solutions and practice the questions and answers.