NCERT Solutions have been structured in logical and easy language for quick revisions. Well-illustrated solutions for CBSE Class 8 Maths Exercise 9.4 are extremely accurate and solved using a step-by-step problem-solving approach. This exercise helps students to learn how to perform the multiplication of two polynomials. Download free NCERT Solutions for Maths Chapter 9 prepared by BYJU’S subject experts and practise offline.
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4
Access Answers to NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4 Page number 148
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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)Â (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)
= a2Â (a + b2) + b(a + b2)
= a3Â + a2Â b2Â + ab + b3
= a3Â + b3Â + a2Â b2Â + 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
= a2b3Â + 3a2Â + 5b3Â + 15 + 5
= a2b3Â + 5b3Â + 3a2Â + 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
= x3Â + 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
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4 explains 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 practise the questions and answers.
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