Mathematics – Class 8 – Chapter 9 – Algebraic Expression and Identities – Exercise 9.4 -NCERT Exercise Solution

1. Multiply the binomials.

(i) (2x + 5) and (4x – 3)

Solution:

= (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) and (3y – 4)

Solution:

= (y – 8) (3y – 4)

= y x 3y – 4y – 8 x 3y + 32

= 3y2 – 4y – 24y + 32

= 3y2 – 28y + 32

(iii) (2.5l – 0.5m) and (2.5l + 0.5m)

Solution:

= (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) and (x + 5)

Solution:

= (a + 3b) (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) and (3pq – 2q2)

Solution:

= (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)

Solution:

= (3/4 a² + 3b²) and 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

= 3a4 – 2a² b² + 12 a² b² – 8b4

= 3a4 + 10a² b² – 8b4

2. Find the product.

(i) (5 – 2x) (3 + x)

Solution:

= (5 – 2x) (3 + x)

= 5 (3 + x) – 2x (3 + x)

=15 + 5x – 6x – 2x2

= 15 – x -2 x2

(ii) (x + 7y) (7x – y)

Solution:

= (x + 7y) (7x – y)

= x(7x-y) + 7y (7x-y)

=7x2 – xy + 49xy – 7y2

= 7x2 – 7y2 + 48xy

(iii) (a2+ b) (a + b2)

Solution:

= (a2+ b) (a + b2)

= a2(a + b2) + b(a + b2)

= a3 + a2b2 + ab + b3

= a3 + b3 + a2b2 + ab

(iv) (p2 – q2) (2p + q)

Solution:

= (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:

= (x2– 5) (x + 5) + 25

= x3 + 5x2 – 5x – 25 + 25

= x3 + 5x2 – 5x

(ii) (a2+ 5) (b3+ 3) + 5

Solution:

= (a2+ 5) (b3+ 3) + 5

= a2b3 + 3a2 + 5b3 + 15 + 5

= a2b3 + 5b3 + 3a2 + 20

(iii)(t + s2)(t2 – s)

Solution:

= (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:

= (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:

= (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:

= (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:

= (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:

= (a + b + c) (a + b – c)

= a2 + ab – ac + ab + b2 – bc + ac + bc – c2

= a2 + b2 – c2 + 2ab

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