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

## 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

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