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

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

Question 1.
Multiply the binomials.

1. (2x + 5) and (4x – 3)
2. (y – 8) and (3y – 4)
3. (2.51 – 0.5m) and (2.51 + 0.5m)
4. (a + 3b) and (x + 5)
5. (2pq + 3q2) and (3pq – 2q2)
6. ($$\frac{3}{4}$$a2 + 3b2) and 4(a2 – $$\frac{2}{3}$$b2)

Solution:
1. (2x + 5) Ć (4x – 3)
= 2x(4x – 3) + 5(4x – 3)
= (2x Ć 4x) – (2x Ć 3) + (5 Ć 4x) – (5 Ć 3)
(8x2) – (6x) + (20x) – (15)
= 8x2 – 6x + 20x – 15
= 8x2 + (-6 + 20)x – 15
= 8x2 + 14x – 15

2. (y – 8) Ć (3y – 4) = y(3y – 4) – 8(3y – 4)
= (3y Ć y) – (4 Ć y) – 8 Ć 3y – 8 Ć (-4)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32

3. (2.5l – 0.5m) Ć (2.5l + 0.5m)
= 2.5l(2.5l + 0.5m) – 0.5m(2.5l + 0.5m)
= (2.5l Ć 2.5l) + (2.5l Ć 0.5m) – (2.5l Ć 0.5m) – (0.5m Ć 0.5m)
= 6.25l2 + 1.25lm – 1.25lm – 0.25m2
= 6.25l2 + (0)lm – 0.25m2 = 6.25l2 – 0.25m2

4. (a + 3b) Ć (x + 5)
= a Ć (x + 5) + 3b Ć (x + 5)
= (a Ć x) + (a Ć 5) + (3b Ć x) + (3b Ć 5)
= ax + 5a + 3bx + 15b

5. (2pq + 3q2) Ć (3pq – 2q2)
= 2pq(3pq – 2q2) + 3q2(3pq – 2q2)
= (2pq Ć 3pq) – (2pq Ć 2q2) + (3q2 Ć 3pq) – (3q2 Ć 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + (-4 + 9)pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4

6. ($$\frac{3}{4}$$a2 + 3b2) Ć 4(a2 – $$\frac{2}{3}$$b2)
= $$\frac{3}{4}$$a2 Ć 4(a2 – $$\frac{2}{3}$$b2) + 3b2 Ć 4(a2 – $$\frac{2}{3}$$b2)
= (3a2 Ć a2) – (3a2 Ć $$\frac{2}{3}$$b2) + (12b2 Ć $$\frac{2}{3}$$b2)
= 3a4 – 2a2b2 + 12a2b2 – 8b4
= 3a4 + (-2 + 12)a2b2 – 8b4
= 3a4 + 10a2b2 – 8b4

Question 2.
Find the product:

1. (5 – 2x)(3 + x)
2. (x + 7y)(7x – y)
3. (a2 + b)(a + b2)
4. (p2 – q2)(2p + q)

Solution:
1. (5 – 2x) Ć (x + 3) = 5(x + 3) – 2x(x + 3)
= 5x + 15 – (2x Ć x) – (2x Ć 3)
= 5x + 15 – 2x2 – 6x
= -2x2 + (-6 + 5)x + 15
= -2x2 – x + 15 = 15 – x – 2x2

2. (x + 7y) Ć (7x – y) = x(7x – y) + 7y(7x – y)
= (x Ć 7x) – (x Ć y) + (7y Ć 7x) – (7y Ć y)
= 7x2 – xy + 49xy – 7y2
= 7x2 + 48xy – 7y2

3. (a2 + b)(a + b2) = a2(a + b2) + b(a + b2)
= (a2 Ć a) + (a2 Ć b2) + (b Ć a) + (b Ć b2)
= a3 + a2b2 + ab + b3

4. (p2 – q2) Ć (2p + q) = p2(2p + q) – q2(2p + q)
= (P2 Ć 2p) + (p2 Ć q) – (q2 Ć 2p) – (q2 Ć q)
= 2p3 + p2q – 2pq2 – q3

Question 3.
Simplify:

1. (x2 – 5)(x + 5) + 25
2. (a2 + 5)(b3 + 3) + 5
3. (t + s2)(t2 – s)
4. (a + b)(c -d) + (a- b)(c + d) + 2(ac + bd)
5. (x + y) (2x + y) + (x + 2y)(x – y)
6. (x + y)(x2 – xy + y2)
7. (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
8. (a + b + c)(a + b – c)

Solution:
1. (x2 – 5)(x + 5) + 25
= x2(x + 5) – 5(x + 5) + 25
= (x2 Ć x) + (x2 Ć 5) – (5 Ć x) – (5 Ć 5) + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x

2. (a2 + 5)(b3 + 3) + 5 = a2(b3 + 3) + 5(b3 + 3) + 5(b3 + 5) + 5
= (a2 Ć b3) + (a2 Ć 3) + (5 Ć b3) + (5 Ć 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a3b3 + 3a2 + 5b3 + 20

3. (t + s2) Ć (t2 – s) = t(t2 – s) + s2(t2 – s)
= (t Ć t2) – (t Ć s) + (s2 Ć t2) + [s2 Ć (-s)]
= t3 – ts + s2t2 – s3
= t3 – st + s2t2 – s3

4. (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) + 2(ac + bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac = 4 ac

5. (x + y)(2x + y) + (x + 2y)(x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
= (x Ć 2x) + (x Ć y) + (y Ć 2x) + (y Ć y) + (x Ć x) – (x Ć y) + (2y Ć x) – (2y Ć y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= (2x2 + x2) + (xy + 2xy – xy + 2xy) + (y2 – 2y2)
= 3x2 + 4xy – y2

6. (x + y)(x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= (x Ć x2) – (x Ć xy) + (x Ć y2) + (y Ć x2) – (y Ć xy) + (y Ć y2)
= x3 – x2y + xy2 + yx2 – xy2 + y3
= x3 + (-x2y + x2y) + (xy2 – xy2) + y3
= x3 + (0 Ć x2) + (0 Ć xy2) + y3
= x3 + 0 + 0 + y3 = x3 + y3

7. (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
= 1.5x(1.5x + 4y +3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
= (1.5x Ć 1.5x) + (1.5x Ć 4y) + (1.5x Ć 3) – (4y Ć 1.5x) – (4y Ć 4y) – (4y Ć 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
= 2.25x2 + (6 – 6)xy + (4.5 – 4.5)x – 16y2 + (12 – 12 )y
= 2.25x2 + (0)xy + (0)x – 16y2 + (o)y
= 2.25x2 + 0 + 0 – 16y2 + 0
= 2.25x2 – 16y2

8. (a + b + c)(a + b – c)
= a(a + b – c) + b(a + b – c) + c(a + b – c) = (a Ć a) + (a Ć b) – (a Ć c) + (b Ć a) + (b Ć b) – (b Ć c) + (c Ć a) + (c Ć b) – (c Ć c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + (ab + ab) + (-ac + ac) + b2 + (-bc+ bc) – c2
= a2 + (2ab) + (0) + b2 + 0 – c2
= a2 + 2ab + b2 – c2
= a2 + b2 – c2 + 2ab