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

Gujarat Board 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) × (4x – 3)
= 2x(4x – 3) + 5(4x – 3)
= (2x × 4x) – (2x × 3) + (5 × 4x) – (5 × 3)
(8x²) – (6x) + (20x) – (15)
= 8x² – 6x + 20x – 15
= 8x² + (-6 + 20)x – 15
= 8x² + 14x – 15

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

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

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

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

6. (\(\frac{3}{4}\)a² + 3b²) × 4(a² – \(\frac{2}{3}\)b²)
= \(\frac{3}{4}\)a² × 4(a² – \(\frac{2}{3}\)b²) + 3b² × 4(a² – \(\frac{2}{3}\)b²)
= (3a² × a²) – (3a² × \(\frac{2}{3}\)b²) + (12b² × \(\frac{2}{3}\)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) × (x + 3) = 5(x + 3) – 2x(x + 3)
= 5x + 15 – (2x × x) – (2x × 3)
= 5x + 15 – 2x² – 6x
= -2x² + (-6 + 5)x + 15
= -2x² – x + 15 = 15 – x – 2x²

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

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

4. (p² – q²) × (2p + q) = p²(2p + q) – q²(2p + q)
= (P² × 2p) + (p² × q) – (q² × 2p) – (q² × 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 × x) + (x² × 5) – (5 × x) – (5 × 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 × b³) + (a² × 3) + (5 × b³) + (5 × 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a³b³ + 3a² + 5b³ + 20

3. (t + s²) × (t² – s) = t(t² – s) + s²(t² – s)
= (t × t²) – (t × s) + (s² × t²) + [s² × (-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 × 2x) + (x × y) + (y × 2x) + (y × y) + (x × x) – (x × y) + (2y × x) – (2y × 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 × x²) – (x × xy) + (x × y²) + (y × x²) – (y × xy) + (y × y²)
= x³ – x²y + xy² + yx² – xy² + y³
= x³ + (-x²y + x²y) + (xy² – xy²) + y³
= x³ + (0 × x²) + (0 × 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 × 1.5x) + (1.5x × 4y) + (1.5x × 3) – (4y × 1.5x) – (4y × 4y) – (4y × 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 × a) + (a × b) – (a × c) + (b × a) + (b × b) – (b × c) + (c × a) + (c × b) – (c × 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