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

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

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

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

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

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