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

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 Textbook Exercise Questions and Answers.

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

1. Multiply the binomials:

Question (i)
(2x + 5) and (4x – 3)
Solution:
= (2x + 5)(4x – 3)
= 2x(4x – 3) + 5 (4x – 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15

Question (ii)
(y – 8) and (3y – 4)
Solution:
= (y -8) (3y – 4)
= y (3y – 4) – 8 (3y – 4)
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32

Question (iii)
(2.51 – 0.5m) and (2.51 + 0.5m)
Solution:
= (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l(2.5l + 0.5m) – 0.5m (2.5l + 0.5m)
= 6.25l² + 1.25lm – 1.25lm – 0.25m²
= 6.25l² – 0.25m²

Question (iv)
(a + 3b) and (x + 5)
Solution:
= (a + 3b) (x + 5)
= a (x + 5) + 3b (x + 5)
= ax + 5a + 3bx + 15b

Question (v)
(2pq + 3q²) and (3pq – 2q²)
Solution:
= (2pq + 3q²) (3pq – 2q²)
= 2pq (3pq – 2q²) + 3q² (3pq – 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 5pq³ – 6q⁴

Question (vi)
(\(\frac {3}{4}\)a² + 3b²) and 4 (a² – \(\frac {2}{3}\)b²)
Solution:
= (\(\frac {3}{4}\)a² + 3b²) (4a² – \(\frac {8}{3}\)b²)
= \(\frac {3}{4}\)a⁴ (4a² – \(\frac {8}{3}\)b²) + 3b² (4a² – \(\frac {8}{3}\)b²)
= 3a₄ – 2a²b² + 12a²b² – 8b₄
= 3a⁴ + 10a²b² – 8b⁴

2. Find the product:

Question (i)
(5 – 2x) (3 + x)
Solution:
= 5 (3 + x) – 2x (3 + x)
= 15 + 5x – 6x – 2x²
= 15 – x – 2x²

Question (ii)
(x + 7y) (7x – y)
Solution:
= x(7x – y) + 7y (7x – y)
= 7x² – xy + 49xy – 7y²
= 7x² + 48xy – 7y²

Question (iii)
(a² + b) (a + b²)
Solution:
= a² (a + b²) + b (a + b²)
= a³ + a²b² + ab + b³

Question (iv)
(p² – q²) (2p + q)
Solution:
= p²(2p + q) – q²(2p + q)
= 2p³ + p²q – 2pq² – q³

3. Simplify:

Question (i)
(x² – 5) (x + 5) + 25
Solution:
= x² (x + 5) – 5 (x + 5) + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x

Question (ii)
(a² + 5) (b³ + 3) + 5
Solution:
= a²(b³ + 3) + 5 (b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20

Question (iii)
(t + s²)(t² – s)
Solution:
= t (t² – s) + s² (t² – s)
= t³ – st + s²t² – s³

Question (iv)
(a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
Solution:
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2 (ac + bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2 bd
= 4 ac

Question (v)
(x + y) (2x + y) + (x + 2y) (x – y)
Solution:
= x (2x + y) + y (2x + y) +x(x – y) + 2y (x – y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²
= 2x² + x² + xy + 2xy – xy + 2xy + y² – 2y²
= 3x² + 4xy – y²

Question (vi)
(x + y) (x² – xy + y²)
Solution:
= x (x² – xy + y²) + y (x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ – x²y + x²y + xy² – xy² + y³
= x³ + y³

Question (vii)
(1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
Solution:
= 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 + 6xy – 6xy + 4.5x – 4.5x – 16y² – 12y + 12y
= 2.25x² – 16y²

Question (viii)
(a + b + c) (a + b – c)
Solution:
= a (a + b – c) + b (a + b – c) + c (a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + ab + ab – ac + ac + b² – bc + bc – c²
= a² + 2ab + b² – c²
= a² + b² – c² + 2 ab