NCERT SOLUTIONS FOR MATHS CLASS 8 ALGEBRAIC EXPRESSIONS AND IDENTITIES CHAPTER 9 Ex. 9.4

Question 1.
Multiply the binomials:
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q²) and (3pq – 2q²)
(vi) (34a² + 3b²) and 4(a² – 23 b²)
Solution:
(i) (2x + 5) × (4x – 3)
= 2x × (4x – 3) + 5 × (4x – 3)
= (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15

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

(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l² + 1.25ml – 1.25ml – 0.25m²
= 6.25l² + 0 – 0.25m²
= 6.25l² – 0.25m²

(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 36 × (x + 5)
= (a × x) + (a × 5) + (36 × x) + (36 × 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq – 2q²)
= 2pq × (3pq – 2q²) + 3q² (3pq – 2q²)
= (2pq × 3pq) – (2pq × 2q²) + (3q² × 3pq) – (3q² × 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 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)
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
= 15 + 5x – 6x – 2x²

(ii) (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

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

(iv) (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:
(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
= x³ + 5x² – 5x + 0
= x³ + 5x² – 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 – st + s²t² – s³
= t³ + s²t² – st – 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) + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd
= 4ac + 0 + 0 + 0
= 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
= 2x² + x² + xy + 2xy – xy + 2xy + y² – 2y²
= 3x² + 4xy – y²

(vi) (x + y)(x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + x²y + xy² – xy² + y³
= x³ – 0 + 0 + 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 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y²
= 2.25x² + 0 + 0 + 0 – 16y²
= 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 + ab + ab – bc + bc – ac + ac + b² – c²
= a² + 2ab + b² – c² + 0 + 0
= a² + 2ab + b² – c²