Exercise 9.4 - Chapter 9 Algebraic Expressions & Identities class 8 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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) 

Answer:

(i) (2x + 5) × (4x − 3) = 2x × (4x − 3) + 5 × (4x − 3)

= 8x² − 6x + 20x − 15

= 8x² + 14x −15 (By adding like terms)

(ii) (y − 8) × (3y − 4) = y × (3y − 4) − 8 × (3y − 4)

= 3y² − 4y − 24y + 32

= 3y² − 28y + 32 (By adding like terms)

(iii) (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

(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq − 2q²) = 2pq × (3pq − 2q²) + 3q² × (3pq − 2q²)

= 6p²q² − 4pq³ + 9pq³ − 6q⁴

= 6p²q² + 5pq³ − 6q⁴

(vi) 

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)

Answer:

(i) (5 − 2x) (3 + x) = 5 (3 + x) − 2x (3 + x)

= 15 + 5x − 6x − 2x²

= 15 − x − 2x²

(ii) (x + 7y) (7x − y) = x (7x − y) + 7y (7x − y)

= 7x² − xy + 49xy − 7y²

= 7x² + 48xy − 7y²

(iii) (a² + b) (a + b²) = a² (a + b²) + b (a + b²)

= a³ + a²b² + ab + b³

(iv) (p² − q²) (2p + q) = p² (2p + q) − q² (2p + 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)

Answer:

(i) (x² − 5) (x + 5) + 25

= x² (x + 5) − 5 (x + 5) + 25

= x³ + 5x² − 5x − 25 + 25

= 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

(iv) (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 + 2ac + 2bd

= (ac + ac + 2ac) + (ad − ad) + (bc − bc) + (2bd − bd − bd)

= 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) + (y² − 2y²) + (xy + 2xy − xy + 2xy)

= 3x² − y² + 4xy

(vi) (x + y) (x² − xy + y²)

= x (x² − xy + y²) + y (x² − xy + y²)

= x³ − x²y + xy² + x²y − xy² + y³

= x³ + y³ + (xy² − xy²) + (x²y − x²y)

= 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

= 2.25 x² + 6xy + 4.5x − 6xy − 16y² − 12y − 4.5x + 12y

= 2.25 x² + (6xy − 6xy) + (4.5x − 4.5x) − 16y² + (12y − 12y)

= 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 + ca + bc − c²

= a² + b² − c² + (ab + ab) + (bc − bc) + (ca − ca)

= a² + b² − c² + 2ab