NCERT solution class 8 chapter 9 Algebraic Expression and Identities exercise 9.4

Exercise 9.1  Exercise 9.2  Exercise 9.3  Exercise 9.4  Exercise 9.5

Exercise 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)

Sol :
(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)

Sol :
(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
Sol :
(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
Sol :
(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)
Sol :
(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)
Sol :
(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)
Sol :
(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)
Sol :
(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
Sol :
(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)
Sol :
(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