NCERT solution class 8 chapter 9 Algebraic Expression and Identities

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) (7xy)

(iii) (a2 + b) (a + b2) (iv) (p2q2) (2p + q)

 

Sol :

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

= 15 + 5x − 6x − 2x2

= 15 − x − 2x2

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

= 7x2xy + 49xy − 7y2

= 7x2 + 48xy − 7y2

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

= a3 + a2b2 + ab + b3

(iv) (p2q2) (2p + q) = p2 (2p + q) − q2 (2p + q)

= 2p3 + p2q − 2pq2q3

 

 

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)

= t3st + s2t2 s3

 

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

Sol :

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

= a (cd) + b (cd) + a (c + d) − b (c + d) + 2 (ac + bd)

= acad + bcbd + ac + adbcbd + 2ac + 2bd

= (ac + ac + 2ac) + (adad) + (bcbc) + (2bdbdbd)

= 4ac

 

(v) (x + y) (2x + y) + (x + 2y) (xy)

Sol :

(v) (x + y) (2x + y) + (x + 2y) (xy)

= x (2x + y) + y (2x + y) + x (xy) + 2y (xy)

= 2x2 + xy + 2xy + y2 + x2xy + 2xy − 2y2

= (2x2 + x2) + (y2 − 2y2) + (xy + 2xyxy + 2xy)

= 3x2y2 + 4xy

 

(vi) (x + y) (x2xy + y2)

Sol :

(vi) (x + y) (x2xy + y2)

= x (x2xy + y2) + y (x2xy + y2)

= x3x2y + xy2 + x2yxy2 + y3

= x3 + y3 + (xy2xy2) + (x2yx2y)

= 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 + bc)

Sol :

(viii) (a + b + c) (a + bc)

= a (a + bc) + b (a + bc) + c (a + bc)

= a2 + abac + ab + b2bc + ca + bcc2

= a2 + b2c2 + (ab + ab) + (bcbc) + (caca)

= a2 + b2c2 + 2ab

1 thought on “NCERT solution class 8 chapter 9 Algebraic Expression and Identities

Leave a Reply

Your email address will not be published. Required fields are marked *