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 9.3

Question 1

Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r (ii) ab, ab (iii) a + b, 7a2b2

(iv) a2 − 9, 4a (v) pq + qr + rp, 0

 

Sol :

(i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr

(ii) (ab) × (ab) = (ab × a) + [ab × (− b)] = a2bab2

(iii) (a + b) × (7a2 b2) = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3

(iv) (a2 − 9) × (4a) = (a2 × 4a) + (− 9) × (4a) = 4a3 − 36a

(v) (pq + qr + rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) = 0

 

Question 2

Complete the table

First expression

Second Expression

Product

(i)

a

b + c + d

(ii)

x + y − 5

5 xy

(iii)

p

6p2 − 7p + 5

(iv)

4p2q2

p2 q2

(v)

a + b + c

abc

Sol :

 

The table can be completed as follows.

First expression

Second Expression

Product

(i)

a

b + c + d

ab + ac + ad

(ii)

x + y − 5

5 xy

5x2y + 5xy2 − 25xy

(iii)

p

6p2 − 7p + 5

6p3 − 7p2 + 5p

(iv)

4p2q2

p2 q2

4p4q2 − 4p2q4

(v)

a + b + c

abc

a2bc + ab2c + abc2

 

Question 3

Find the product.

(i) (a2) × (2a22) × (4a26)

(ii)

(iii)

(iv) x × x2 × x3 × x4

 

Sol :

(i) (a2) × (2a22) × (4a26) = 2 × 4 ×a2 × a22 × a26 = 8a50

(ii)

(iii)

(iv) x × x2 × x3 × x4 = x10

 

Question 4

(a) Simplify 3x (4x −5) + 3 and find its values for (i) x = 3, (ii) .

(b) a (a2 + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = − 1.

 

Sol :

(a) 3x (4x − 5) + 3 = 12x2 − 15x + 3

(i) For x = 3, 12x2 − 15x + 3 = 12 (3)2 − 15(3) + 3

= 108 − 45 + 3

= 66

(ii) For

(b)a (a2 + a + 1) + 5 = a3 + a2 + a + 5

(i) For a = 0, a3 + a2 + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a3 + a2 + a + 5 = (1)3 + (1)2 + 1 + 5

= 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a3 + a2 + a + 5 = (−1)3 + (−1)2 + (−1) + 5

= − 1 + 1 − 1 + 5 = 4

 

Question 5

(a) Add: p (pq), q (q ­­­r) and r (r ­− p)

(b) Add: 2x (zxy) and 2y (zyx)

(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l)

(d) Subtract: 3a (a + b + c) − 2b (ab + c) from 4c (− a + b + c)

 

Sol :

(a) First expression = p (pq) = p2pq

Second expression = q (q ­­­r) = q2qr

Third expression = r (r ­− p) = r2pr

Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p2 + q2 + r2pqqrrp.

(b) First expression = 2x (zxy) = 2xz − 2x2 − 2xy

Second expression = 2y (zyx) = 2yz − 2y2 − 2yx

Adding the two expressions, we obtain

Therefore, the sum of the given expressions is − 2x2 − 2y2 − 4xy + 2yz + 2zx.

(c) 3l (l − 4m + 5n) = 3l2 − 12lm + 15ln

4l (10n − 3m + 2l) = 40ln − 12lm + 8l2

Subtracting these expressions, we obtain

Therefore, the result is 5l2 + 25ln.

(d) 3a (a + b + c) − 2b (ab + c) = 3a2 +3ab + 3ac − 2ba + 2b2 − 2bc

= 3a2 + 2b2 + ab + 3ac − 2bc

4c (− a + b + c) = − 4ac + 4bc + 4c2

Subtracting these expressions, we obtain

Therefore, the result is −3a2 −2b2 + 4c2ab + 6bc − 7ac.

Leave a Reply

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