NCERT Solutions of Algebraic Expressions and Identities Class 8
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Algebraic Expressions

Exercise 2

1. Find the product of the following pairs of monomials.

(i) 4, 7p

Answer: 4 x 7 p = 28p

(ii) – 4p, 7p

Answer: - 4p x 7p = -28p2

(iii) – 4p, 7pq

Answer: - 4p x 7pq = -28p2q

(iv) 4p3, – 3p

Answer: 4p3q x - 3p = -12p4q

(v) 4p, 0

Answer: 4p x 0 = 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x2 5y2; (4x, 3x2); (3mn, 4np)

Answer: Area = Length x breadth

(i) p x q = pq

(ii)10m x 5n = 50mn

(iii) 20x2 x 5y2 = 100x2y2

(iv) 4x x 3x2 = 12x3

(v)3mn x 4np = 12mn2p

3. Complete the following table of products:

algebra 1

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a8 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2(iv) a, 2b, 3c

Answer: Volume = length x breadth x height

(i) 5a x 3a2 x 7a8 = 105a11

(ii) 2p x 4q x 8r = 64pqr

(iii) xy x 2x2y x 2xy2 = 4x4y4

(iv) a x 2b x 3c = 6abc

5. Obtain the product of

(i) xy, yz, zx (ii) a, – a2 a3(iii) 2, 4y, 8y2 16y3(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp

Answer: (i) x2y2z2

(ii) –a5

(iii) 1024y6

(iv) 36a2b2c2

(v) –m3n2p

Exercise 3

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

(i) 4p, q + r

Answer: 4p(q + r) = 4pq + 4pr

(ii) ab, a – b

Answer: ab(a - b) = a2b - ab2

(iii) a + b, 7a²b²

Answer: (a + b) (7a2b2) = 7a3b2 + 7a2b3

(iv) a2– 9, 4a

Answer: (a2 - 9)(4a) = 4a3 - 36a2

(v) pq + qr + rp, 0

Answer: (pq + qr + rp) x 0 = 0

2. Find the product.

(i) a2 x (2a22) x (4a26)

Answer: As you know

am x an x ao = am+n+o

So, we get

a2 x (2a22) x (4a26) = 8a48

algebra 2

(iv) x x x2 x x3 x x8

= x14

3. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2

(i) putting x=3 in the equation we get

12x2 - 15x + 3

=108-45+3 = 66

(ii) putting x=1/2 in the equation we get

algebra 3

(b) Simplify a (a2+ a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1.

Answer: a(a2a+1)

=a3+a2+a

(i) putting a= 0 in the equation we get

03+02+0=0

(ii) putting a=1 in the equation we get

13 + 12 + 1 = 1 + 1 + 1 = 3

(iii) putting a = -1 in the equation we get

-13+12 -1 = -1 + 1 + 1 = 1

5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)

Answer: (p2 - pq) + (q2 - qr) + (r2 - pr)

= p2 + q2 + r2 - pq - qr - pr

(b) Add: 2x (z – x – y) and 2y (z – y – x)

Answer: (2xz - 2x2 - 2xy) + (2yz - 2y2 - 2xy)

= 2xz - 4xy + 2yz - 2x2 - 2y2

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

Answer: (40ln - 12lm + 8l2) - (3l2 - 12lm + 15ln)

= 40ln - 12lm + 8l2 - 3l2 - 12lm + 15ln

= 55ln - 24lm + 5l2

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

= (-4ac + 4bc + 4c2) - (3a2 + 3ab + 3ac)

= -4ac + 4bc + 4c2 - 3a2 - 3ab - 3ac

= -7ac + 4bc + 4c2 - 3a2 - 3ab


NCERT Solutions of Algebraic Expressions and Identities Class 8
<< Solution of NCERT Exercise 9.1Solution of NCERT Exercise 9.4 >>

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