NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

Class 8 Mathematics by VP Classes

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Class 8 : NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

The document NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.
All you need of Class 8 at this link: Class 8

Exercise 9.2

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

(i) 4, 7p (ii) –4p, 7p (iii) –4p, 7pq
(iv) 4p3, –3p (v) 4p, 0

Solution: 

(i) 4 and 7p

4 * 7p = (4 * 7) * p = 28p

(ii) –4p and 7p

–4p * 7p = {(–4) * 7} * p * p = –28p2

(iii) –4p and 7pq

–4p * 7pq = (–4 * 7) * p * qp = –28 * p2q = –28p2q

(iv) 4p3 and –3p

4p3 * (–3p) = {4 * (–3)}p3 * p = –12 * p4 = –12p4

(v) 4p and 0 ⇒ 4p * 0 = 0

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

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

Solution:

(i) Length = p Breadth = q}             ∴ Area of the rectangle = p * q = pq

(ii) Length = 10m Breadth = 5n}     ∴ Area = 10m * 5n = 10 * 5 * m * n = 50mn

(iii) Length = 20x2 Breadth = 5y2}   ∴ Area = 20x2 * 5y= 20 * 5 * x2 * y2 = 100x2y2

(iv) Length = 4x Breadth = 3*2}       ∴ Area = 4x * 3 * x= 4 * 3 * x *x2= 12*3

(v) Length = 3mn Breadth = 4np}    ∴ Area = 3mn * 4np = (3 * 4) * m * n * n * p = 12 mn2p

Question 3. Complete the table of products.

First monomial → 

Second monomial  NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

2*–5y3*2–4*y7*2y–9x2y2

2x

–5y

3x3

–4xy

7x2y

–9x2y2

4*2

–15x2y

 

Solution: We have

NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

First monomial → Second monomial NCERT Solutions (Part- 2)- Algebraic Expressions and Identities Class 8 Notes | EduRev

2x

-5y

3x2

-4xy

7x2y

–9x2y2

2x

4x2

-10xy

6x3

-8x2y

14x3y

-18x3y2

-5y

-10xy

25y2

-15x2y

20xy2

-35x2y2

45x2y3

3x3

6x3

-15x2y

9x4

-12x3y

21x4y

-27x4y2

-4xy

-8x2y

20xy2

-12x3y

16x2y2

-28x3y2

36x3y3

7x2y

14x3y

-35x2y2

21x4y

-28x3y2

49x4y2

-63x4y3

-9x2y2

-18x3y2

45x2y3

-27x4y2

36x3y3

-63x4y3

81x4y4

 

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

(i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2 (iv) a, 2b, 3c

Solution: Volume of a rectangular box = Length * Breadth * Height

(i) Length = 5a; Breadth = 3a2, Height = 7a4

∴ Volume = Length × Breadth × Height

= 5a * 3a2 * 7a4

= (5 * 3 * 7) * a * a2 * a4 = 105 * a7 = 105a7

(ii) Length = 2p, Breadth = 4q, Height = 8r

∴  Volume = Length * Breadth * Height

= 2p * 4q * 8r

= (2 * 4 * 8) * p * q * r = 64 * pqr = 64pqr

(iii) Length = xy, Breadth = 2x2y, Height = 2xy2

∴ Volume = Length* Breadth* Height

= xy* 2x2y* 2xy2

= (1* 2* 2)* xy* x2y* xy2

= 4* x4y4 = 4x4y4

(iv) Length = a, Breadth = 2b, Height = 3c

∴ Volume = Length* Breadth* Height

= a* 2b* 3c

= (1* 2* 3)* a* b* c

= 6* abc = 6abc

Question 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

Solution:

(i) xy *yz *zx = (1 *1 *1) *x *y *y *z *z *x

= 1 *(x2 *y2 *z2) = x2y2z2

(ii) a *(–a2) *a3 = [1 *(–1) *1] *a *a2 *a3

= (–1) *a6 = –a6

(iii) 2 *4y *8y2 *16y3 = (2 *4 *8 *16) *y *y2 *y3

= 1024 *y6 = 1024y6

(iv) a *2b *3c *6abc = (1 *2 *3 *6) *a *b *c *abc

= 36 *a2b2c2 = 36a2b2c2

(v) m *(–mn) *mnp = [1 *(–1) *1] *m *mn *mnp

= (–1)m3n2p = –m3n2p

Multiplying a Monomial by a Binomial

Question: Find the product

(i) 2x(3x + 5xy)   (ii) a2(2ab – 5c)

Solution: 

(i) 2x(3x + 5xy) = 2x * 3x + 2x * 5xy

= (2 * 3) * x * x + (2 * 5) * x * xy

= 6 * x2 + 10 * x2y

= 6x2 + 10x2y

(ii) a2(2ab – 5c) = a2 * 2ab + a2 * (–5c)

= (1 * 2) * a2 * ab + [1 * (–5)] * a2 * c

= 2 * a3b + (–5) * a2c

= 2a3b – 5a2c

Question: Find the product: (4p2 + 5p + 7) * 3p

Solution:

(4p2 + 5p + 7) * 3p = (4p2 * 3p) + (5p * 3p) + (7 * 3p)

= [(4 * 3) * p2 * p] + [(5 * 3) * p * p] + (7 * 3) * p

= 12 * p3 + 15 * p2 + 21 * p

= 12p3 + 15p2 + 21p

Example 1. Simplify: 2x(x – 1) + 5 and find its value at x = –1

Solution: 

2x(x – 1) + 5 = [2x * x] + [2x * (–1)] + 5

= [(2 * 1) * x * x] + [2 * (–1) * x] + 5

= [2 * x2] + [–2 * x] + 5

= 2x2 – 2x + 5

For x = –1, 2x2 – 2x + 5 = 2(–1)2 – 2(–1) + 5

= (2 * 1) – (–2) + 5

= 2 + 2 + 5 = 9

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