# NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

NCERT Intext Question Page No. 139
Question 1.
Write two terms which are like
(i) 7xy
(ii) 4mn2
(iii) 21
(i) Two terms like 7xy are: -3xy and 8xy.
(ii) Two terms like 4mn2 are: -6mn2 and 2n2m.
(iii) Two terms like 21 are: -51 and -7b.

NCERT Intext Question Page No. 143
Question 1.
Find 4x × 5y × 7z. First find 4x × 5y and multiply it by 7z; or first find 5y × 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
We have:
4x × 5y × 7z = (4x × 5y) × 7z
= 20xy × 7z = 140xyz
Also 4x × 5y × 7z = 4x × (5y × 7z)
= 4x × 35yz = 140xyz
We observe that
(4x × 5y) × 7x = 4 × (5y × 7z)
The product of monomials is associative, i.e. the order in which we multiply the monomials does not matter.

NCERT Intext Question Page No. 144
Question 1.
Find the product
(i) 2x (3x + 5xy)
(ii) a2 (2ab – 5c)
(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 x 2ab + a2 2 – 5c
= (1 × 2) × a2 × ab + [1 × (-5)] × a2 × c
= 2 × a3b + (-5) × a2c = 2a3b – 5a2c

NCERT Intext Question Page No. 145
Question 1.
Find the product: (4p2 + 5p + 7) × 3p
(4p2 + 5p + 7) × 3p
= (4p2 × 3p) + (5p + 3p) + (7 × 3p)
= [(4 × 3) × p2 × p2] + [(5 × 3) × p × p] + (7 × 3) × p
= 12 × p3 + 15 × p2 + 21 2 p
= 12p3 + 15p2 + 21p

NCERT Intext Question Page No. 149
Question 1.
Verify Identity (IV), for a = 2, b = 3, x = 5.
We have
(x + a)(x + b) = x2 + (a + b)x + ab
Putting a = 2, b = 3 and x = 5 in the identity:
LHS= (x + a) (x + b)
= (5 + 2) (5 + 3)
= 7 × 8 = 56
RHS= x2 + (a + b)x + ab
= (5)2 + (2 + 3) x (2 x 3)
= 25 × (5) × 5 + 6
= 25 × (25) × 6
∴ LHS = RHS
∴ The given identity is true for the given values.

Question 2.
Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity (I)?
When a = b (each = y)
(x + a)(x + b) = x2 + (a + b)x + ab becomes
(x + y)(x + y) = x2 + (y + y)x + (y + y)
= x2 + (2y)x + y2
= x2 + 2xy + y2
= Yes, it is the same as Identity I

Question 3.
Consider, the special case of Identity (IV) with a = -c and b -c. What do you get? Is it related to Identity (II)?
Identity IV is given by
(x + a)(x + b) = x2 + (a + b)x + ab
Replacing ‘a by (-c) and ‘b’ by (-c), we have (x – c)(x – c)
= x2 + [(-c) + (-c)]x + [(-c) × (-c)]
= x2 + [-2c]x + (c2)] = x2 – 2cx + c2
which is same as Identity II.

Question 4.
Consider the special case of Identity (IV) with b = -a. What do you get? Is it related to Identity (III)?