NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3

Class VIII Mathematics NCERT textbook Ex. 8.3 Algebraic Expressions and Identities in Hindi and English Medium updated for academic session 2026-27. The entire content on Tiwari Academy is free to use without any formal registration. We never call or message to user for any reason. If user need help, he can contact us for the problem. Contents in PDF, images and videos are placed on related pages in easy to use mode.

Before taking up the product of an algebraic expressions, let us look at two simple rules.
(i) The product of two factors with like signs is positive, and the product of two factors with unlike signs is negative.
(ii) If x is a variable and m, n are positive integers, then (x3 X x5) = x(3 + 5) = x8
Thus, (x6 X x4) = x(6 + 4) = x10 , etc.

Rule: Product of two monomials = (product of their numerical coefficients) x (product of their variable parts)

Find the product of: (i) 6xy and -3x²y² (ii) 7ab², -4a²b and -5abc
We have:
(i) (6xy) x (-3x² y³) = {6 x (-3)} x {xy X x² y³}
= -18 {x³ y⁴} = -18x³ y4
(ii) (7ab²) x (–4a² b) x {–5abc} = {7 x (–4) x (–5) x {ab² x a² b x abc}
140a⁴ b⁴ c = 140a⁴ b⁴ c.

Rule
Multiply each term of the polynomial by the monomial, using the distributive law:
a x (b + c) = a x b + a x c

Find each of the following products: (i) 5a² b² (3a² – 4ab + 6b²)
We have:
(i) 5a² b² (3a² – 4ab + 6b²) = (5a² b²) x (3a) + (5a² b²) x (–4ab) + (5a² b²) x (6b²)
15a³b² – 20a³b³ + 30a²b⁴

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.
(a+ b) x (c + d) = a x (c + d) + b x (c + d)
= (a x c + a x d) + (b x c + b x d) = ac + ad + bc +bd.
This method is known as the horizontal method.

Multiply (3x + 5y) and (5x – 7y).
We have:
(3x + 5y) X (5x – 7y) = 3x X (5x – 7y) + 5y X (5x – 7y) = (3x X 5x – 3x X 7y) + (5y X 5x – 5y X 7y)
= (15x² – 21xy) + (25xy – 35y²) = 15x² + 4xy – 35y²