Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 CBSE Part 4

In this page we have NCERT Solutions for Algebraic Expressions and Identities Class 8 Mathematics Chapter 9 for EXERCISE 4 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Multiply the binomials.
i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
Answer: Let  ( a+b) (c+d) to be done
then
( a+b) (c+d)= a(c+d) + b( c+d)
=a x  c+a x  d+b x  c+b x d
We will use the same concept in all the question below
i) (2x + 5)(4x - 3)
2x x 4x - 2x x 3 + 5 x 4x - 5 x 3
8x² - 6x + 20x -15
8x² + 14x -15
ii) ( y - 8)(3y - 4)
= y x 3y - 4y - 8 x 3y + 32
= 3y2 - 4y - 24y + 32
= 3y2 - 28y + 32
iii) (2.5l - 0.5m)(2.5l + 0.5)
Using (a+b)(a-b) = a2 - b2
We get = 6.25l2 - 0.25m2
iv) ax + 5a + 3bx + 15b
v) 2pq x 3pq - 2pq x 2q2 + 3q2 x 3pq - 3q2 x 2q2
= 6p2q2 - 4pq3 + 9pq3 - 6q4 = 6p2q2 - 5pq3 - 6p4
Question 2
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
iii) (a2+ b) (a + b2)
(iv) (p2– q2) (2p + q)
Answer: i) 15 + 5x - 6x - 2x2
= 15 - x - x 2
ii) 7x2 - xy + 49xy - 7y2
= 7x2 - 7y2 + 48xy
iii) a2 x a + ax b + a x b + b3
= a3 + a2b + ab + b3
= a3 + b3 + a2b + ab
iv) 2p3 + p2q - 2pq2 - q3
= 2p3 - q3 + p2q - 2pq2

Question 3
Simplify.
(i) (x2– 5) (x + 5) + 25
(ii) (a2+ 5) (b3+ 3) + 5
(iii) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(iv) (x + y)(2x + y) + (x + 2y)(x – y)
(v) (x + y)(x2– xy + y2)
(vi) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(vii) (a + b + c)(a + b – c)
i) x3 + 5x2 - 5x - 25 + 25
= x3 + 5x2 -5x
ii) a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 5b3 + 3a2 + 20
iii) (ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)
= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd
= 4ac
iv) 2x2 + xy + 2xy + y2 + x2 - xy + 2xy - 2y2
= 3x2 + 4xy - y2
v) x3 - x2y + xy2 + x2y - xy2 + y3
= x3 + y3
vi) 2.25x2 + 6xy + 4.5x - 6xy - 16y2 - 12y - 4.5x + 12y
= 2.25x2 - 16y2
vii) a2 + ab - ac + ab + b2 - bc + ac + bc - c2
= a2 + b2 - c2 + 2ab