NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.4

NCERT Solutions for Class 8 Maths Chapter 8

Algebraic Expressions and Identities Exercise 8.4

Ex 8.4 Class 8 Maths 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)
(vi) (34a2 + 3b2) and 4(a2 – 23 b2)
Solution:
(i) (2x + 5) × (4x – 3)
= 2x × (4x – 3) + 5 × (4x – 3)
= (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3)
= 8x2 – 6x + 20x – 15
= 8x2 + 14x – 15

(ii) (y – 8) × (3y – 4)
= y × (3y – 4) – 8 × (3y – 4)
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32

(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l2 + 1.25ml – 1.25ml – 0.25m2
= 6.25l2 + 0 – 0.25m2
= 6.25l2 – 0.25m2

(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 36 × (x + 5)
= (a × x) + (a × 5) + (36 × x) + (36 × 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) × (3pq – 2q2)
= 2pq × (3pq – 2q2) + 3q2 (3pq – 2q2)
= (2pq × 3pq) – (2pq × 2q2) + (3q2 × 3pq) – (3q2 × 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Q1

Ex 8.4 Class 8 Maths 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)
Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
= 15 + 5x – 6x – 2x2

(ii) (x + 7y) (7x – y)
= x(7x – y) + 7y(7x – y)
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
= 7x2 – xy + 49xy – 7y2
= 7x2 + 48xy – 7y2

(iii) (a2 + b) (a + b2)
= a2 (a + b2) + b(a + b2)
= (a2 × a) + (a2 × b2) + (b × a) + (b × b2)
= a3 + a2b2 + ab + b3

(iv) (p2 – q2)(2p + q)
= p2(2p + q) – q2(2p + q)
= (p2 × 2p) + (p2 × q) – (q2 × 2p) – (q2 × q)
= 2p3 + p2q – 2pq2 – q3

Ex 8.4 Class 8 Maths Question 3.
Simplify:
(i) (x2 – 5) (x + 5) + 25
(ii) (a2 + 5)(b3 + 3) + 5
(iii) (t + s2) (t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y)(x2 – xy + y2)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c) (a + b – c)
Solution:
(i) (x2 – 5) (x + 5) + 25
= x2(x + 5) + 5(x + 5) + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x + 0
= x3 + 5x2 – 5x

(ii) (a2 + 5)(b3 + 3) + 5
= a2(b3 + 3) + 5(b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20

(iii) (t + s2) (t2 – s)
= t(t2 – s) + s2(t2 – s)
= t3 – st + s2t2 – s3
= t3 + s2t2 – st – s3

(iv) (a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 2x2 + x2 + xy + 2xy – xy + 2xy + y2 – 2y2
= 3x2 + 4xy – y2

(vi) (x + y)(x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + x2y + xy2 – xy2 + y3
= x3 – 0 + 0 + y3
= x3 + y3

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x.+ 12y
= 1.5x (1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
= 2.25x2 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y2
= 2.25x2 + 0 + 0 + 0 – 16y2
= 2.25x2 – 16y2

(viii) (a + b + c) (a + b – c)
= a(a + b – c) + b(a + b – c) + c(a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + ab + ab – bc + bc – ac + ac + b2 – c2
= a2 + 2ab + b2 – c2 + 0 + 0
= a2 + 2ab + b2 – c2