Class 8 Maths Exercise 9.4 Easy Solutions NCERT focuses on dividing algebraic expressions. In this exercise, students learn how to divide polynomials by monomials and binomials using basic algebraic rules. The step-by-step NCERT solutions provided here simplify the division process and help students understand the concepts more easily. With these easy-to-follow explanations, students can build their skills in dividing algebraic expressions confidently.
Class 8 Maths Exercise 9.4 Easy Solutions NCERT
Ex 9.4 Class 8 Maths Question 1.
Multiply the binomials:
(i) (2x + 5) and (4x β 3)
(ii) (y β 8) and (3y β 4)
(iii) (2.5l β 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq β 2q2)
(vi) (34a2 + 3b2) and 4(a2 β 23 b2)
Solution:
(i) (2x + 5) Γ (4x β 3)
= 2x Γ (4x β 3) + 5 Γ (4x β 3)
= (2x Γ 4x) β (3 Γ 2x) + (5 Γ 4x) β (5 Γ 3)
NCERT Books
= 8x2 β 6x + 20x β 15
= 8x2 + 14x β 15
(ii) (y β 8) Γ (3y β 4)
= y Γ (3y β 4) β 8 Γ (3y β 4)
= (y Γ 3y) β (y Γ 4) β (8 Γ 3y) + (-8 Γ -4)
= 3y2 β 4y β 24y + 32
= 3y2 β 28y + 32
(iii) (2.5l β 0.5m) Γ (2.5l + 0.5m)
= (2.5l Γ 2.5l) + (2.5l Γ 0.5m) β (0.5m Γ 2.5l) β (0.5m Γ 0.5m)
= 6.25l2 + 1.25ml β 1.25ml β 0.25m2
= 6.25l2 + 0 β 0.25m2
= 6.25l2 β 0.25m2
(iv) (a + 3b) Γ (x + 5)
= a Γ (x + 5) + 36 Γ (x + 5)
= (a Γ x) + (a Γ 5) + (36 Γ x) + (36 Γ 5)
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2) Γ (3pq β 2q2)
= 2pq Γ (3pq β 2q2) + 3q2 (3pq β 2q2)
= (2pq Γ 3pq) β (2pq Γ 2q2) + (3q2 Γ 3pq) β (3q2 Γ 2q2)
= 6p2q2 β 4pq3 + 9pq3 β 6q4
= 6p2q2 + 5pq3 β 6q4

Ex 9.4 Class 8 Maths Question 2.
Find the product:
(i) (5 β 2x) (3 + x)
(ii) (x + 7y) (7x β y)
(iii) (a2 + b) (a + b2)
(iv) (p2 β q2)(2p + q)
Solution:
(i) (5 β 2x) (3 + x)
= 5(3 + x) β 2x(3 + x)
= (5 Γ 3) + (5 Γ x) β (2x Γ 3) β (2x Γ x)
= 15 + 5x β 6x β 2x2
(ii) (x + 7y) (7x β y)
= x(7x β y) + 7y(7x β y)
= (x Γ 7x) β (x Γ y) + (7y Γ 7x) β (7y Γ y)
= 7x2 β xy + 49xy β 7y2
= 7x2 + 48xy β 7y2
(iii) (a2 + b) (a + b2)
= a2 (a + b2) + b(a + b2)
= (a2 Γ a) + (a2 Γ b2) + (b Γ a) + (b Γ b2)
= a3 + a2b2 + ab + b3
(iv) (p2 β q2)(2p + q)
= p2(2p + q) β q2(2p + q)
= (p2 Γ 2p) + (p2 Γ q) β (q2 Γ 2p) β (q2 Γ q)
= 2p3 + p2q β 2pq2 β q3
Ex 9.4 Class 8 Maths Question 3.
Simplify:
(i) (x2 β 5) (x + 5) + 25
(ii) (a2 + 5)(b3 + 3) + 5
(iii) (t + s2) (t2 β s)
(iv) (a + b) (c β d) + (a β b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x β y)
(vi) (x + y)(x2 β xy + y2)
(vii) (1.5x β 4y)(1.5x + 4y + 3) β 4.5x + 12y
(viii) (a + b + c) (a + b β c)
Solution:
(i) (x2 β 5) (x + 5) + 25
= x2(x + 5) + 5(x + 5) + 25
= x3 + 5x2 β 5x β 25 + 25
= x3 + 5x2 β 5x + 0
= x3 + 5x2 β 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 β s)
= t(t2 β s) + s2(t2 β s)
= t3 β st + s2t2 β s3
= t3 + s2t2 β st β s3
(iv) (a + b)(c β d) + (a β b) (c + d) + 2(ac + bd)
= a(c β d) + b(c β d) + a(c + d) β b(c + d) + 2ac + 2bd
= ac β ad + bc β bd + ac + ad β bc β bd + 2ac + 2bd
= ac + ac + 2ac + bc β bc β ad + ad β bd β bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac
(v) (x + y) (2x + y) + (x + 2y) (x β y)
= x(2x + y) + y(2x + y) + x(x β y) + 2y(x β y)
= 2x2 + xy + 2xy + y2 + x2 β xy + 2xy β 2y2
= 2x2 + x2 + xy + 2xy β xy + 2xy + y2 β 2y2
= 3x2 + 4xy β y2
(vi) (x + y)(x2 β xy + y2)
= x(x2 β xy + y2) + y(x2 β xy + y2)
= x3 β x2y + x2y + xy2 β xy2 + y3
= x3 β 0 + 0 + y3
= x3 + y3
(vii) (1.5x β 4y)(1.5x + 4y + 3) β 4.5x.+ 12y
= 1.5x (1.5x + 4y + 3) β 4y(1.5x + 4y + 3) β 4.5x + 12y
= 2.25x2 + 6xy + 4.5x β 6xy β 16y2 β 12y β 4.5x + 12y
= 2.25x2 + 6xy β 6xy + 4.5x β 4.5x + 12y β 12y β 16y2
= 2.25x2 + 0 + 0 + 0 β 16y2
= 2.25x2 β 16y2
(viii) (a + b + c) (a + b β c)
= a(a + b β c) + b(a + b β c) + c(a + b β c)
= a2 + ab β ac + ab + b2 β bc + ac + bc β c2
= a2 + ab + ab β bc + bc β ac + ac + b2 β c2
= a2 + 2ab + b2 β c2 + 0 + 0
= a2 + 2ab + b2 β c2





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