Class 8 Maths Exercise 9.3 Easy Solutions NCERT focuses on multiplying algebraic expressions. This exercise helps students understand how to multiply terms with variables, apply the distributive property, and simplify expressions correctly.
The step-by-step NCERT solutions provided here make these concepts easy to understand and apply. With these clear explanations, students can practice and improve their skills in multiplying algebraic expressions with confidence.
Class 8 Maths Exercise 9.3 Easy Solutions NCERT
Ex 9.3 Class 8 Maths Question 1.
Carry out the multiplication of the expressions in each of the following pairs:
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a2b2
(iv) a2 – 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) 4p × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
(ii) ab, a – b = ab × (a – b) = (ab × a) – (ab × b) = a2b – ab2
(iii) (a + b) × 7a2b2 = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3
(iv) (a2 – 9) × 4a = (a2 × 4a) – (9 × 4a) = 4a3 – 36a
(v) (pq + qr + rp) × 0 = 0
[∵ Any number multiplied by 0 is = 0]
Ex 9.3 Class 8 Maths Question 2.
Complete the table.
| S.No. | First Expression | Second Expression | Product |
| (i) | a | b + c + d | – |
| (ii) | x + y – 5 | 5xy | – |
| (iii) | p | 6p2 – 7p + 5 | – |
| (iv) | 4p2q2 | p2 – q2 | – |
| (v) | a + b + c | abc | – |
Solution:
(i) a × (b + c + d) = (a × b) + (a × c) + (a × d) = ab + ac + ad
(ii) (x + y – 5) (5xy) = (x × 5xy) + (y × 5xy) – (5 × 5xy) = 5x2y + 5xy2 – 25xy
Completed Table:
| S.No. | First Expression | Second Expression | Product |
| (i) | a | b + c + d | ab + ac + ad |
| (ii) | x + y – 5 | 5xy | 5x2y + 5xy2 – 25xy |
| (iii) | p | 6p2 – 7p + 5 | 6p3 – 7p2 + 5p |
| (iv) | 4p2q2 | p2 – q2 | 4p4q2 – 4p2q4 |
| (v) | a + b + c | abc | a2bc + ab2c + abc2 |
Ex 9.3 Class 8 Maths Question 3.
Find the products.
Solution:
Ex 9.3 Class 8 Maths Question 4.
(a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = 12.
(b) Simplify: a(a2 + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1
Solution:
(a) We have 3x(4x – 5) + 3 = 4x × 3x – 5 × 3x + 3 = 12x2 – 15x + 3
(i) For x = 3, we have
12 × (3)2 – 15 × 3 + 3 = 12 × 9 – 45 + 3 = 108 – 42 = 66
(b) We have a(a2 + a + 1) + 5
= (a2 × a) + (a × a) + (1 × a) + 5
= a3 + a2 + a + 5
(i) For a = 0, we have
= (0)3 + (0)2 + (0) + 5 = 5
(ii) For a = 1, we have
= (1)3 + (1)2 + (1) + 5 = 1 + 1 + 1 + 5 = 8
(iii) For a = -1, we have
= (-1)3 + (-1)2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4
Ex 9.3 Class 8 Maths Question 5.
(a) Add: p(p – q), q(q – r) and r(r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
Solution:
(a) p(p – q) + q(q – r) + r(r – p)
= (p × p) – (p × q) + (q × q) – (q × r) + (r × r) – (r × p)
= p2 – pq + q2 – qr + r2 – rp
= p2 + q2 + r2 – pq – qr – rp
(b) 2x(z – x – y) + 2y(z – y – x)
= (2x × z) – (2x × x) – (2x × y) + (2y × z) – (2y × y) – (2y × x)
= 2xz – 2x2 – 2xy + 2yz – 2y2 – 2xy
= -2x2 – 2y2 + 2xz + 2yz – 4xy
= -2x2 – 2y2 – 4xy + 2yz + 2xz
(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l × 10n) – (4l × 3m) + (4l × 2l) – (3l × l) – (3l × -4m) – (3l × 5n)
= 40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln
= (40ln – 15ln) + (-12lm + 12lm) + (8l2 – 3l2)
= 25ln + 0 + 5l2
= 25ln + 5l2
= 5l2 + 25ln
(d) [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]
= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)
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