NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

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
(iii) p × (6p² – 7p + 5) = (p × 6p²) – (p × 7p) + (p × 5) = 6p³ – 7p² + 5p
(iv) 4p²q² × (p² – q²) = 4p²q² × p² – 4p²q² × q² = 4p⁴q² – 4p²q⁴
(v) (a + b + c) × (abc) = (a × abc) + (b × abc) + (c × abc) = a²bc + ab²c + abc²

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 = \(\frac { 1 }{ 2 }\).
(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)
= -4ac + 4bc + 4c² – 3a² – 3ab – 3ac + 2ab – 2b² + 2bc
= -3a² – 2b² + 4c² – ab + 6bc – 7ac

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

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