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

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.

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

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) – (ab × b)
= a2b – ab2

(iii) (a + b) (7a2b2)
= (a × 7a2b2) + (b × 7a2b2)
= 7a3b2 + 7a2b3

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

(iv) (a2 – 9) × 4a
= (a2 × 4a) – (9 × 4a)
= 4a3 – 36a

(v) (pq + qr + rp) × 0
= (pq × 0) + (qr × 0) + (rp × 0)
= 0 + 0 + 0
= 0

Question 2.
Complete the table.
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Q2
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
= 6p3 – 7p2 + 5p

(iv) 4p2q2 × (p2 – q2)
= (4p2q2 × p2) + (4p2q2 × -q2)
= 4p4q2 + (-4p2q4)
= 4p4q2 – 4p2q4

(v) (a + b + c) × abc
= (a × abc) + (b × abc) + (c × abc)
= a2bc + ab2c + abc2

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

Question 3.
Find the product.
(i) (a2) × (2a22) × (4a26)
(ii) \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^{2} y^{2}\right)\)
(iii) \(\left(-\frac{10}{3} \mathrm{pq}^{3}\right) \times\left(\frac{6}{5} \mathrm{p}^{3} \mathrm{q}\right)\)
(iv) x × x2 × x3 × x4
Solution:
(i) (a2) × (2a22) × (4a26)
= (1 × 2 × 4) × (a2 × a22 × a26)
= 8 × (a2+22+26)
= 8a50
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Q3
(iv) x × x2 × x3 × x4
= x1+2+3+4
= x10

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) 3x (4x – 5) + 3
= (3x × 4x) + (3x × -5) + 3
= 12x2 – 15x + 3
(i) when x = 3
12x2 – 15x + 3
= 12 × (3)2 – 15(3) + 3
= 12 × 9 – 45 + 3
= 108 – 45 + 3
= 111 – 45
= 66

(ii) When x = \(\frac{1}{2}\)
12x2 – 15x + 3
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Q4

(b) a(a2 + a + 1) + 5
= (a × a2) + (a × a) + (a × 1) + 5
= a3 + a2 + a + 5

(i) When a = 0
a3 + a2 + a + 5
= (0)3 + (0)2 + 0 + 5
= 0 + 0 + 0 + 5
= 5

(ii) When a = 1
a3 + a2 + a + 5
= (13) + (12) + 1 + 5
= 1 + 1 + 1 + 5
= 8

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

(iii) When a = -1
a3 + a2 + a + 5
= (-1)3 + (-1)2 + (-1) + 5
= (-1) + 1 – 1 + 5
= -1 + 1 – 1 + 5
= 4

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) + 2yz + 2y (-y) + 2y(-x)
= 2xz – 2x2 – 2xy + 2yz – 2y2 – 2xy
= -2x2 – 2y2 – 4xy + 2yz + 2xz

(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l × 10n) + (4l × -3m) + (4l × 2l) + (-3l × 1) + (-3l × -4m) + (-3l × 5n)
= 40ln + (-12lm) + 8l2 + (-3l2) + 12lm + (-15ln)
= 40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln
= 40ln – 15ln – 12lm + 12lm + 8l2 – 3l2
= 25ln + 5l2
= 5l2 + 25ln

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

(d) Simplify the I part we get 3a (a + b + c) – 2b(a – b + c)
= (3a × a) + (3a × b) + (3a × c) – [(2b × a) + 2b(-b) + (2b) × c]
= 3a2 + 3ab + 3ac – (2ab – 2b2 + 2bc)
= 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc
= 3a2 + 2b2 + 3ab – 2ab + 3ac – 2bc
= 3a2 + 2b2 + ab + 3ac – 2bc
Simplify the 2nd part 4c × (-a + b + c) = (4c × – a) + (4c × b) + (4c × c)
According to the given question = -4ac + 4bc + 4c2
2nd part – 1st part
= -4ac + 4bc + 4c2 – (3a2 + 2b2 + ab + 3ac – 2bc)
= -4ac + 4bc + 4c2 – 3a2 – 2b2 – ab – 3ac + 2bc
= -4ac – 3ac + 4bc + 2bc + 4c2 – 3a2 – 2b2 – ab
= -7ac + 6bc + 4c2 – 3a2 – 2b2 – ab
= -3a2 – 2b2 + 4c2 – ab + 6bc – 7ac

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