NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities

NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 8 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 8 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 8 Mathematics are an important part of exams for Class 8 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 8 Mathematics and also download more latest study material for all subjects. Chapter 9 Algebraic expressions and identities is an important topic in Class 8, please refer to answers provided below to help you score better in exams

Chapter 9 Algebraic expressions and identities Class 8 Mathematics NCERT Solutions

Class 8 Mathematics students should refer to the following NCERT questions with answers for Chapter 9 Algebraic expressions and identities in Class 8. These NCERT Solutions with answers for Class 8 Mathematics will come in exams and help you to score good marks

Chapter 9 Algebraic expressions and identities NCERT Solutions Class 8 Mathematics

Exercise 9.1

Q.1) Identify the terms, their coefficients for each of the following expressions:
i) 5๐‘ฅ๐‘ฆ๐‘ง2 โˆ’ 3๐‘ง๐‘ฆ ii) 1 + ๐‘ฅ + ๐‘ฅ2 iii) 4๐‘ฅ2๐‘ฆ2 โˆ’ 4๐‘ฅ2๐‘ฆ2๐‘ง2 + ๐‘ง2
iv) 3 โˆ’ ๐‘๐‘ž + ๐‘ž๐‘Ÿ โˆ’ ๐‘Ÿ๐‘   v) ๐‘ฅ/2 + ๐‘ฆ/2 โˆ’ ๐‘ฅ๐‘ฆ vi) 0.3๐‘Ž โˆ’ 06๐‘Ž๐‘ + 0.5๐‘
Sol.1) I) Terms : 5๐‘ฅ๐‘ฆ๐‘ง2 And โˆ’3๐‘ง๐‘ฆ
Coefficient in 5๐‘ฅ๐‘ฆ๐‘ง2 Is 5 and in โˆ’3๐‘ง๐‘ฆ is โˆ’3.
ii) terms : 1, ๐‘ฅ and ๐‘ฅ2
Coefficient of ๐‘ฅ and coefficient of ๐‘ฅ2 in 1.
iii) terms : 4๐‘ฅ2๐‘ฆ2, โˆ’4๐‘ฅ2๐‘ฆ2๐‘ง2 and ๐‘ง2
Coefficient in 4๐‘ฅ2๐‘ฆ2 is 4, coefficient of โˆ’4๐‘ฅ2๐‘ฆ2๐‘ง2 is โˆ’4 and coefficient of ๐‘ง2 is 1.
iv) terms : 3, โˆ’๐‘๐‘ž, ๐‘ž๐‘Ÿ, โˆ’๐‘Ÿ๐‘
Coefficient of โ€“ ๐‘๐‘ž is โˆ’1, coeficient of ๐‘ž๐‘Ÿ is 1 and coefficient of โ€“ ๐‘Ÿ๐‘ is โˆ’1.
v) terms: ๐‘ฅ/2, ๐‘ฆ/2 and โ€“ ๐‘ฅ๐‘ฆ 
coefficient of ๐‘ฅ/2 is 1/2, coefficient of ๐‘ฆ/2 is โˆ’1 and coefficient of โ€“ ๐‘ฅ๐‘ฆ is โˆ’1
vi) Terms: 0.3๐‘Ž, โˆ’06๐‘Ž๐‘ and 0.5๐‘
coefficient of 0.3๐‘Ž is 0.3, coeficient of โˆ’0.6๐‘Ž๐‘ is โˆ’0.6 and coefficient of 0.5๐‘ is 0.5.

Q.2) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories:
i) ๐‘ฅ + ๐‘ฆ ii) 1000 iii) ๐‘ฅ + ๐‘ฅ2 + ๐‘ฅ3 + ๐‘ฅ4 iv) 7 + ๐‘ฆ + 5๐‘ฅ
v) 2๐‘ฆ โ€“ 3๐‘ฆ2 vi) 2๐‘ฆ โ€“ 3๐‘ฆ2 + 4๐‘ฆ3 vii) 5๐‘ฅ โ€“ 4๐‘ฆ + 3๐‘ฅ๐‘ฆ viii) 4๐‘ง โ€“ 15๐‘ง2
ix) ๐‘Ž๐‘ + ๐‘๐‘ + ๐‘๐‘‘ + ๐‘‘๐‘Ž x) ๐‘๐‘ž๐‘Ÿ xi) ๐‘2๐‘ž + ๐‘๐‘ž2 xii) 2๐‘ + 2๐‘ž
Sol.2) (i) Since ๐‘ฅ + ๐‘ฆ contains two terms. Therefore it is binomial.
(ii) Since 1000 contains one term. Therefore it is monomial.
(iii) Since ๐‘ฅ + ๐‘ฅ2 + ๐‘ฅ3 + ๐‘ฅ4 contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(iv) Since 7 + ๐‘ฆ + 5๐‘ฅ contains three terms. Therefore it is trinomial.
(v) Since 2๐‘ฆ โ€“ 3๐‘ฆ2 contains two terms. Therefore it is binomial.
(vi) Since 2๐‘ฆ โ€“ 3๐‘ฆ2 + 4๐‘ฆ3 contains three terms. Therefore it is trinomial.
(vii) Since 5๐‘ฅ โ€“ 4๐‘ฆ + 3๐‘ฅ๐‘ฆ contains three terms. Therefore it is trinomial.
(viii) Since 4๐‘ง โ€“ 15๐‘ง2 contains two terms. Therefore it is binomial. -415 x z
(ix) Since ๐‘Ž๐‘ + ๐‘๐‘ + ๐‘๐‘‘ + ๐‘‘๐‘Ž contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(x) Since ๐‘๐‘ž๐‘Ÿ contains one term. Therefore it is monomial.
(xi) Since ๐‘2๐‘ž + ๐‘๐‘ž2 contains two terms. Therefore it is binomial.
(xii) Since 2๐‘ + 2๐‘ž contains two terms. Therefore it is binomial.

Q.3) Add the following.
(i) ๐‘Ž๐‘ โ€“ ๐‘๐‘, ๐‘๐‘ โ€“ ๐‘๐‘Ž, ๐‘๐‘Ž โ€“ ๐‘Ž๐‘ (ii) ๐‘Ž โ€“ ๐‘ + ๐‘Ž๐‘, ๐‘ โ€“ ๐‘ + ๐‘๐‘, ๐‘ โ€“ ๐‘Ž + ๐‘Ž๐‘

(iii) 2๐‘ ๐‘ž โ€“ 3๐‘๐‘ž + 4, 5 + 7๐‘๐‘ž โ€“ 3๐‘ ๐‘ž (iv) ๐‘™ + ๐‘š , ๐‘š + ๐‘› , ๐‘› + ๐‘™ , 2๐‘™๐‘š + 2๐‘š๐‘› + 2๐‘›๐‘™
Sol.3) (i) ๐‘Ž๐‘ โ€“ ๐‘๐‘, ๐‘๐‘ โ€“ ๐‘๐‘Ž, ๐‘๐‘Ž โ€“ ๐‘Ž๐‘
๐‘Ž๐‘ โˆ’ ๐‘๐‘
+๐‘๐‘ โˆ’ ๐‘๐‘Ž
โˆ’๐‘Ž๐‘ + ๐‘๐‘Ž
0 + 0 + 0
Hence, the sum is 0.

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-2

Q.4) (a) Subtract 4๐‘Ž โˆ’ 7๐‘Ž๐‘ + 3๐‘ + 12 from 12๐‘Ž โˆ’ 9๐‘Ž๐‘ + 5๐‘ โˆ’ 3
(b) Subtract 3๐‘ฅ๐‘ฆ + 5๐‘ฆ๐‘ง โˆ’ 7๐‘ง๐‘ฅ from 5๐‘ฅ๐‘ฆ โˆ’ 2๐‘ฆ๐‘ง โˆ’ 2๐‘ง๐‘ฅ + 10๐‘ฅ๐‘ฆ๐‘ง
(c) Subtract 4๐‘2๐‘ž โˆ’ 3๐‘๐‘ž + 5๐‘๐‘ž2 โˆ’ 8๐‘ + 7๐‘ž โˆ’ 10 from 18 โˆ’ 3๐‘ โˆ’ 11๐‘ž + 5๐‘๐‘ž โˆ’ 2๐‘๐‘ž2 + 5๐‘2๐‘ž
Sol.4)

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Exercise 9.2

Q.1) Find the product of the following pairs of monomials:
(i) 4 , 7๐‘ (ii) โ€“ 4๐‘, 7๐‘ (iii) โˆ’4๐‘, 7๐‘๐‘ž (iv) 4๐‘3, โˆ’3๐‘ (v) 4๐‘, 0
Sol.1) (i) 4 , 7๐‘
4 ร— 7 ๐‘ = 28๐‘
(ii) โ€“ 4๐‘, 7๐‘
4๐‘ ร— 7๐‘ = โˆ’28๐‘2
(iii) โˆ’4๐‘, 7๐‘๐‘ž
4๐‘ ร— 7๐‘๐‘ž = โˆ’28๐‘2๐‘ž
(iv) 4๐‘3, โˆ’3๐‘
4๐‘3๐‘ž ร— โˆ’ 3๐‘ = โˆ’12๐‘4๐‘ž
(v) 4๐‘, 0
4๐‘ ร— 0 = 0

Q.2) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively: 
(๐‘, ๐‘ž); (10๐‘š, 5๐‘›); (20๐‘ฅ2, 5๐‘ฆ2); (4๐‘ฅ, 3๐‘ฅ2); (3๐‘š๐‘›, 4๐‘›๐‘)
Sol.2) Area = Length ร— breadth
(i) ๐‘ ร— ๐‘ž = ๐‘๐‘ž
(ii) 10๐‘š ร— 5๐‘› = 50๐‘š๐‘›
(iii) 20๐‘ฅ2 ร— 5๐‘ฆ2 = 100๐‘ฅ2๐‘ฆ2
(iv) 4๐‘ฅ ร— 3๐‘ฅ2 = 12๐‘ฅ3
(v) 3๐‘š๐‘› ร— 4๐‘›๐‘ = 12๐‘š๐‘›2๐‘

Q.3) Complete the following table of products:

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-4

Q.4) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5๐‘Ž, 3๐‘Ž2, 7๐‘Ž4 (ii) 2๐‘, 4๐‘ž, 8๐‘Ÿ (iii) ๐‘ฅ๐‘ฆ, 2๐‘ฅ2๐‘ฆ, 2๐‘ฅ๐‘ฆ2 (iv) ๐‘Ž, 2๐‘, 3๐‘
Sol.4) We know that,
Volume = Length ร— Breadth ร— Height
(i) Volume = 5๐‘Ž ร— 3๐‘Ž2 ร— 7๐‘Ž4 = 5 ร— 3 ร— 7 ร— ๐‘Ž ร— ๐‘Ž2 ร— ๐‘Ž4 = 105 ๐‘Ž7
(ii) Volume = 2๐‘ ร— 4๐‘ž ร— 8๐‘Ÿ = 2 ร— 4 ร— 8 ร— ๐‘ ร— ๐‘ž ร— ๐‘Ÿ = 64๐‘๐‘ž๐‘Ÿ
(iii) Volume = ๐‘ฅ๐‘ฆ ร— 2๐‘ฅ2๐‘ฆ ร— 2๐‘ฅ๐‘ฆ2 = 2 ร— 2 ร— ๐‘ฅ๐‘ฆ ร— ๐‘ฅ2๐‘ฆ ร— ๐‘ฅ๐‘ฆ2 = 4๐‘ฅ4๐‘ฆ4
(iv) Volume = ๐‘Ž ร— 2๐‘ ร— 3๐‘ = 2 ร— 3 ร— ๐‘Ž ร— ๐‘ ร— ๐‘ = 6๐‘Ž๐‘๐‘

Q.5) Obtain the product of
(i) ๐‘ฅ๐‘ฆ, ๐‘ฆ๐‘ง, ๐‘ง๐‘ฅ (ii) ๐‘Ž, โˆ’ ๐‘Ž2, ๐‘Ž3 (iii) 2, 4๐‘ฆ, 8๐‘ฆ2, 16๐‘ฆ3
(iv) ๐‘Ž, 2๐‘, 3๐‘, 6๐‘Ž๐‘๐‘ (v) ๐‘š, โˆ’ ๐‘š๐‘›, ๐‘š๐‘›๐‘
Sol.5) (i) ๐‘ฅ๐‘ฆ ร— ๐‘ฆ๐‘ง ร— ๐‘ง๐‘ฅ = ๐‘ฅ2๐‘ฆ2๐‘ง2
(ii) ๐‘Ž ร— (โˆ’ ๐‘Ž2) ร— ๐‘Ž3 = โˆ’ ๐‘Ž6
(iii) 2 ร— 4๐‘ฆ ร— 8๐‘ฆร— 16๐‘ฆ3 = 2 ร— 4 ร— 8 ร— 16 ร— ๐‘ฆ ร— ๐‘ฆ2 ร— ๐‘ฆ3 = 1024 ๐‘ฆ6
(iv) ๐‘Ž ร— 2๐‘ ร— 3๐‘ ร— 6๐‘Ž๐‘๐‘ = 2 ร— 3 ร— 6 ร— ๐‘Ž ร— ๐‘ ร— ๐‘ ร— ๐‘Ž๐‘๐‘ = 36๐‘Ž2 ๐‘2๐‘2
(v) ๐‘š ร— (โˆ’ ๐‘š๐‘›) ร— ๐‘š๐‘›๐‘ = โˆ’ ๐‘š3๐‘›2๐‘

Exercise 9.3

Q.1) Carry out the multiplication of the expressions in each of the following pairs.
(i) 4๐‘, ๐‘ž + ๐‘Ÿ (ii) ๐‘Ž๐‘, ๐‘Ž โˆ’ ๐‘ (iii) ๐‘Ž + ๐‘, 7๐‘Ž2๐‘2
(iv) ๐‘Ž2 โ€“ 9, 4๐‘Ž (v) ๐‘๐‘ž + ๐‘ž๐‘Ÿ + ๐‘Ÿ๐‘, 0
Sol.1) (i) (4๐‘) ร— (๐‘ž + ๐‘Ÿ) = (4๐‘ ร— ๐‘ž) + (4๐‘ ร— ๐‘Ÿ) = 4๐‘๐‘ž + 4๐‘๐‘Ÿ
(ii) (๐‘Ž๐‘) ร— (๐‘Ž โˆ’ ๐‘) = (๐‘Ž๐‘ ร— ๐‘Ž) + [๐‘Ž๐‘ ร— (โˆ’ ๐‘)] = ๐‘Ž2๐‘ โˆ’ ๐‘Ž๐‘2
(iii) (๐‘Ž + ๐‘) ร— (7๐‘Ž2 ๐‘2) = (๐‘Ž ร— 7๐‘Ž2 ๐‘2) + (๐‘ ร— 7๐‘Ž2 ๐‘2) = 7๐‘Ž3๐‘2 + 7๐‘Ž2๐‘3
(iv) (๐‘Ž2 โˆ’ 9) ร— (4๐‘Ž) = (๐‘Ž2 ร— 4๐‘Ž) + (โˆ’ 9) ร— (4๐‘Ž) = 4๐‘Ž3 โˆ’ 36๐‘Ž
(v) (๐‘๐‘ž + ๐‘ž๐‘Ÿ + ๐‘Ÿ๐‘) ร— 0 = (๐‘๐‘ž ร— 0) + (๐‘ž๐‘Ÿ ร— 0) + (๐‘Ÿ๐‘ ร— 0) = 0

Q.2) Complete the table

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-5

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-6

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-7

Q.4) (a) Simplify 3๐‘ฅ (4๐‘ฅ โˆ’ 5) + 3 and find its values for (i) ๐‘ฅ = 3, (ii) ๐‘ฅ = 1/2
(b) ๐‘Ž (๐‘Ž2 + ๐‘Ž + 1) + 5 and find its values for (i) ๐‘Ž = 0, (ii) ๐‘Ž = 1, (iii) ๐‘Ž = โˆ’ 1.
Sol.4) (a) 3๐‘ฅ (4๐‘ฅ โˆ’ 5) + 3 = 12๐‘ฅ2 โˆ’ 15๐‘ฅ + 3
(i) For ๐‘ฅ = 3, 12๐‘ฅ2 โˆ’ 15๐‘ฅ + 3 = 12 (3)2 โˆ’ 15(3) + 3
= 108 โˆ’ 45 + 3
= 66
(ii) For ๐‘ฅ = 1/2, 12๐‘ฅ2 โˆ’ 15๐‘ฅ + 3 = 12 (1/2)2 โˆ’ 15 (1/2) + 3 
= 6 โˆ’ 15/2
= 12 โˆ’ 15/2 = โˆ’3/2

(b) ๐‘Ž (๐‘Ž2 + ๐‘Ž + 1) + 5 = ๐‘Ž3 + ๐‘Ž2 + ๐‘Ž + 5
(i) For ๐‘Ž = 0, ๐‘Ž3 + ๐‘Ž2 + ๐‘Ž + 5 = 0 + 0 + 0 + 5 = 5
(ii) For ๐‘Ž = 1, ๐‘Ž3 + ๐‘Ž2 + ๐‘Ž + 5 = (1)3 + (1)2 + 1 + 5
= 1 + 1 + 1 + 5 = 8
(iii) For ๐‘Ž = โˆ’1, ๐‘Ž3 + ๐‘Ž2 + ๐‘Ž + 5 = (โˆ’1)3 + (โˆ’1)2 + (โˆ’1) + 5
= โˆ’ 1 + 1 โˆ’ 1 + 5 = 4

Q.5) (a) Add: ๐‘ (๐‘ โˆ’ ๐‘ž), ๐‘ž (๐‘ž โˆ’ ๐‘Ÿ) and ๐‘Ÿ (๐‘Ÿ โˆ’ ๐‘)
(b) Add: 2๐‘ฅ (๐‘ง โˆ’ ๐‘ฅ โˆ’ ๐‘ฆ) and 2๐‘ฆ (๐‘ง โˆ’ ๐‘ฆ โˆ’ ๐‘ฅ)
(c) Subtract: 3๐‘™ (๐‘™ โˆ’ 4๐‘š + 5๐‘›) from 4๐‘™ (10๐‘› โˆ’ 3๐‘š + 2๐‘™)
(d) Subtract: 3๐‘Ž (๐‘Ž + ๐‘ + ๐‘) โˆ’ 2๐‘ (๐‘Ž โˆ’ ๐‘ + ๐‘) from 4๐‘ (โˆ’ ๐‘Ž + ๐‘ + ๐‘)
Sol.5) a) ๐‘(๐‘ โˆ’ ๐‘ž) + ๐‘ž(๐‘ž โˆ’ ๐‘Ÿ) + ๐‘Ÿ(๐‘Ÿ โˆ’ ๐‘)
= ๐‘2 โˆ’ ๐‘๐‘ž + ๐‘ž2 โˆ’ ๐‘ž๐‘Ÿ + ๐‘Ÿ2 โˆ’ ๐‘Ÿ๐‘
= ๐‘2 + ๐‘ž2 + ๐‘Ÿ2 โˆ’ ๐‘๐‘ž โˆ’ ๐‘ž๐‘Ÿ โˆ’ ๐‘Ÿ๐‘

b) 2๐‘ฅ(๐‘ง โˆ’ ๐‘ฅ โˆ’ ๐‘ฆ) + 2๐‘ฆ(๐‘ง โˆ’ ๐‘ฆ โˆ’ ๐‘ฅ)
= 2๐‘ฅ๐‘ง โˆ’ 2๐‘ฅ2 โˆ’ 2๐‘ฅ๐‘ฆ + 2๐‘ฆ๐‘ง โˆ’ 2๐‘ฆ2 โˆ’ 2๐‘ฅ๐‘ฆ
= 2๐‘ฅ๐‘ง โˆ’ 2๐‘ฅ๐‘ฆ โˆ’ 2๐‘ฅ๐‘ฆ + 2๐‘ฆ๐‘ง โˆ’ 2๐‘ฅ2 โˆ’ 2๐‘ฆ2
= โˆ’2๐‘ฅ2 โˆ’ 2๐‘ฆ2 โˆ’ 4๐‘ฅ๐‘ฆ + 2๐‘ฆ๐‘ง + 2๐‘ง๐‘ฅ

c) 4๐‘™ (10๐‘› โˆ’ 3๐‘š + 2๐‘™) โˆ’ 3๐‘™ (๐‘™ โ€“ 4๐‘š + 5๐‘›)
= 40๐‘™๐‘› โˆ’ 12๐‘™๐‘š + 8๐‘™2 โˆ’ 3๐‘™2 + 12๐‘™๐‘š โˆ’ 15๐‘™๐‘›
= 8๐‘™2 โˆ’ 3๐‘™2 โˆ’ 12๐‘™๐‘š + 12๐‘™๐‘š + 40๐‘™๐‘› โˆ’ 15๐‘™๐‘›
= 5๐‘™2 + 25๐‘™๐‘›

d) 4๐‘(โˆ’๐‘Ž + ๐‘ + ๐‘) โˆ’ [3๐‘Ž(๐‘Ž + ๐‘ + ๐‘) โˆ’ 2๐‘(๐‘Ž โˆ’ ๐‘ + ๐‘)]
= โˆ’4๐‘Ž๐‘ + 4๐‘๐‘ + 4๐‘2 โˆ’ [3๐‘Ž2 + 3๐‘Ž๐‘ + 3๐‘Ž๐‘ โˆ’ 2๐‘Ž๐‘ + 2๐‘2 โˆ’ 2๐‘๐‘
= โˆ’4๐‘Ž๐‘ + 4๐‘๐‘ + 4๐‘2 โˆ’ [3๐‘Ž2 + 2๐‘2 + 3๐‘Ž๐‘ โˆ’ 2๐‘๐‘ + 3๐‘Ž๐‘ โˆ’ 2๐‘Ž๐‘]
= โˆ’4๐‘Ž๐‘ + 4๐‘๐‘ + 4๐‘2 โˆ’ [3๐‘Ž2 + 2๐‘2 + ๐‘Ž๐‘ + 3๐‘Ž๐‘ โˆ’ 2๐‘๐‘]
= โˆ’4๐‘Ž๐‘ + 4๐‘๐‘ + 4๐‘2 โˆ’ 3๐‘Ž2 โˆ’ 2๐‘2 โˆ’ ๐‘Ž๐‘ โˆ’ 3๐‘Ž๐‘ + 2๐‘๐‘
= โˆ’3๐‘Ž2 โˆ’ 2๐‘2 + 4๐‘2 โˆ’ ๐‘Ž๐‘ + 4๐‘๐‘ + 2๐‘๐‘ โˆ’ 4๐‘Ž๐‘ โˆ’ 3๐‘Ž๐‘
= โˆ’3๐‘Ž2 โˆ’ 2๐‘2 + 4๐‘2 โˆ’ ๐‘Ž๐‘ + 6๐‘๐‘ โˆ’ 7๐‘Ž๐‘

Exercise 9.4

Q.1) Multiply the binomials.
(i) (2๐‘ฅ + 5) and (4๐‘ฅ โˆ’ 3) (ii) (๐‘ฆ โˆ’ 8) and (3๐‘ฆ โˆ’ 4)
(iii) (2.5๐‘™ โˆ’ 0.5๐‘š) and (2.5๐‘™ + 0.5๐‘š) (iv) (๐‘Ž + 3๐‘) and (๐‘ฅ + 5)
(v) (2๐‘๐‘ž + 3๐‘ž2)and (3๐‘๐‘ž โˆ’ 2๐‘ž2) (vi) (3/4 ๐‘Ž2 + 3๐‘2) and 4 (๐‘Ž2 โˆ’2/3 ๐‘2)
Sol.1) (i) (2๐‘ฅ + 5) ร— (4๐‘ฅ โˆ’ 3)
= 2๐‘ฅ ร— (4๐‘ฅ โˆ’ 3) + 5 ร— (4๐‘ฅ โˆ’ 3)
= 8๐‘ฅ2 โˆ’ 6๐‘ฅ + 20๐‘ฅ โˆ’ 15
= 8๐‘ฅ2 + 14๐‘ฅ โˆ’ 15                 (By adding like terms)

(ii) (๐‘ฆ โˆ’ 8) ร— (3๐‘ฆ โˆ’ 4)
= ๐‘ฆ ร— (3๐‘ฆ โˆ’ 4) โˆ’ 8 ร— (3๐‘ฆ โˆ’ 4)
= 3๐‘ฆ2 โˆ’ 4๐‘ฆ โˆ’ 24๐‘ฆ + 32
= 3๐‘ฆ2 โˆ’ 28๐‘ฆ + 32                  (By adding like terms)

(iii) (2.5๐‘™ โˆ’ 0.5๐‘š) ร— (2.5๐‘™ + 0.5๐‘š)
= 2.5๐‘™ ร— (2.5๐‘™ + 0.5๐‘š) โˆ’ 0.5๐‘š (2.5๐‘™ + 0.5๐‘š)
= 6.25๐‘™2 + 1.25๐‘™๐‘š โˆ’ 1.25๐‘™๐‘š โˆ’ 0.25๐‘š2
= 6.25๐‘™โˆ’ 0.25๐‘š2

(iv) (๐‘Ž + 3๐‘) ร— (๐‘ฅ + 5)
= ๐‘Ž ร— (๐‘ฅ + 5) + 3๐‘ ร— (๐‘ฅ + 5)
= ๐‘Ž๐‘ฅ + 5๐‘Ž + 3๐‘๐‘ฅ + 15๐‘

(v) (2๐‘๐‘ž + 3๐‘ž2) ร— (3๐‘๐‘ž โˆ’ 2๐‘ž2)
= 2๐‘๐‘ž ร— (3๐‘๐‘ž โˆ’ 2๐‘ž2) + 3๐‘ž2 ร— (3๐‘๐‘ž โˆ’ 2๐‘ž2)
= 6๐‘2๐‘ž2 โˆ’ 4๐‘๐‘ž3 + 9๐‘๐‘ž3 โˆ’ 6๐‘ž4
= 6๐‘2๐‘ž2 + 5๐‘๐‘ž2 โˆ’ 6๐‘ž4

Q.2) Find the product.
(i) (5 โˆ’ 2๐‘ฅ) (3 + ๐‘ฅ) (ii) (๐‘ฅ + 7๐‘ฆ) (7๐‘ฅ โˆ’ ๐‘ฆ)
(iii) (๐‘Ž2 + ๐‘) (๐‘Ž + ๐‘2) (iv) (๐‘2 โˆ’ ๐‘ž2) (2๐‘ + ๐‘ž)
Sol.2) (i) (5 โˆ’ 2๐‘ฅ) (3 + ๐‘ฅ)
= 5 (3 + ๐‘ฅ) โˆ’ 2๐‘ฅ (3 + ๐‘ฅ)
= 15 + 5๐‘ฅ โˆ’ 6๐‘ฅ โˆ’ 2๐‘ฅ2
= 15 โˆ’ ๐‘ฅ โˆ’ 2๐‘ฅ2

(ii) (๐‘ฅ + 7๐‘ฆ) (7๐‘ฅ โˆ’ ๐‘ฆ)
= ๐‘ฅ (7๐‘ฅ โˆ’ ๐‘ฆ) + 7๐‘ฆ (7๐‘ฅ โˆ’ ๐‘ฆ)
= 7๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฆ + 49๐‘ฅ๐‘ฆ โˆ’ 7๐‘ฆ2
= 7๐‘ฅ2 + 48๐‘ฅ๐‘ฆ โˆ’ 7๐‘ฆ2

(iii) (๐‘Ž2 + ๐‘) (๐‘Ž + ๐‘2)
= ๐‘Ž2(๐‘Ž + ๐‘2) + ๐‘ (๐‘Ž + ๐‘2)
= ๐‘Ž3 + ๐‘Ž2๐‘2 + ๐‘Ž๐‘ + ๐‘3

(iv) (๐‘2 โˆ’ ๐‘ž2) (2๐‘ + ๐‘ž)
= ๐‘2(2๐‘ + ๐‘ž) โˆ’ ๐‘ž2(2๐‘ + ๐‘ž)
= 2๐‘3 + ๐‘2๐‘ž โˆ’ 2๐‘๐‘ž2 โˆ’ ๐‘ž3

Q.3) Simplify.
(i) (๐‘ฅ2 โˆ’ 5) (๐‘ฅ + 5) + 25
(ii) (๐‘Ž2 + 5) (๐‘3 + 3) + 5
(iii) (๐‘ก + ๐‘ 2) (๐‘ก2 โˆ’ ๐‘ )
(iv) (๐‘Ž + ๐‘) (๐‘ โˆ’ ๐‘‘) + (๐‘Ž โˆ’ ๐‘) (๐‘ + ๐‘‘) + 2 (๐‘Ž๐‘ + ๐‘๐‘‘)
(v) (๐‘ฅ + ๐‘ฆ) (2๐‘ฅ + ๐‘ฆ) + (๐‘ฅ + 2๐‘ฆ) (๐‘ฅ โˆ’ ๐‘ฆ)
(vi) (๐‘ฅ + ๐‘ฆ) (๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฆ + ๐‘ฆ2)
(vii) (1.5๐‘ฅ โˆ’ 4๐‘ฆ) (1.5๐‘ฅ + 4๐‘ฆ + 3) โˆ’ 4.5๐‘ฅ + 12๐‘ฆ
(viii) (๐‘Ž + ๐‘ + ๐‘) (๐‘Ž + ๐‘ โˆ’ ๐‘)
Sol.3) (i) (๐‘ฅ2 โˆ’ 5) (๐‘ฅ + 5) + 25
= ๐‘ฅ2 (๐‘ฅ + 5) โˆ’ 5 (๐‘ฅ + 5) + 25
= ๐‘ฅ3 + 5๐‘ฅ2 โˆ’ 5๐‘ฅ โˆ’ 25 + 25
= ๐‘ฅ3 + 5๐‘ฅ2 โˆ’ 5๐‘ฅ

(ii) (๐‘Ž2 + 5) (๐‘3 + 3) + 5
= ๐‘Ž2(๐‘3 + 3) + 5 (๐‘3 + 3) + 5
= ๐‘Ž2๐‘3 + 3๐‘Ž2 + 5๐‘3 + 15 + 5
= ๐‘Ž2๐‘3 + 3๐‘Ž2 + 5๐‘3 + 20

(iii) (๐‘ก + ๐‘ 2) (๐‘ก2 โˆ’ ๐‘ )
= ๐‘ก (๐‘ก2 โˆ’ ๐‘ ) + ๐‘ 2 (๐‘ก2 โˆ’ ๐‘ )
= ๐‘ก3 โˆ’ ๐‘ ๐‘ก + ๐‘ 2๐‘ก2 โˆ’ ๐‘ 3

(iv) (๐‘Ž + ๐‘) (๐‘ โˆ’ ๐‘‘) + (๐‘Ž โˆ’ ๐‘) (๐‘ + ๐‘‘) + 2 (๐‘Ž๐‘ + ๐‘๐‘‘)
= ๐‘Ž (๐‘ โˆ’ ๐‘‘) + ๐‘ (๐‘ โˆ’ ๐‘‘) + ๐‘Ž (๐‘ + ๐‘‘) โˆ’ ๐‘ (๐‘ + ๐‘‘) + 2 (๐‘Ž๐‘ + ๐‘๐‘‘)
= ๐‘Ž๐‘ โˆ’ ๐‘Ž๐‘‘ + ๐‘๐‘ โˆ’ ๐‘๐‘‘ + ๐‘Ž๐‘ + ๐‘Ž๐‘‘ โˆ’ ๐‘๐‘ โˆ’ ๐‘๐‘‘ + 2๐‘Ž๐‘ + 2๐‘๐‘‘
= (๐‘Ž๐‘ + ๐‘Ž๐‘ + 2๐‘Ž๐‘) + (๐‘Ž๐‘‘ โˆ’ ๐‘Ž๐‘‘) + (๐‘๐‘ โˆ’ ๐‘๐‘) + (2๐‘๐‘‘ โˆ’ ๐‘๐‘‘ โˆ’ ๐‘๐‘‘)
= 4๐‘Ž๐‘

(v) (๐‘ฅ + ๐‘ฆ) (2๐‘ฅ + ๐‘ฆ) + (๐‘ฅ + 2๐‘ฆ) (๐‘ฅ โˆ’ ๐‘ฆ)
= ๐‘ฅ (2๐‘ฅ + ๐‘ฆ) + ๐‘ฆ (2๐‘ฅ + ๐‘ฆ) + ๐‘ฅ (๐‘ฅ โˆ’ ๐‘ฆ) + 2๐‘ฆ (๐‘ฅ โˆ’ ๐‘ฆ)
= 2๐‘ฅ2 + ๐‘ฅ๐‘ฆ + 2๐‘ฅ๐‘ฆ + ๐‘ฆ2 + ๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฆ + 2๐‘ฅ๐‘ฆ โˆ’ 2๐‘ฆ2
= (2๐‘ฅ2 + ๐‘ฅ2) + (๐‘ฆ2 โˆ’ 2๐‘ฆ2) + (๐‘ฅ๐‘ฆ + 2๐‘ฅ๐‘ฆ โˆ’ ๐‘ฅ๐‘ฆ + 2๐‘ฅ๐‘ฆ)
= 3๐‘ฅ2 โˆ’ ๐‘ฆ2 + 4๐‘ฅ๐‘ฆ

(vi) (๐‘ฅ + ๐‘ฆ) (๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฆ + ๐‘ฆ2)
= ๐‘ฅ (๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฆ + ๐‘ฆ2) + ๐‘ฆ (๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฆ + ๐‘ฆ2)
= ๐‘ฅ3 โˆ’ ๐‘ฅ2๐‘ฆ + ๐‘ฅ๐‘ฆ2 + ๐‘ฅ2๐‘ฆ โˆ’ ๐‘ฅ๐‘ฆ2 + ๐‘ฆ3
= ๐‘ฅ3 + ๐‘ฆ3 + (๐‘ฅ๐‘ฆ2 โˆ’ ๐‘ฅ๐‘ฆ2) + (๐‘ฅ2๐‘ฆ โˆ’ ๐‘ฅ2๐‘ฆ)
= ๐‘ฅ3 + ๐‘ฆ3

(vii) (1.5๐‘ฅ โˆ’ 4๐‘ฆ) (1.5๐‘ฅ + 4๐‘ฆ + 3) โˆ’ 4.5๐‘ฅ + 12๐‘ฆ
= 1.5๐‘ฅ (1.5๐‘ฅ + 4๐‘ฆ + 3) โˆ’ 4๐‘ฆ (1.5๐‘ฅ + 4๐‘ฆ + 3) โˆ’ 4.5๐‘ฅ + 12๐‘ฆ
= 2.25 ๐‘ฅ2 + 6๐‘ฅ๐‘ฆ + 4.5๐‘ฅ โˆ’ 6๐‘ฅ๐‘ฆ โˆ’ 16๐‘ฆ2 โˆ’ 12๐‘ฆ โˆ’ 4.5๐‘ฅ + 12๐‘ฆ
= 2.25 ๐‘ฅ2 + (6๐‘ฅ๐‘ฆ โˆ’ 6๐‘ฅ๐‘ฆ) + (4.5๐‘ฅ โˆ’ 4.5๐‘ฅ) โˆ’ 16๐‘ฆ2 + (12๐‘ฆ โˆ’ 12๐‘ฆ)
= 2.25๐‘ฅ2 โˆ’ 16๐‘ฆ2

(viii) (๐‘Ž + ๐‘ + ๐‘) (๐‘Ž + ๐‘ โˆ’ ๐‘)
= ๐‘Ž (๐‘Ž + ๐‘ โˆ’ ๐‘) + ๐‘ (๐‘Ž + ๐‘ โˆ’ ๐‘) + ๐‘ (๐‘Ž + ๐‘ โˆ’ ๐‘)
= ๐‘Ž2 + ๐‘Ž๐‘ โˆ’ ๐‘Ž๐‘ + ๐‘Ž๐‘ + ๐‘2 โˆ’ ๐‘๐‘ + ๐‘๐‘Ž + ๐‘๐‘ โˆ’ ๐‘2
= ๐‘Ž2 + ๐‘2 โˆ’ ๐‘2 + (๐‘Ž๐‘ + ๐‘Ž๐‘) + (๐‘๐‘ โˆ’ ๐‘๐‘) + (๐‘๐‘Ž โˆ’ ๐‘๐‘Ž)
= ๐‘Ž22 + ๐‘2 โˆ’ ๐‘2 + 2๐‘Ž๐‘

Exercise 9.5

Q.1) Use a suitable identity to get each of the following products.
(i) (๐‘ฅ + 3) (๐‘ฅ + 3) (ii) (2๐‘ฆ + 5) (2๐‘ฆ + 5) (iii) (2๐‘Ž โˆ’ 7) (2๐‘Ž โˆ’ 7)
(iv) (3๐‘Ž โˆ’1/2) (3๐‘Ž โˆ’ 1/2) (v) (1.1๐‘š โˆ’ 0.4) (1.1 ๐‘š + 0.4) (vi) (๐‘Ž2 + ๐‘2) (โˆ’ ๐‘Ž2 + ๐‘2)
(vii) (6๐‘ฅ โˆ’ 7) (6๐‘ฅ + 7) (viii) (โˆ’ ๐‘Ž + ๐‘) (โˆ’ ๐‘Ž + ๐‘) (ix) (๐‘ฅ/2 + 3๐‘ฆ/4) (๐‘ฅ/2 + 3๐‘ฆ/4)
(x) (7๐‘Ž โˆ’ 9๐‘) (7๐‘Ž โˆ’ 9๐‘)
Sol.1) The products will be as follows.
(i) (๐‘ฅ + 3) (๐‘ฅ + 3)
= (๐‘ฅ + 3)2
= (๐‘ฅ)2 + 2(๐‘ฅ) (3) + (3)2 [(๐‘Ž + ๐‘)2
= ๐‘Ž2 + 2๐‘Ž๐‘ + ๐‘2]
= ๐‘ฅ2 + 6๐‘ฅ + 9

(ii) (2๐‘ฆ + 5) (2๐‘ฆ + 5) = (2๐‘ฆ + 5)2
= (2๐‘ฆ)2 + 2(2๐‘ฆ) (5) + (5)2 [(๐‘Ž + ๐‘)2
= ๐‘Ž2 + 2๐‘Ž๐‘ + ๐‘2]
= 4๐‘ฆ+ 20๐‘ฆ + 25

(iii) (2๐‘Ž โˆ’ 7) (2๐‘Ž โˆ’ 7) = (2๐‘Ž โˆ’ 7)2
= (2๐‘Ž)2 โˆ’ 2(2๐‘Ž) (7) + (7)2 [(๐‘Ž โˆ’ ๐‘)2
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2]
= 4๐‘Ž2 โˆ’ 28๐‘Ž + 49

(iv) (3๐‘Ž โˆ’1/2) (3๐‘Ž โˆ’ 1/2)
= (3๐‘Ž โˆ’ 12)2
= (3๐‘Ž)2 โˆ’ 2 ร— 3๐‘Ž ร— 1/2 + (1/2)2
[(๐‘Ž โˆ’ ๐‘)2
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2]
= 9๐‘Ž2 โˆ’ 3๐‘Ž + 14

(v) (1.1๐‘š โˆ’ 0.4) (1.1 ๐‘š + 0.4)
= (1.1๐‘š)2 โˆ’ (0.4)2 [(๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘) = ๐‘Ž2 โˆ’ ๐‘22]
= 1.21๐‘š2 โˆ’ 0.16

(vi) (๐‘Ž2 + ๐‘2) (โˆ’ ๐‘Ž2 + ๐‘2)
= (๐‘2 + ๐‘Ž2) (๐‘2 โˆ’ ๐‘Ž2)
= (๐‘2)2 โˆ’ (๐‘Ž2)2 [(๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘) = ๐‘Žโˆ’ ๐‘2]
= ๐‘4 โˆ’ ๐‘Ž4

(vii) (6๐‘ฅ โˆ’ 7) (6๐‘ฅ + 7)
= (6๐‘ฅ)2 โˆ’ (7)2 [(๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘) = ๐‘Ž2 โˆ’ ๐‘2]
= 36๐‘ฅ2 โ€“ 49

(viii) (โˆ’ ๐‘Ž + ๐‘) (โˆ’ ๐‘Ž + ๐‘)
= (โˆ’ ๐‘Ž + ๐‘)2
= (โˆ’ ๐‘Ž)2 + 2(โˆ’ ๐‘Ž) (๐‘) + (๐‘)2            [(๐‘Ž + ๐‘)2= ๐‘Ž2 + 2๐‘Ž๐‘ + ๐‘2]
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-9

(x) (7๐‘Ž โˆ’ 9๐‘) (7๐‘Ž โˆ’ 9๐‘)
= (7๐‘Ž โˆ’ 9๐‘)2
= (7๐‘Ž)2 โˆ’ 2(7๐‘Ž)(9๐‘) + (9๐‘)2 [(๐‘Ž โˆ’ ๐‘)2
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2]
= 49๐‘Ž2 โˆ’ 126๐‘Ž๐‘ + 81๐‘2

Q.2) Use the identity (๐‘ฅ + ๐‘Ž) (๐‘ฅ + ๐‘) = ๐‘ฅ2 + (๐‘Ž + ๐‘)๐‘ฅ + ๐‘Ž๐‘ to find the following products.
(i) (๐‘ฅ + 3) (๐‘ฅ + 7) (ii) (4๐‘ฅ + 5) (4๐‘ฅ + 1) (iii) (4๐‘ฅ โˆ’ 5) (4๐‘ฅ โˆ’ 1)
(iv) (4๐‘ฅ + 5) (4๐‘ฅ โˆ’ 1) (v) (2๐‘ฅ + 5๐‘ฆ) (2๐‘ฅ + 3๐‘ฆ) (vi) (2๐‘Ž2 + 9) (2๐‘Ž2 + 5)
(vii) (๐‘ฅ๐‘ฆ๐‘ง โˆ’ 4) (๐‘ฅ๐‘ฆ๐‘ง โˆ’ 2)
Sol.2) The products will be as follows.
(i) (๐‘ฅ + 3) (๐‘ฅ + 7)
= ๐‘ฅ2 + (3 + 7) ๐‘ฅ + (3) (7)
= ๐‘ฅ2 + 10๐‘ฅ + 21

(ii) (4๐‘ฅ + 5) (4๐‘ฅ + 1)
= (4๐‘ฅ)2 + (5 + 1) (4๐‘ฅ) + (5) (1)
= 16๐‘ฅ2 + 24๐‘ฅ + 5

(iii) (4๐‘ฅ โ€“ 5)(4๐‘ฅ โ€“ 1)
= (4๐‘ฅ)2 + (โˆ’5 โˆ’ 1)4๐‘ฅ + (โˆ’5) ร— (โˆ’1)
= 16๐‘ฅ2 + (โˆ’6) ร— 4๐‘ฅ + 5 = 16๐‘ฅ2 โˆ’ 24๐‘ฅ + 5

(iv) (4๐‘ฅ + 5)(4๐‘ฅ โ€“ 1)
= (4๐‘ฅ)2 + {5 ร— (โˆ’1)} ร— 4๐‘ฅ + 5 ร— (โˆ’1)
= 16๐‘ฅ2 + (5 โˆ’ 1) ร— 4๐‘ฅ โˆ’ 5
= 16๐‘ฅ2 + 4 ร— 4๐‘ฅ โˆ’ 5
= 16๐‘ฅ2 + 16๐‘ฅ โˆ’ 5

(v) (2๐‘ฅ + 5๐‘ฆ) (2๐‘ฅ + 3๐‘ฆ)
= (2๐‘ฅ)2 + (5๐‘ฆ + 3๐‘ฆ) (2๐‘ฅ) + (5๐‘ฆ) (3๐‘ฆ)
= 4๐‘ฅ2 + 16๐‘ฅ๐‘ฆ + 15๐‘ฆ2

(vi) (2๐‘Ž2 + 9) (2๐‘Ž2 + 5)
= (2๐‘Ž2)2 + (9 + 5) (2๐‘Ž2) + (9) (5)
= 4๐‘Ž4 + 28๐‘Ž2 + 45

(vii) (๐‘ฅ๐‘ฆ๐‘ง โˆ’ 4) (๐‘ฅ๐‘ฆ๐‘ง โˆ’ 2)
= (๐‘ฅ๐‘ฆ๐‘ง)2 + (โˆ’4 โˆ’ 2) ร— ๐‘ฅ๐‘ฆ๐‘ง + (โˆ’4) ร— (โˆ’2)
= ๐‘ฅ2๐‘ฆ2๐‘ง2 โˆ’ 6๐‘ฅ๐‘ฆ๐‘ง + 8

Q.3) Find the following squares by suing the identities.
(i) (๐‘ โˆ’ 7)2
(ii) (๐‘ฅ๐‘ฆ + 3๐‘ง)2
(iii) (6๐‘ฅ2 โˆ’ 5๐‘ฆ)2 (iv) (2/3 ๐‘š + 3/2 ๐‘›)2
(v) (0.4๐‘ โˆ’ 0.5๐‘ž)2
(vi) (2๐‘ฅ๐‘ฆ + 5๐‘ฆ)2
Sol.3) (i) (๐‘ โˆ’ 7)22
= (๐‘)2 โˆ’ 2(๐‘) (7) + (7)2 [(๐‘Ž โˆ’ ๐‘)2
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2
= ๐‘2 โˆ’ 14๐‘ + 49

(ii) (๐‘ฅ๐‘ฆ + 3๐‘ง)22
= (๐‘ฅ๐‘ฆ)2 + 2(๐‘ฅ๐‘ฆ) (3๐‘ง) + (3๐‘ง)2 [(๐‘Ž + ๐‘)2
= ๐‘Ž2 + 2๐‘Ž๐‘ + ๐‘2]
= ๐‘ฅ2๐‘ฆ2 + 6๐‘ฅ๐‘ฆ๐‘ง + 9๐‘ง2

(iii) (6๐‘ฅ2 โˆ’ 5๐‘ฆ)2
= (6๐‘ฅ2)2 โˆ’ 2(6๐‘ฅ2) (5๐‘ฆ) + (5๐‘ฆ)2 [(๐‘Ž โˆ’ ๐‘)2
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + ๐‘2]
= 36๐‘ฅ2 โˆ’ 60๐‘ฅ2๐‘ฆ + 25๐‘ฆ2

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities-8

(v) (0.4๐‘ โˆ’ 0.5๐‘ž)2
= (0.4๐‘)2 โˆ’ 2 (0.4๐‘) (0.5๐‘ž) + (0.5๐‘ž)2[(๐‘Ž โˆ’ ๐‘)2
= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + b2]
= 0.16๐‘2 โˆ’ 0.4๐‘๐‘ž + 0.25q2
(vi) (2๐‘ฅ๐‘ฆ + 5๐‘ฆ)2
= (2๐‘ฅ๐‘ฆ)2 + 2(2๐‘ฅ๐‘ฆ) (5๐‘ฆ) + (5๐‘ฆ)2 [(๐‘Ž + ๐‘)2
= ๐‘Ž2 + 2๐‘Ž๐‘ + b2]
= 4x2๐‘ฆ2 + 20๐‘ฅ๐‘ฆ2 + 25๐‘ฆ2

Q.4) Simplify.
(i) (๐‘Ž2 โˆ’ b2)2
(ii) (2๐‘ฅ + 5)2 โˆ’ (2๐‘ฅ โˆ’ 5)2
(iii) (7๐‘š โˆ’ 8๐‘›)2 + (7๐‘š + 8๐‘›)2
(iv) (4๐‘š + 5๐‘›)2 + (5๐‘š + 4๐‘›)2
(v) (2.5๐‘ โˆ’ 1.5๐‘ž)2 โˆ’ (1.5๐‘ โˆ’ 2.5๐‘ž)2 (vi) (๐‘Ž๐‘ + ๐‘๐‘)2 โˆ’ 2๐‘Žb2๐‘
(vii) (m2 โˆ’ ๐‘›2๐‘š)2 + 2๐‘š3๐‘›2
Sol.4) (i) (๐‘Ž2 โˆ’ b2)2
= (๐‘Ž2)2 โˆ’ 2(๐‘Ž2) (b2) + (b2)2                                                     [(a โˆ’ b)2 = ๐‘Ž2 โˆ’ 2ab + b2 ]
= a4 โˆ’ ๐‘Ž2b2 + b4
(ii) (2x +5)2 โˆ’ (2x โˆ’ 5)2
= (2x)2 + 2(2x) (5) + (5)2 โˆ’ [(2x)2 โˆ’ 2(2x) (5) + (5)2]              [(a โˆ’ b)2 = ๐‘Ž2 โˆ’ 2ab + b2]
[(a + b)2 = ๐‘Ž2 + 2ab + b2]
= 4x2 + 20x + 25 โˆ’ [4x2 โˆ’ 20x + 25]
= 4x2 + 20x + 25 โˆ’ 4x2 + 20x โˆ’ 25 = 40x

(iii) (7m โˆ’ 8n)2 + (7m + 8n)2
= (7m)2 โˆ’ 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2       [(a โˆ’ b)2 = ๐‘Ž2 โˆ’ 2ab + b2 and
(a + b)2 = ๐‘Ž2 + 2ab + b2]
= 49m2 โˆ’ 112mn + 64๐‘›2 + 49m2 + 112mn + 64๐‘›2
= 98m2 + 128๐‘›2

(iv) (4m + 5n)2 + (5m + 4n)2
= (4m)2 + 2(4m) (5n) + (5n)2 + (5m)2 + 2(5m) (4n) + (4n)2        [(a + b)2 = ๐‘Ž2 + 2ab + b2]
= 16m2 + 40mn + 25๐‘›2 + 25m2 + 40mn + 16๐‘›2
= 41m2 + 80mn + 41๐‘›2

(v) (2.5p โˆ’ 1.5q)2 โˆ’ (1.5p โˆ’ 2.5q)2
= (2.5p)2 โˆ’ 2(2.5p) (1.5q) + (1.5q)2 โˆ’ [(1.5p)2 โˆ’ 2(1.5p)(2.5q) + (2.5q)2]              [(a โˆ’ b)2 = ๐‘Ž2 โˆ’ 2ab + b2 ]
= 6.25๐‘2 โˆ’ 7.5pq + 2.25q2 โˆ’ [2.25๐‘2 โˆ’ 7.5pq + 6.25q2]
= 6.25๐‘2 โˆ’ 7.5pq + 2.25q2 โˆ’ 2.25๐‘2 + 7.5pq โˆ’ 6.25q2]
= 4๐‘2 โˆ’ 4q2

(vi) (ab + bc)2 โˆ’ 2ab2c
= (ab)2 + 2(ab)(bc) + (bc)2 โˆ’ 2ab2c                         [(a + b)2 = ๐‘Ž2 + 2ab + b2 ]
= ๐‘Ž2b2 + 2ab2c + b2c2 โˆ’ 2ab2c
= ๐‘Ž2b2 + b2c2

(vii) (m2 โˆ’ ๐‘›2m)2 + 2๐‘š3๐‘›2
= (m2)2 โˆ’ 2(m2) (๐‘›2m) + (๐‘›2m)2 + 2๐‘š3๐‘›2              [(a โˆ’ b)2 = ๐‘Ž2 โˆ’ 2ab + b2 ]
= m4 โˆ’ 2๐‘š3๐‘›2 + n4m2 + 2๐‘š3๐‘›2
= m4 + n4m2

Q.5) `Show that
(i) (3๐‘ฅ + 7)2 โˆ’ 84๐‘ฅ = (3๐‘ฅ โˆ’ 7)2
(ii) (9๐‘ โˆ’ 5๐‘ž)2 + 180๐‘๐‘ž = (9๐‘ + 5๐‘ž)2

(iv) (4๐‘๐‘ž + 3๐‘ž)2 โˆ’ (4๐‘๐‘ž โˆ’ 3๐‘ž)2 = 48๐‘q2
(v) (๐‘Ž โˆ’ ๐‘) (๐‘Ž + ๐‘) + (๐‘ โˆ’ ๐‘) (๐‘ + ๐‘) + (๐‘ โˆ’ ๐‘Ž) (๐‘ + ๐‘Ž) = 0
Sol.5)
(i) L.H.S = (3x + 7)2 โˆ’ 84x
= (3x)2 + 2(3x)(7) + (7)2 โˆ’ 84x
= 9x2 + 42x + 49 โˆ’ 84x
= 9x2 โˆ’ 42x + 49
R.H.S = (3x โˆ’ 7)2 = (3x)2 โˆ’ 2(3x)(7) +(7)2
= 9x2 โˆ’ 42x + 49
L.H.S = R.H.S
(ii) L.H.S = (9p โˆ’ 5q)2 + 180pq
= (9p)2 โˆ’ 2(9p)(5q) + (5q)2 โˆ’ 180pq
= 81p2 โˆ’ 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
R.H.S = (9p + 5q)2
= (9p)2 + 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2q2
L.H.S = R.H.S

""NCERT-Solutions-Class-8-Mathematics-Algebraic-expressions-and-identities

(iv) L.H.S = (4pq + 3q)2 โˆ’ (4pq โˆ’ 3q)2
= (4pq)2 + 2(4pq)(3q) + (3q)2 โˆ’ [(4pq)2 โˆ’ 2(4pq) (3q) + (3q)2]
= 16p2q2 + 24pq2 + 9q2 โˆ’ [16p2q2 โˆ’ 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 โˆ’16p2q2 + 24pq2 โˆ’ 9q2
= 48pq2 = R.H.S

(v) L.H.S = (a โˆ’ b) (a + b) + (b โˆ’ c) (b + c) + (c โˆ’ a) (c + a)
= (a2 โˆ’ b2) + (b2 โˆ’ c2) + (c2 โˆ’ a2) = 0 = R.H.S

Q.6) Using identities, evaluate.
(i) 712 (ii) 992 (iii) 1022 (iv) 9982 (v) (5.2)2 (vi) 297 ร— 303
(vii) 78 ร— 82 (viii) 8.92 (ix) 1.05 ร— 9.5
Sol.6) (i) 712 = (70 + 1)2
= (70)2 + 2(70) (1) + (1)2              [(๐‘Ž + ๐‘)2= a2 + 2๐‘Ž๐‘ + b2 ]
= 4900 + 140 + 1 = 5041

(ii) 992 = (100 โˆ’ 1)2
= (100)2 โˆ’ 2(100) (1) + (1)2           [(๐‘Ž โˆ’ ๐‘)= ๐‘Ž2 โˆ’ 2๐‘Ž๐‘ + b2 ]
= 10000 โˆ’ 200 + 1 = 9801

(๐‘–๐‘–๐‘–)1022 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2             [(๐‘Ž + ๐‘)= a2 + 2๐‘Ž๐‘ + b2 ]
= 10000 + 400 + 4 = 10404

(๐‘–๐‘ฃ)9982 = (1000 โˆ’ 2)2
= (1000)2 โˆ’ 2(1000)(2) + (2)2           [(๐‘Ž โˆ’ ๐‘)= a2 โˆ’ 2๐‘Ž๐‘ + b2 ]
= 1000000 โˆ’ 4000 + 4 = 996004

(๐‘ฃ) (5.2)2 = (5.0 + 0.2)2
= (5.0)2 + 2(5.0) (0.2) + (0.2)2         [(๐‘Ž + ๐‘)2 = a2 + 2๐‘Ž๐‘ + b2 ]
= 25 + 2 + 0.04 = 27.04

(๐‘ฃ๐‘–) 297 ร— 303 = (300 โˆ’ 3) ร— (300 + 3)
= (300)2 โˆ’ (3)2                                  [(๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘) = a2 โˆ’ b2
= 90000 โˆ’ 9 = 89991

(๐‘ฃ๐‘–๐‘–) 78 ร— 82 = (80 โˆ’ 2) (80 + 2)
= (80)2 โˆ’ (2)2                                    [(๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘) = a2 โˆ’ b2]
= 6400 โˆ’ 4 = 6396

(๐‘ฃ๐‘–๐‘–๐‘–) 8.92 = (9.0 โˆ’ 0.1)2
= (9.0)2 โˆ’ 2(9.0) (0.1) + (0.1)2         [(๐‘Ž โˆ’ ๐‘)2 = a2 โˆ’ 2๐‘Ž๐‘ + b2]
= 81 โˆ’ 1.8 + 0.01 = 79.21

(๐‘–๐‘ฅ) 1.05 ร— 9.5 = 1.05 ร— 0.95 ร— 10
= (1 + 0.05) (1 โˆ’ 0.05) ร— 10
= [(1)2 โˆ’ (0.05)2] ร— 10
= [1 โˆ’ 0.0025] ร— 10                         [(๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘) = a2 โˆ’ b2]
= 0.9975 ร— 10 = 9.975

Q.7) Using a2 โˆ’ b2 = (๐‘Ž + ๐‘) (๐‘Ž โˆ’ ๐‘), find
(i) 512 โ€“ 492 (ii) (1.02)2 โ€“ (0.98)2 (iii) 1532 โˆ’ 1472 (iv) 12.12 โˆ’ 7.92
Sol.7) (๐‘–) 512 โˆ’ 492
= (51 + 49) (51 โˆ’ 49)
= (100) (2) = 200

(๐‘–๐‘–)(1.02)2 โˆ’ (0.98)2
= (1.02 + 0.98) (1.02 โˆ’ 0.98)
= (2) (0.04) = 0.08

(๐‘–๐‘–๐‘–)1532 โˆ’ 1472
= (153 + 147) (153 โˆ’ 147)
= (300) (6) = 1800

(๐‘–๐‘ฃ)12.12 โˆ’ 7.92
= (12.1 + 7.9) (12.1 โˆ’ 7.9)
= (20.0) (4.2) = 84

Q.8) Using (๐‘ฅ + ๐‘Ž) (๐‘ฅ + ๐‘) = ๐‘ฅ2 + (๐‘Ž + ๐‘) ๐‘ฅ + ๐‘Ž๐‘, find
(i) 103 ร— 104 (ii) 5.1 ร— 5.2 (iii) 103 ร— 98 (iv) 9.7 ร— 9.8
Sol.8) (๐‘–) 103 ร— 104
= (100 + 3) (100 + 4)
= (100)2 + (3 + 4) (100) + (3) (4)
= 10000 + 700 + 12 = 10712

(๐‘–๐‘–) 5.1 ร— 5.2
= (5 + 0.1) (5 + 0.2)
= (5)2 + (0.1 + 0.2) (5) + (0.1) (0.2)
= 25 + 1.5 + 0.02 = 26.52

(๐‘–๐‘–๐‘–) 103 ร— 98
= (100 + 3) (100 โˆ’ 2)
= (100)2 + [3 + (โˆ’ 2)] (100) + (3) (โˆ’ 2)
= 10000 + 100 โˆ’ 6
= 10094

(๐‘–๐‘ฃ) 9.7 ร— 9.8
= (10 โˆ’ 0.3) (10 โˆ’ 0.2)
= (10)2 + [(โˆ’ 0.3) + (โˆ’ 0.2)] (10) + (โˆ’ 0.3) (โˆ’ 0.2)
= 100 + (โˆ’ 0.5)10 + 0.06 = 100.06 โˆ’ 5 = 95.06

 

NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities

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