Mathematics NCERT Grade 8, Chapter 9: Algebraic Expressions and Identities - The chapters give an insight into concepts related to algebraic expressions.
• Expressions are formed from variables and constants
Section 9.2 gives details about terms, factors, and coefficients.
Terms like monomials, binomials, trinomials and polynomials are discussed in detail with examples.
• Expression that contains only one term is called a monomial.
• Expression that contains two terms is called a binomial.
• An expression containing three terms is a trinomial.
• An expression containing, one or more terms with non-zero coefficient (with variables having non-negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one​ term.
Emphasis is given to important topics like Addition and subtraction of algebraic expressions. Like and unlike terms are discussed under section 9.4.
• Like terms are formed from the same variables and the powers of these variables are the same, too. Coefficients of like terms need not be the same.
The discussion moves further to the next topic-multiplication of algebraic expressions.
• There are a number of situations in which we need to multiply algebraic expressions: for example, in finding the area of a rectangle, the sides of which are given as expressions.
This topic contains various sub-parts:
• Multiplying a Monomial by a Monomial
• Multiplying two monomials
• Multiplying three or more monomials
• Multiplying a Monomial by a Polynomial
• Multiplying a monomial by a binomial
• Multiplying a monomial by a trinomial
• Multiplying a Polynomial by a Polynomial
• Multiplying a binomial by a binomial
• Multiplying a binomial by a trinomial
​Ever wondered what is an identity? This chapter will clear all doubts of students relating to the same concept- Identity.
• An identity is an equality, which is true for all values of the variables in equality. On the other hand, an equation is true only for certain values of its variables
The chapter also discusses Standard identities and also the concept of Applying Identities.
Unsolved exercises, questions in different patterns will make the chapter more comprehensible to students.
All important points of the chapter are cited in the end under the title- What Have We Discussed?

#### Question 1:

Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz23zy

(ii) 1 + x + x2

(iii) 4x2y24x2y2z2 + z2

(iv) 3 − pq + qrrp

(v)

(vi) 0.3a − 0.6ab + 0.5b

The terms and the respective coefficients of the given expressions are as follows.

 - Terms Coefficients (i) 5xyz2 − 3zy 5 − 3 (ii) 1 x x2 1 1 1 (iii) 4x2y2 − 4x2y2z2 z2 4 − 4 1 (iv) 3 − pq qr − rp 3 −1 1 −1 (v) − xy − 1 (vi) 0.3a − 0.6ab 0.5b 0.3 − 0.6 0.5

#### Question 2:

Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y − 3y2, 2y − 3y2 + 4y3, 5x − 4y + 3xy, 4z − 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q

The given expressions are classified as

Monomials: 1000, pqr

Binomials: x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q

Trinomials: 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 4y + 3xy

Polynomials that do not fit in any of these categories are

x + x2 + x3 + x4, ab + bc + cd + da

#### Question 3:

(i) abbc, bcca, caab

(ii) ab + ab, bc + bc, ca + ac

(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2

(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are as follows.

(i)

Thus, the sum of the given expressions is 0.

(ii)

Thus, the sum of the given expressions is ab + bc + ac.

(iii)

Thus, the sum of the given expressions is −p2q2 + 4pq + 9.

(iv)

Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl).

#### Question 4:

(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3

(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz

(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11q + 5pq − 2pq2 + 5p2q

The given expressions in separate rows, with like terms one below the other and then the subtraction of these expressions is as follows.

(a)

(b)

(c)

#### Question 1:

Find the product of the following pairs of monomials.

(i) 4, 7p (ii) − 4p, 7p (iii) − 4p, 7pq

(iv) 4p3, − 3p (v) 4p, 0

The product will be as follows.

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) − 4p × 7p = − 4 × p × 7 × p = (− 4 × 7) × (p × p) = − 28 p2

(iii) − 4p × 7pq = − 4 × p × 7 × p × q = (− 4 × 7) × (p × p × q) = − 28p2q

(iv) 4p3 × − 3p = 4 × (− 3) × p × p × p × p = − 12 p4

(v) 4p × 0 = 4 × p × 0 = 0

#### Question 2:

Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)

We know that,

Area of rectangle = Length × Breadth

Area of 1st rectangle = p × q = pq

Area of 2nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn

Area of 3rd rectangle = 20x2 × 5y2 = 20 × 5 × x2 × y2 = 100 x2y2

Area of 4th rectangle = 4x × 3x2 = 4 × 3 × x × x2 = 12x3

Area of 5th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn2p

#### Question 3:

Complete the table of products.

 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 … … … … … − 5y … … − 15x2y … … … 3x2 … … … … … … − 4xy … … … … … … 7x2y … … … … … … − 9x2y2 … … … … … …

The table can be completed as follows.

 2x − 5y 3x2 − 4xy 7x2y − 9x2y2 2x 4x2 − 10xy 6x3 − 8x2y 14x3y − 18x3y2 − 5y − 10xy 25 y2 − 15x2y 20xy2 − 35x2y2 45x2y3 3x2 6x3 − 15x2y 9x4 − 12x3y 21x4y − 27x4y2 − 4xy − 8x2y 20xy2 − 12x3y 16x2y2 − 28x3y2 36x3y3 7x2y 14x3y − 35x2y2 21x4y − 28x3y2 49x4y2 − 63x4y3 − 9x2y2 − 18x3y2 45 x2y3 − 27x4y2 36x3y3 − 63x4y3 81x4y4

#### Question 4:

Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

We know that,

Volume = Length × Breadth × Height

(i) Volume = 5a × 3a2 × 7a4 = 5 × 3 × 7 × a × a2 × a4 = 105 a7

(ii) Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr

(iii) Volume = xy × 2x2y × 2xy2 = 2 × 2 × xy ×x2y × xy2 = 4x4y4

(iv) Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc

##### Video Solution for algebraic expressions and identities (Page: 144 , Q.No.: 4)

NCERT Solution for Class 8 maths - algebraic expressions and identities 144 , Question 4

#### Question 5:

Obtain the product of

(i) xy, yz, zx (ii) a, − a2, a3 (iii) 2, 4y, 8y2, 16y3

(iv) a, 2b, 3c, 6abc (v) m, − mn, mnp

(i) xy × yz × zx = x2y2z2

(ii) a × (− a2) × a3 = − a6

(iii) 2 × 4y × 8y2 × 16y3 = 2 × 4 × 8 × 16 × y × y2 × y3 = 1024 y6

(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a2b2c2

(v) m × (− mn) × mnp = − m3n2p

#### Question 1:

Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r (ii) ab, ab (iii) a + b, 7a2b2

(iv) a2 − 9, 4a (v) pq + qr + rp, 0

(i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr

(ii) (ab) × (ab) = (ab × a) + [ab × (− b)] = a2b ab2

(iii) (a + b) × (7a2 b2) = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3

(iv) (a2 − 9) × (4a) = (a2 × 4a) + (− 9) × (4a) = 4a3 − 36a

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

#### Question 2:

Complete the table

 --- First expression Second Expression Product (i) a b + c + d - (ii) x + y − 5 5 xy - (iii) p 6p2 − 7p + 5 - (iv) 4p2q2 p2 − q2 - (v) a + b + c abc -

The table can be completed as follows.

 - First expression Second Expression Product (i) a b + c + d ab + ac + ad (ii) x + y − 5 5 xy 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

#### Question 3:

Find the product.

(i) (a2) × (2a22) × (4a26)

(ii)

(iii)

(iv) x × x2 × x3 × x4

(i) (a2) × (2a22) × (4a26) = 2 × 4 ×a2 × a22 × a26 = 8a50

(ii)

(iii)

(iv) x × x2 × x3 × x4 = x10

#### Question 4:

(a) Simplify 3x (4x −5) + 3 and find its values for (i) x = 3, (ii) .

(b) a (a2 + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = − 1.

(a) 3x (4x − 5) + 3 = 12x2 − 15x + 3

(i) For x = 3, 12x2 − 15x + 3 = 12 (3)2 − 15(3) + 3

= 108 − 45 + 3

= 66

(ii) For

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

(i) For a = 0, a3 + a2 + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a3 + a2 + a + 5 = (1)3 + (1)2 + 1 + 5

= 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a3 + a2 + a + 5 = (−1)3 + (−1)2 + (−1) + 5

= − 1 + 1 − 1 + 5 = 4

#### Question 1:

Multiply the binomials.

(i) (2x + 5) and (4x − 3) (ii) (y − 8) and (3y − 4)

(iii) (2.5l − 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)

(v) (2pq + 3q2) and (3pq − 2q2)

(vi)

(i) (2x + 5) × (4x − 3) = 2x × (4x − 3) + 5 × (4x − 3)

= 8x2 − 6x + 20x − 15

= 8x2 + 14x −15 (By adding like terms)

(ii) (y − 8) × (3y − 4) = y × (3y − 4) − 8 × (3y − 4)

= 3y2 − 4y − 24y + 32

= 3y2 − 28y + 32 (By adding like terms)

(iii) (2.5l − 0.5m) × (2.5l + 0.5m) = 2.5l × (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)

= 6.25l2 + 1.25lm − 1.25lm − 0.25m2

= 6.25l2 − 0.25m2

(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) × (3pq − 2q2) = 2pq × (3pq − 2q2) + 3q2 × (3pq − 2q2)

= 6p2q2 − 4pq3 + 9pq3 − 6q4

= 6p2q2 + 5pq3 − 6q4

(vi)

#### Question 2:

Find the product.

(i) (5 − 2x) (3 + x) (ii) (x + 7y) (7xy)

(iii) (a2 + b) (a + b2) (iv) (p2q2) (2p + q)

(i) (5 − 2x) (3 + x) = 5 (3 + x) − 2x (3 + x)

= 15 + 5x − 6x − 2x2

= 15 − x − 2x2

(ii) (x + 7y) (7xy) = x (7xy) + 7y (7xy)

= 7x2xy + 49xy − 7y2

= 7x2 + 48xy − 7y2

(iii) (a2 + b) (a + b2) = a2 (a + b2) + b (a + b2)

= a3 + a2b2 + ab + b3

(iv) (p2q2) (2p + q) = p2 (2p + q) − q2 (2p + q)

= 2p3 + p2q − 2pq2q3

#### Question 3:

Simplify.

(i) (x2 − 5) (x + 5) + 25

(ii) (a2 + 5) (b3 + 3) + 5

(iii) (t + s2) (t2 − s)

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

(v) (x + y) (2x + y) + (x + 2y) (xy)

(vi) (x + y) (x2xy + y2)

(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y

(viii) (a + b + c) (a + bc)

(i) (x2 − 5) (x + 5) + 25

= x2 (x + 5) − 5 (x + 5) + 25

= x3 + 5x2 − 5x − 25 + 25

= x3 + 5x2 − 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 − s)

= t (t2 s) + s2 (t2 − s)

= t3st + s2t2 s3

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

= a (cd) + b (cd) + a (c + d) − b (c + d) + 2 (ac + bd)

= acad + bcbd + ac + adbcbd + 2ac + 2bd

= (ac + ac + 2ac) + (adad) + (bcbc) + (2bdbdbd)

= 4ac

(v) (x + y) (2x + y) + (x + 2y) (xy)

= x (2x + y) + y (2x + y) + x (xy) + 2y (xy)

= 2x2 + xy + 2xy + y2 + x2xy + 2xy − 2y2

= (2x2 + x2) + (y2 − 2y2) + (xy + 2xyxy + 2xy)

= 3x2y2 + 4xy

(vi) (x + y) (x2xy + y2)

= x (x2xy + y2) + y (x2xy + y2)

= x3x2y + xy2 + x2yxy2 + y3

= x3 + y3 + (xy2xy2) + (x2yx2y)

= x3 + y3

(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y

= 1.5x (1.5x + 4y + 3) − 4y (1.5x + 4y + 3) − 4.5x + 12y

= 2.25 x2 + 6xy + 4.5x − 6xy − 16y2 − 12y − 4.5x + 12y

= 2.25 x2 + (6xy − 6xy) + (4.5x − 4.5x) − 16y2 + (12y − 12y)

= 2.25x2 − 16y2

(viii) (a + b + c) (a + bc)

= a (a + bc) + b (a + bc) + c (a + bc)

= a2 + abac + ab + b2bc + ca + bcc2

= a2 + b2c2 + (ab + ab) + (bcbc) + (caca)

= a2 + b2c2 + 2ab

#### Question 1:

Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)

(iii) (2a ­− 7) (2a − 7) (iv)

(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a2 + b2) (− a2 + b2)

(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)

(ix) (x) (7a − 9b) (7a − 9b)

The products will be as follows.

(i) (x + 3) (x + 3) = (x + 3)2

= (x)2 + 2(x) (3) + (3)2 [(a + b)2 = a2 + 2ab + b2]

= x2 + 6x + 9

(ii) (2y + 5) (2y + 5) = (2y + 5)2

= (2y)2 + 2(2y) (5) + (5)2 [(a + b)2 = a2 + 2ab + b2]

= 4y2 + 20y + 25

(iii) (2a ­− 7) (2a − 7) = (2a − 7)2

= (2a)2 − 2(2a) (7) + (7)2 [(ab)2 = a2 − 2ab + b2]

= 4a2 − 28a + 49

(iv)

[(ab)2 = a2 − 2ab + b2]

(v) (1.1m − 0.4) (1.1 m + 0.4)

= (1.1m)2 − (0.4)2 [(a + b) (ab) = a2b2]

= 1.21m2 − 0.16

(vi) (a2 + b2) (− a2 + b2) = (b2 + a2) (b2a2)

= (b2)2 − (a2)2 [(a + b) (ab) = a2b2]

= b4a4

(vii) (6x − 7) (6x + 7) = (6x)2 − (7)2 [(a + b) (ab) = a2b2]

= 36x2 − 49

(viii) (− a + c) (− a + c) = (− a + c)2

= (− a)2 + 2(− a) (c) + (c)2 [(a + b)2 = a2 + 2ab + b2]

= a2 − 2ac + c2

(ix)

[(a + b)2 = a2 + 2ab + b2]

(x) (7a − 9b) (7a − 9b) = (7a − 9b)2

= (7a)2 − 2(7a)(9b) + (9b)2 [(a b)2 = a2 − 2ab + b2]

= 49a2 − 126ab + 81b2

#### Question 2:

Use the identity (x + a) (x + b) = x2 + (a + b)x + ab to find the following products.

(i) (x + 3) (x + 7) (ii) (4x +5) (4x + 1)

(iii) (4x − 5) (4x − 1) (iv) (4x + 5) (4x − 1)

(v) (2x +5y) (2x + 3y) (vi) (2a2 +9) (2a2 + 5)

(vii) (xyz − 4) (xyz − 2)

The products will be as follows.

(i) (x + 3) (x + 7) = x2 + (3 + 7) x + (3) (7)

= x2 + 10x + 21

(ii) (4x + 5) (4x + 1) = (4x)2 + (5 + 1) (4x) + (5) (1)

= 16x2 + 24x + 5

(iii)

= 16x2 − 24x + 5

(iv)

= 16x2 + 16x − 5

(v) (2x +5y) (2x + 3y) = (2x)2 + (5y + 3y) (2x) + (5y) (3y)

= 4x2 + 16xy + 15y2

(vi) (2a2 +9) (2a2 + 5) = (2a2)2 + (9 + 5) (2a2) + (9) (5)

= 4a4 + 28a2 + 45

(vii) (xyz − 4) (xyz − 2)

=

= x2y2z2 − 6xyz + 8

#### Question 3:

Find the following squares by suing the identities.

(i) (b − 7)2 (ii) (xy + 3z)2 (iii) (6x2 − 5y)2

(iv) (v) (0.4p − 0.5q)2 (vi) (2xy + 5y)2

(i) (b − 7)2 = (b)2 − 2(b) (7) + (7)2 [(ab)2 = a2 − 2ab + b2]

= b2 − 14b + 49

(ii) (xy + 3z)2 = (xy)2 + 2(xy) (3z) + (3z)2 [(a + b)2 = a2 + 2ab + b2]

= x2y2 + 6xyz + 9z2

(iii) (6x2 − 5y)2 = (6x2)2 − 2(6x2) (5y) + (5y)2 [(ab)2 = a2 − 2ab + b2]

= 36x4 − 60x2y + 25y2

(iv) [(a + b)2 = a2 + 2ab + b2]

(v) (0.4p − 0.5q)2 = (0.4p)2 − 2 (0.4p) (0.5q) + (0.5q)2

[(ab)2 = a2 − 2ab + b2]

= 0.16p2 − 0.4pq + 0.25q2

(vi) (2xy + 5y)2 = (2xy)2 + 2(2xy) (5y) + (5y)2

[(a + b)2 = a2 + 2ab + b2]

= 4x2y2 + 20xy2 + 25y2

#### Question 4:

Simplify.

(i) (a2b2)2 (ii) (2x +5)2 − (2x − 5)2

(iii) (7m − 8n)2 + (7m + 8n)2 (iv) (4m + 5n)2 + (5m + 4n)2

(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2

(vi) (ab + bc)2 − 2ab2c (vii) (m2n2m)2 + 2m3n2

(i) (a2b2)2 = (a2)2 − 2(a2) (b2) + (b2)2 [(ab)2 = a2 − 2ab + b2 ]

= a4 − 2a2b2 + b4

(ii) (2x +5)2 − (2x − 5)2 = (2x)2 + 2(2x) (5) + (5)2 − [(2x)2 − 2(2x) (5) + (5)2]

[(ab)2 = a2 − 2ab + b2]

[(a + b)2 = a2 + 2ab + b2]

= 4x2 + 20x + 25 − [4x2 − 20x + 25]

= 4x2 + 20x + 25 − 4x2 + 20x − 25 = 40x

(iii) (7m − 8n)2 + (7m + 8n)2

= (7m)2 − 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2

[(ab)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2]

= 49m2 − 112mn + 64n2 + 49m2 + 112mn + 64n2

= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2

= (4m)2 + 2(4m) (5n) + (5n)2 + (5m)2 + 2(5m) (4n) + (4n)2

[ (a + b)2 = a2 + 2ab + b2]

= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2

= 41m2 + 80mn + 41n2

(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2

= (2.5p)2 − 2(2.5p) (1.5q) + (1.5q)2 − [(1.5p)2 − 2(1.5p)(2.5q) + (2.5q)2]

[(ab)2 = a2 − 2ab + b2 ]

= 6.25p2 − 7.5pq + 2.25q2 − [2.25p2 − 7.5pq + 6.25q2]

= 6.25p2 − 7.5pq + 2.25q2 − 2.25p2 + 7.5pq − 6.25q2]

= 4p2 − 4q2

(vi) (ab + bc)2 − 2ab2c

= (ab)2 + 2(ab)(bc) + (bc)2 − 2ab2c [(a + b)2 = a2 + 2ab + b2 ]

= a2b2 + 2ab2c + b2c2 − 2ab2c

= a2b2 + b2c2

(vii) (m2n2m)2 + 2m3n2

= (m2)2 − 2(m2) (n2m) + (n2m)2 + 2m3n2 [(ab)2 = a2 − 2ab + b2 ]

= m4 − 2m3n2 + n4m2 + 2m3n2

= m4 + n4m2

#### Question 5:

Show that

(i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9p − 5q)2 + 180pq = (9p + 5q)2

(iii)

(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2

(v) (ab) (a + b) + (bc) (b + c) + (ca) (c + a) = 0

(i) L.H.S = (3x + 7)2 − 84x

= (3x)2 + 2(3x)(7) + (7)2 − 84x

= 9x2 + 42x + 49 − 84x

= 9x2 − 42x + 49

R.H.S = (3x − 7)2 = (3x)2 − 2(3x)(7) +(7)2

= 9x2 − 42x + 49

L.H.S = R.H.S

(ii) L.H.S = (9p − 5q)2 + 180pq

= (9p)2 − 2(9p)(5q) + (5q)2 − 180pq

= 81p2 − 90pq + 25q2 + 180pq

= 81p2 + 90pq + 25q2

R.H.S = (9p + 5q)2

= (9p)2 + 2(9p)(5q) + (5q)2

= 81p2 + 90pq + 25q2

L.H.S = R.H.S

(iii) L.H.S =

(iv) L.H.S = (4pq + 3q)2 − (4pq − 3q)2

= (4pq)2 + 2(4pq)(3q) + (3q)2 − [(4pq)2 − 2(4pq) (3q) + (3q)2]

= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]

= 16p2q2 + 24pq2 + 9q2 −16p2q2 + 24pq2 − 9q2

= 48pq2 = R.H.S

(v) L.H.S = (ab) (a + b) + (bc) (b + c) + (ca) (c + a)

= (a2b2) + (b2c2) + (c2a2) = 0 = R.H.S.

#### Question 6:

Using identities, evaluate.

(i) 712 (ii) 992 (iii) 1022 (iv) 9982

(v) (5.2)2 (vi) 297 × 303 (vii) 78 × 82

(viii) 8.92 (ix) 1.05 × 9.5

(i) 712 = (70 + 1)2

= (70)2 + 2(70) (1) + (1)2 [(a + b)2 = a2 + 2ab + b2 ]

= 4900 + 140 + 1 = 5041

(ii) 992 = (100 − 1)2

= (100)2 − 2(100) (1) + (1)2 [(ab)2 = a2 − 2ab + b2 ]

= 10000 − 200 + 1 = 9801

(iii) 1022 = (100 + 2)2

= (100)2 + 2(100)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2 ]

= 10000 + 400 + 4 = 10404

(iv) 9982 = (1000 − 2)2

= (1000)2 − 2(1000)(2) + (2)2 [(ab)2 = a2 − 2ab + b2 ]

= 1000000 − 4000 + 4 = 996004

(v) (5.2)2 = (5.0 + 0.2)2

= (5.0)2 + 2(5.0) (0.2) + (0.2)2 [(a + b)2 = a2 + 2ab + b2 ]

= 25 + 2 + 0.04 = 27.04

(vi) 297 × 303 = (300 − 3) × (300 + 3)

= (300)2 − (3)2 [(a + b) (ab) = a2b2]

= 90000 − 9 = 89991

(vii) 78 × 82 = (80 − 2) (80 + 2)

= (80)2 − (2)2 [(a + b) (ab) = a2b2]

= 6400 − 4 = 6396

(viii) 8.92 = (9.0 − 0.1)2

= (9.0)2 − 2(9.0) (0.1) + (0.1)2 [(ab)2 = a2 − 2ab + b2 ]

= 81 − 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10

= (1 + 0.05) (1− 0.05) ×10

= [(1)2 − (0.05)2] × 10

= [1 − 0.0025] × 10 [(a + b) (ab) = a2b2]

= 0.9975 × 10 = 9.975

##### Video Solution for algebraic expressions and identities (Page: 152 , Q.No.: 6)

NCERT Solution for Class 8 maths - algebraic expressions and identities 152 , Question 6

#### Question 7:

Using a2 b2 = (a + b) (ab), find

(i) 512 − 492 (ii) (1.02)2 − (0.98)2 (iii) 1532 − 1472

(iv) 12.12 − 7.92

(i) 512 − 492 = (51 + 49) (51 − 49)

= (100) (2) = 200

(ii) (1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 ­− 0.98)

= (2) (0.04) = 0.08

(iii) 1532 − 1472 = (153 + 147) (153 − 147)

= (300) (6) = 1800

(iv) 12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)

= (20.0) (4.2) = 84

##### Video Solution for algebraic expressions and identities (Page: 152 , Q.No.: 7)

NCERT Solution for Class 8 maths - algebraic expressions and identities 152 , Question 7

#### Question 8:

Using (x + a) (x + b) = x2 + (a + b) x + ab, find

(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8

(i) 103 × 104 = (100 + 3) (100 + 4)

= (100)2 + (3 + 4) (100) + (3) (4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 × 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

(iii) 103 × 98 = (100 + 3) (100 − 2)

= (100)2 + [3 + (− 2)] (100) + (3) (− 2)

= 10000 + 100 − 6

= 10094

(iv) 9.7 × 9.8 = (10 − 0.3) (10 − 0.2)

= (10)2 + [(− 0.3) + (− 0.2)] (10) + (− 0.3) (− 0.2)

= 100 + (− 0.5)10 + 0.06 = 100.06 − 5 = 95.06

##### Video Solution for algebraic expressions and identities (Page: 152 , Q.No.: 8)

NCERT Solution for Class 8 maths - algebraic expressions and identities 152 , Question 8

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