NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities

Exercise 9.1

Q.1) Identify the terms, their coefficients for each of the following expressions:
i) 5𝑥𝑦𝑧² − 3𝑧𝑦 ii) 1 + 𝑥 + 𝑥² iii) 4𝑥²𝑦2 − 4𝑥²𝑦²𝑧² + 𝑧²
iv) 3 − 𝑝𝑞 + 𝑞𝑟 − 𝑟𝑝   v) 𝑥/2 + 𝑦/2 − 𝑥𝑦 vi) 0.3𝑎 − 06𝑎𝑏 + 0.5𝑏
Sol.1) I) Terms : 5𝑥𝑦𝑧² And −3𝑧𝑦
Coefficient in 5𝑥𝑦𝑧² Is 5 and in −3𝑧𝑦 is −3.
ii) terms : 1, 𝑥 and 𝑥²
Coefficient of 𝑥 and coefficient of 𝑥² in 1.
iii) terms : 4𝑥²𝑦², −4𝑥²𝑦²𝑧² and 𝑧²
Coefficient in 4𝑥²𝑦² is 4, coefficient of −4𝑥²𝑦²𝑧² is −4 and coefficient of 𝑧² is 1.
iv) terms : 3, −𝑝𝑞, 𝑞𝑟, −𝑟𝑝
Coefficient of – 𝑝𝑞 is −1, coeficient of 𝑞𝑟 is 1 and coefficient of – 𝑟𝑝 is −1.
v) terms: 𝑥/2, 𝑦/2 and – 𝑥𝑦 
coefficient of 𝑥/2 is 1/2, coefficient of 𝑦/2 is −1 and coefficient of – 𝑥𝑦 is −1
vi) Terms: 0.3𝑎, −06𝑎𝑏 and 0.5𝑏
coefficient of 0.3𝑎 is 0.3, coeficient of −0.6𝑎𝑏 is −0.6 and coefficient of 0.5𝑏 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) 𝑥 + 𝑦 ii) 1000 iii) 𝑥 + 𝑥² + 𝑥³ + 𝑥⁴ iv) 7 + 𝑦 + 5𝑥
v) 2𝑦 – 3𝑦² vi) 2𝑦 – 3𝑦² + 4𝑦³ vii) 5𝑥 – 4𝑦 + 3𝑥𝑦 viii) 4𝑧 – 15𝑧²
ix) 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑑 + 𝑑𝑎 x) 𝑝𝑞𝑟 xi) 𝑝²𝑞 + 𝑝𝑞² xii) 2𝑝 + 2𝑞
Sol.2) (i) Since 𝑥 + 𝑦 contains two terms. Therefore it is binomial.
(ii) Since 1000 contains one term. Therefore it is monomial.
(iii) Since 𝑥 + 𝑥² + 𝑥³ + 𝑥⁴ contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(iv) Since 7 + 𝑦 + 5𝑥 contains three terms. Therefore it is trinomial.
(v) Since 2𝑦 – 3𝑦² contains two terms. Therefore it is binomial.
(vi) Since 2𝑦 – 3𝑦² + 4𝑦³ contains three terms. Therefore it is trinomial.
(vii) Since 5𝑥 – 4𝑦 + 3𝑥𝑦 contains three terms. Therefore it is trinomial.
(viii) Since 4𝑧 – 15𝑧² contains two terms. Therefore it is binomial. -415 x z
(ix) Since 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑑 + 𝑑𝑎 contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(x) Since 𝑝𝑞𝑟 contains one term. Therefore it is monomial.
(xi) Since 𝑝²𝑞 + 𝑝𝑞² contains two terms. Therefore it is binomial.
(xii) Since 2𝑝 + 2𝑞 contains two terms. Therefore it is binomial.

Q.3) Add the following.
(i) 𝑎𝑏 – 𝑏𝑐, 𝑏𝑐 – 𝑐𝑎, 𝑐𝑎 – 𝑎𝑏 (ii) 𝑎 – 𝑏 + 𝑎𝑏, 𝑏 – 𝑐 + 𝑏𝑐, 𝑐 – 𝑎 + 𝑎𝑐
(iii) 2𝑝 𝑞 – 3𝑝𝑞 + 4, 5 + 7𝑝𝑞 – 3𝑝 𝑞 (iv) 𝑙 + 𝑚 , 𝑚 + 𝑛 , 𝑛 + 𝑙 , 2𝑙𝑚 + 2𝑚𝑛 + 2𝑛𝑙
Sol.3) (i) 𝑎𝑏 – 𝑏𝑐, 𝑏𝑐 – 𝑐𝑎, 𝑐𝑎 – 𝑎𝑏
𝑎𝑏 − 𝑏𝑐
+𝑏𝑐 − 𝑐𝑎
−𝑎𝑏 + 𝑐𝑎
0 + 0 + 0
Hence, the sum is 0.

Q.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
Sol.4)

Exercise 9.2

Q.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
Sol.1) (i) 4 , 7𝑝
4 × 7 𝑝 = 28𝑝
(ii) – 4𝑝, 7𝑝
4𝑝 × 7𝑝 = −28𝑝²
(iii) −4𝑝, 7𝑝𝑞
4𝑝 × 7𝑝𝑞 = −28𝑝²𝑞
(iv) 4𝑝³, −3𝑝
4𝑝³𝑞 × − 3𝑝 = −12𝑝⁴𝑞
(v) 4𝑝, 0
4𝑝 × 0 = 0

Q.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)
Sol.2) Area = Length × breadth
(i) 𝑝 × 𝑞 = 𝑝𝑞
(ii) 10𝑚 × 5𝑛 = 50𝑚𝑛
(iii) 20𝑥² × 5𝑦2 = 100𝑥²𝑦²
(iv) 4𝑥 × 3𝑥² = 12𝑥³
(v) 3𝑚𝑛 × 4𝑛𝑝 = 12𝑚𝑛²𝑝

Q.3) Complete the following table of products:

Q.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
Sol.4) 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

Q.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
Sol.5) (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 = 36𝑎² 𝑏²𝑐²
(v) 𝑚 × (− 𝑚𝑛) × 𝑚𝑛𝑝 = − 𝑚³𝑛²𝑝

Exercise 9.3

Q.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
Sol.1) (i) (4𝑝) × (𝑞 + 𝑟) = (4𝑝 × 𝑞) + (4𝑝 × 𝑟) = 4𝑝𝑞 + 4𝑝𝑟
(ii) (𝑎𝑏) × (𝑎 − 𝑏) = (𝑎𝑏 × 𝑎) + [𝑎𝑏 × (− 𝑏)] = 𝑎²𝑏 − 𝑎𝑏²
(iii) (𝑎 + 𝑏) × (7𝑎² 𝑏²) = (𝑎 × 7𝑎² 𝑏²) + (𝑏 × 7𝑎² 𝑏²) = 7𝑎³𝑏² + 7𝑎²𝑏³
(iv) (𝑎² − 9) × (4𝑎) = (𝑎² × 4𝑎) + (− 9) × (4𝑎) = 4𝑎³ − 36𝑎
(v) (𝑝𝑞 + 𝑞𝑟 + 𝑟𝑝) × 0 = (𝑝𝑞 × 0) + (𝑞𝑟 × 0) + (𝑟𝑝 × 0) = 0

Q.2) Complete the table

Q.4) (a) Simplify 3𝑥 (4𝑥 − 5) + 3 and find its values for (i) 𝑥 = 3, (ii) 𝑥 = 1/2
(b) 𝑎 (𝑎² + 𝑎 + 1) + 5 and find its values for (i) 𝑎 = 0, (ii) 𝑎 = 1, (iii) 𝑎 = − 1.
Sol.4) (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 𝑥 = 1/2, 12𝑥² − 15𝑥 + 3 = 12 (1/2)² − 15 (1/2) + 3 
= 6 − 15/2
= 12 − 15/2 = −3/2

(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

Q.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)
Sol.5) a) 𝑝(𝑝 − 𝑞) + 𝑞(𝑞 − 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)
= 2𝑥𝑧 − 2𝑥² − 2𝑥𝑦 + 2𝑦𝑧 − 2𝑦² − 2𝑥𝑦
= 2𝑥𝑧 − 2𝑥𝑦 − 2𝑥𝑦 + 2𝑦𝑧 − 2𝑥² − 2𝑦²
= −2𝑥² − 2𝑦² − 4𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥

c) 4𝑙 (10𝑛 − 3𝑚 + 2𝑙) − 3𝑙 (𝑙 – 4𝑚 + 5𝑛)
= 40𝑙𝑛 − 12𝑙𝑚 + 8𝑙² − 3𝑙² + 12𝑙𝑚 − 15𝑙𝑛
= 8𝑙² − 3𝑙² − 12𝑙𝑚 + 12𝑙𝑚 + 40𝑙𝑛 − 15𝑙𝑛
= 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 + 2𝑏² + 3𝑎𝑏 − 2𝑏𝑐 + 3𝑎𝑐 − 2𝑎𝑏]
= −4𝑎𝑐 + 4𝑏𝑐 + 4𝑐² − [3𝑎² + 2𝑏² + 𝑎𝑏 + 3𝑎𝑐 − 2𝑏𝑐]
= −4𝑎𝑐 + 4𝑏𝑐 + 4𝑐² − 3𝑎² − 2𝑏² − 𝑎𝑏 − 3𝑎𝑐 + 2𝑏𝑐
= −3𝑎² − 2𝑏² + 4𝑐² − 𝑎𝑏 + 4𝑏𝑐 + 2𝑏𝑐 − 4𝑎𝑐 − 3𝑎𝑐
= −3𝑎² − 2𝑏² + 4𝑐² − 𝑎𝑏 + 6𝑏𝑐 − 7𝑎𝑐

Exercise 9.4

Q.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) (3/4 𝑎² + 3𝑏²) and 4 (𝑎² −2/3 𝑏²)
Sol.1) (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 + 5𝑝𝑞² − 6𝑞⁴

Q.2) Find the product.
(i) (5 − 2𝑥) (3 + 𝑥) (ii) (𝑥 + 7𝑦) (7𝑥 − 𝑦)
(iii) (𝑎² + 𝑏) (𝑎 + 𝑏²) (iv) (𝑝² − 𝑞²) (2𝑝 + 𝑞)
Sol.2) (i) (5 − 2𝑥) (3 + 𝑥)
= 5 (3 + 𝑥) − 2𝑥 (3 + 𝑥)
= 15 + 5𝑥 − 6𝑥 − 2𝑥²
= 15 − 𝑥 − 2𝑥²

(ii) (𝑥 + 7𝑦) (7𝑥 − 𝑦)
= 𝑥 (7𝑥 − 𝑦) + 7𝑦 (7𝑥 − 𝑦)
= 7𝑥² − 𝑥𝑦 + 49𝑥𝑦 − 7𝑦²
= 7𝑥² + 48𝑥𝑦 − 7𝑦²

(iii) (𝑎² + 𝑏) (𝑎 + 𝑏²)
= 𝑎²(𝑎 + 𝑏²) + 𝑏 (𝑎 + 𝑏²)
= 𝑎³ + 𝑎²𝑏² + 𝑎𝑏 + 𝑏³

(iv) (𝑝² − 𝑞²) (2𝑝 + 𝑞)
= 𝑝²(2𝑝 + 𝑞) − 𝑞²(2𝑝 + 𝑞)
= 2𝑝³ + 𝑝²𝑞 − 2𝑝𝑞² − 𝑞³

Q.3) Simplify.
(i) (𝑥² − 5) (𝑥 + 5) + 25
(ii) (𝑎² + 5) (𝑏³ + 3) + 5
(iii) (𝑡 + 𝑠²) (𝑡² − 𝑠)
(iv) (𝑎 + 𝑏) (𝑐 − 𝑑) + (𝑎 − 𝑏) (𝑐 + 𝑑) + 2 (𝑎𝑐 + 𝑏𝑑)
(v) (𝑥 + 𝑦) (2𝑥 + 𝑦) + (𝑥 + 2𝑦) (𝑥 − 𝑦)
(vi) (𝑥 + 𝑦) (𝑥² − 𝑥𝑦 + 𝑦²)
(vii) (1.5𝑥 − 4𝑦) (1.5𝑥 + 4𝑦 + 3) − 4.5𝑥 + 12𝑦
(viii) (𝑎 + 𝑏 + 𝑐) (𝑎 + 𝑏 − 𝑐)
Sol.3) (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)
= 𝑡³ − 𝑠𝑡 + 𝑠²𝑡² − 𝑠3

(iv) (𝑎 + 𝑏) (𝑐 − 𝑑) + (𝑎 − 𝑏) (𝑐 + 𝑑) + 2 (𝑎𝑐 + 𝑏𝑑)
= 𝑎 (𝑐 − 𝑑) + 𝑏 (𝑐 − 𝑑) + 𝑎 (𝑐 + 𝑑) − 𝑏 (𝑐 + 𝑑) + 2 (𝑎𝑐 + 𝑏𝑑)
= 𝑎𝑐 − 𝑎𝑑 + 𝑏𝑐 − 𝑏𝑑 + 𝑎𝑐 + 𝑎𝑑 − 𝑏𝑐 − 𝑏𝑑 + 2𝑎𝑐 + 2𝑏𝑑
= (𝑎𝑐 + 𝑎𝑐 + 2𝑎𝑐) + (𝑎𝑑 − 𝑎𝑑) + (𝑏𝑐 − 𝑏𝑐) + (2𝑏𝑑 − 𝑏𝑑 − 𝑏𝑑)
= 4𝑎𝑐

(v) (𝑥 + 𝑦) (2𝑥 + 𝑦) + (𝑥 + 2𝑦) (𝑥 − 𝑦)
= 𝑥 (2𝑥 + 𝑦) + 𝑦 (2𝑥 + 𝑦) + 𝑥 (𝑥 − 𝑦) + 2𝑦 (𝑥 − 𝑦)
= 2𝑥² + 𝑥𝑦 + 2𝑥𝑦 + 𝑦² + 𝑥² − 𝑥𝑦 + 2𝑥𝑦 − 2𝑦²
= (2𝑥² + 𝑥²) + (𝑦² − 2𝑦²) + (𝑥𝑦 + 2𝑥𝑦 − 𝑥𝑦 + 2𝑥𝑦)
= 3𝑥² − 𝑦² + 4𝑥𝑦

(vi) (𝑥 + 𝑦) (𝑥² − 𝑥𝑦 + 𝑦²)
= 𝑥 (𝑥² − 𝑥𝑦 + 𝑦²) + 𝑦 (𝑥² − 𝑥𝑦 + 𝑦²)
= 𝑥³ − 𝑥²𝑦 + 𝑥𝑦² + 𝑥²𝑦 − 𝑥𝑦² + 𝑦³
= 𝑥³ + 𝑦³ + (𝑥𝑦² − 𝑥𝑦²) + (𝑥²𝑦 − 𝑥²𝑦)
= 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 + 𝑏 − 𝑐)
= 𝑎 (𝑎 + 𝑏 − 𝑐) + 𝑏 (𝑎 + 𝑏 − 𝑐) + 𝑐 (𝑎 + 𝑏 − 𝑐)
= 𝑎² + 𝑎𝑏 − 𝑎𝑐 + 𝑎𝑏 + 𝑏² − 𝑏𝑐 + 𝑐𝑎 + 𝑏𝑐 − 𝑐²
= 𝑎² + 𝑏² − 𝑐² + (𝑎𝑏 + 𝑎𝑏) + (𝑏𝑐 − 𝑏𝑐) + (𝑐𝑎 − 𝑐𝑎)
= 𝑎2² + 𝑏² − 𝑐² + 2𝑎𝑏

Exercise 9.5

Q.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) (3𝑎 −1/2) (3𝑎 − 1/2) (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/2 + 3𝑦/4) (𝑥/2 + 3𝑦/4)
(x) (7𝑎 − 9𝑏) (7𝑎 − 9𝑏)
Sol.1) 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 [(𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏2]
= 4𝑎² − 28𝑎 + 49

(iv) (3𝑎 −1/2) (3𝑎 − 1/2)
= (3𝑎 − 12)²
= (3𝑎)² − 2 × 3𝑎 × 1/2 + (1/2)²
[(𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏²]
= 9𝑎² − 3𝑎 + 14

(v) (1.1𝑚 − 0.4) (1.1 𝑚 + 0.4)
= (1.1𝑚)² − (0.4)² [(𝑎 + 𝑏) (𝑎 − 𝑏) = 𝑎² − 𝑏2²]
= 1.21𝑚² − 0.16

(vi) (𝑎² + 𝑏²) (− 𝑎² + 𝑏²)
= (𝑏² + 𝑎²) (𝑏² − 𝑎²)
= (𝑏²)² − (𝑎²)² [(𝑎 + 𝑏) (𝑎 − 𝑏) = 𝑎² − 𝑏²]
= 𝑏⁴ − 𝑎⁴

(vii) (6𝑥 − 7) (6𝑥 + 7)
= (6𝑥)² − (7)² [(𝑎 + 𝑏) (𝑎 − 𝑏) = 𝑎² − 𝑏²]
= 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

(x) (7𝑎 − 9𝑏) (7𝑎 − 9𝑏)
= (7𝑎 − 9𝑏)²
= (7𝑎)² − 2(7𝑎)(9𝑏) + (9𝑏)² [(𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏²]
= 49𝑎² − 126𝑎𝑏 + 81𝑏²

Q.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)
Sol.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) (4𝑥 – 5)(4𝑥 – 1)
= (4𝑥)² + (−5 − 1)4𝑥 + (−5) × (−1)
= 16𝑥² + (−6) × 4𝑥 + 5 = 16𝑥² − 24𝑥 + 5

(iv) (4𝑥 + 5)(4𝑥 – 1)
= (4𝑥)² + {5 × (−1)} × 4𝑥 + 5 × (−1)
= 16𝑥² + (5 − 1) × 4𝑥 − 5
= 16𝑥² + 4 × 4𝑥 − 5
= 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)
= (𝑥𝑦𝑧)² + (−4 − 2) × 𝑥𝑦𝑧 + (−4) × (−2)
= 𝑥²𝑦²𝑧² − 6𝑥𝑦𝑧 + 8

Q.3) Find the following squares by suing the identities.
(i) (𝑏 − 7)²
(ii) (𝑥𝑦 + 3𝑧)²
(iii) (6𝑥² − 5𝑦)² (iv) (2/3 𝑚 + 3/2 𝑛)²
(v) (0.4𝑝 − 0.5𝑞)²
(vi) (2𝑥𝑦 + 5𝑦)²
Sol.3) (i) (𝑏 − 7)2²
= (𝑏)² − 2(𝑏) (7) + (7)² [(𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏²
= 𝑏² − 14𝑏 + 49

(ii) (𝑥𝑦 + 3𝑧)2²
= (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 [(𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏²]
= 36𝑥² − 60𝑥²𝑦 + 25𝑦²

(v) (0.4𝑝 − 0.5𝑞)²
= (0.4𝑝)² − 2 (0.4𝑝) (0.5𝑞) + (0.5𝑞)²[(𝑎 − 𝑏)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

Q.4) Simplify.
(i) (𝑎² − b²)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) (m² − 𝑛²𝑚)² + 2𝑚³𝑛²
Sol.4) (i) (𝑎² − b²)²
= (𝑎²)² − 2(𝑎²) (b²) + (b²)²                                                     [(a − b)² = 𝑎² − 2ab + b² ]
= a⁴ − 𝑎²b² + b⁴
(ii) (2x +5)² − (2x − 5)²
= (2x)² + 2(2x) (5) + (5)² − [(2x)² − 2(2x) (5) + (5)²]              [(a − b)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       [(a − b)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]              [(a − b)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) (m² − 𝑛²m)² + 2𝑚³𝑛²
= (m²)² − 2(m²) (𝑛²m) + (𝑛²m)² + 2𝑚³𝑛²              [(a − b)² = 𝑎² − 2ab + b² ]
= m⁴ − 2𝑚³𝑛² + n⁴m² + 2𝑚³𝑛²
= m⁴ + n⁴m²

Q.5) `Show that
(i) (3𝑥 + 7)² − 84𝑥 = (3𝑥 − 7)²
(ii) (9𝑝 − 5𝑞)² + 180𝑝𝑞 = (9𝑝 + 5𝑞)²

(iv) (4𝑝𝑞 + 3𝑞)² − (4𝑝𝑞 − 3𝑞)² = 48𝑝q²
(v) (𝑎 − 𝑏) (𝑎 + 𝑏) + (𝑏 − 𝑐) (𝑏 + 𝑐) + (𝑐 − 𝑎) (𝑐 + 𝑎) = 0
Sol.5) (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 + 25q²q²
L.H.S = R.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 = (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)
= (a² − b²) + (b² − c²) + (c² − a²) = 0 = R.H.S

Q.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
Sol.6) (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           [(𝑎 − 𝑏)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           [(𝑎 − 𝑏)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) (𝑎 − 𝑏) = a² − b²] 
= 90000 − 9 = 89991

(𝑣𝑖𝑖) 78 × 82 = (80 − 2) (80 + 2)
= (80)² − (2)²                                    [(𝑎 + 𝑏) (𝑎 − 𝑏) = a² − b²]
= 6400 − 4 = 6396

(𝑣𝑖𝑖𝑖) 8.9² = (9.0 − 0.1)²
= (9.0)² − 2(9.0) (0.1) + (0.1)²         [(𝑎 − 𝑏)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) (𝑎 − 𝑏) = a² − b²]
= 0.9975 × 10 = 9.975

Q.7) Using a2 − b2 = (𝑎 + 𝑏) (𝑎 − 𝑏), find
(i) 51² – 49² (ii) (1.02)² – (0.98)² (iii) 153² − 147² (iv) 12.1² − 7.9²
Sol.7) (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

Q.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
Sol.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