Vedanta Excel in Mathematics Book - 8 Final (2078)

Unit Algebraic Expressions

8

8.1 Algebraic terms and expressions - Looking back
Classroom - Exercise

Let’s say and write the answer as quickly as possible.
1. a) In 5x2, the coefficient is ................., base is ................., exponent is .....................

b) In 4a3, the coefficient is ................., base is ................., exponent is ......................
2. a) Are 3x and 3x2 like or unlike terms? .......................................................................

b) Are y3 and 7y3 like or unlike terms? .......................................................................
3. a) How many terms are there in the expression 2xy? ..................................................

b) How many terms are there in the expression 2x + y? .............................................
c) How many terms are there in the expression 2 – x + y? .........................................

2x, 3x2, 5y – 7a2b2, etc. are a few examples of algebraic terms. An algebraic term is the
product form of any number and variables.

There are three important parts in an algebraic term. These parts are coefficient, base,
and exponent.
3x2 Exponent 7y3 Exponent
Base Base
Coefficient Coefficient

The sum or difference of two or more algebraic terms form an algebraic expression. For
example 3x + 2 is a binomial expression, a2 – 4ab + b2 is a trinomial expression, and
so on. However, an algebraic term itself is a monomial expression. 2x, y2, –3p3, etc. are
monomial expressions.

8.2 Polynomials and degree of polynomials

x + y, 2x2 + 4xy – 3y2 , a3 + a2 – 2a + 5, 2 x4 + 2x2 – 7, etc. are the examples of
polynomials. In these polynomials, the powers of variables are whole numbers.

The algebraic expressions in which powers of variables are whole numbers are known
as polynomials.

On the other hand, if the power of variable of an expression is not a whole number, the

expression is not a polynomial. For example, in x +1x , the power of x in 1 is – 1 which
x
1
is not a whole number. So x + x is not a polynomial.

Similarly, 1 + 7, 3 – 4x + 5, etc. are not polynomials.
x x2

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Algebraic Expressions

Let’s consider an algebraic term 3x2.
In 3x2, the variable x is multiplied 2 times. So, the degree of 3x2 is 2.
Similarly, in 2y3, the variable y is multiplied 3 times. So, the degree of 4y3 is 3.

Thus, in a polynomial with a monomial expression, the exponent of the variable is the
degree of the polynomial.

In the case of a term that contains two or more variables, the sum of the powers of
each variable is the degree of the term. For example,
In 3x3y2, the sum of the powers of x and y is (3 + 2) = 5. So, the degree of 3x3y2 is 5.
In 7abc, the sum of the powers of a, b, and c is (1 + 1 + 1) = 3. So, the degree of 7abc is 3.

In the case of a polynomial with multinomial expression, the highest power of its any
term is the degree of the polynomial. For example,

In 2x2 + 3x + 4, the highest power of x is 2. So, its degree is 2.
In 4y5 – 5y3 – y, the highest power of y is 5. So, its degree is 5.

In 3p2q2 – pq + 5, the highest power of pq is (2 + 2) = 4. So, its degree is 4.

8.3 Evaluation of algebraic expressions

Let’s take an algebraic expression 2(l + b) and evaluate it when l = 5 and b = 4.
Here, if l = 5 and b = 4, then 2(l + b) = 2(5 + 4) = 18.
Thus, evaluation is the process of finding the value of an algebraic expression by
replacing the variables with numbers.

Worked-out examples

Example 1: Which of the following expressions are polynomials? Write with reason.

(i) x2 + 1 (ii) y2 + 1 (iii) 3 x – 2 (iv) 2 x + 5
2 y2
Solution:

(i) x2 + 1 is a polynomial because the power of x is 2 which is a whole number.
2
1 1
(ii) y2 + y2 is not a polynomial because the power of y in y2 is –2 which is not a whole

number.

(iii) 3x – 2 is a polynomial because the power of x is 1 which is a whole number.

(iv) 2 x + 5 is not a polynomials because the power of x is 1 which is not a whole number.
2

Example 2: Write the degrees of the following polynomials.

(i) 3x2 (ii) 2a2bc (iii) 2x3 – 4x2 + 7x (iv) x3y2 + x2y2 – xy
Solution:

(i) The degree of 3x2 is 2.

(ii) The degree of 2a2bc is 2 + 1 + 1 = 4

(iii) The degree of 3y3 – 2y2 + 5x is 3. The highest power of the variable y is 3

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Algebraic Expressions

(iv) In x3y2 + x2y2 – xy, the sum of the powers of x3y2= 3 + 2 = 5 and it is the highest
sum of the powers. Therefore, it is 5th degree polynomial.

Example 3: If x = 3 and y = 2, evaluate x2 – 2xy + y2.
Solution:

Here, when x = 3 and y = 2, then x2 – 2xy + y2 = 32 – 2 × 3 × 2 + 22 = 9 – 12 + 4 = 1

EXERCISE 8.1

General Section - Classwork

1. Let’s say and write the polynomial expressions in the table.

a) 2x + 3 b) 3y2 – 2y + 4 Polynomials Expressions

c) a + 1 d) 5x + 1
a f) x + 2 – 3
x2
e) 2 – 3x + 7

2. Let’s say and write the degrees of these polynomials.

a) 3x ................. b) 2p2 ................. c) 4a2b .................

d) 7y – 1 ................. e) 2x3 – 3x2 + 6 ................. f) 9xy – 7x + 2y ................

3. Let’s tell and write the values of the expressions as quickly as possible.

a) If l = 4 and b= 3, then (i) l × b = …….. (ii) 2 (l + b) = ……..

b) If l = 5, then (i) l2 = …….. (ii) l3 = …….. (iii) 6l2 = ……..

Creative Section

4. Let’s answer the following questions.

a) Why is x2 a polynomial but 2 is not a polynomial?
2 x2

b) Why is 2x not a polynomial but 2 x is a polynomial?

c) Define the degree of a polynomial with examples.

5. Find the degrees of the following polynomials.

a) 2xyz b) 5a2bc c) 3y2 – 4y + 8

d) 7x5 – 2x4 + 5x3 – 7x2 – 3x e) x2 + y2 + z2 – 3xyz f) x3y3 – 2x2y2 + 7xy – 5

6. a) If l = 5 and b = 3, find the values of (i) l × b (ii) 2(l + b)

b) If l = 10, b = 8 and h = 5, find the values of

(i) l × b × h (ii) 2h (l + b) (iii) 2 (lb + bh + lh)

c) If r = 7, S = 22 , find the values of (i) Sr2 (ii) 2Sr
7

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Algebraic Expressions

d) If a = 3 and b = 2, find the values of (ii) a2 – 2ab + b2 and (a –b)2
(i) a2 + 2ab + b2 and (a + b)2 (iv) a3 – 3a2b + 3ab2 – b3 and (a – b)3
(iii) a3 + 3a2b + 3ab2 + b3 and (a + b)3 (vi) (a + b)2 – 2ab
(v) (a + b) (a –b) and (a2 – b2)

7. a) Find the value of P ×T× R when P = 200, T = 2 and R = 5
100

b) Find the value of P 1 + R T
100
when P = 100, R = 10, T = 2

c) Find the value of (i) 2Srh (ii) 2Sr(r + h) (iii) Sr2h when S = 22 , r = 7 and h = 5
7

8.4 Special products and formulae
(i) The product of (a + b) and (a + b) : Square of binomials

Let’s multiply (a + b) by (a + b)

(a + b) u (a + b)= a (a + b) + b (a + b)

or, (a + b)2 = a2 + ab + ab + b2

or, = a2 + 2ab + b2 Da bC
a a2 ab
Thus, (a + b)2 = (a2 + 2ab + b2) b ab
a+b
Also, a2 + b2 = (a + b)2 – 2ab b2 b

When the length a of the smaller square is increased
by b, each side of the bigger square ABCD is (a + b)

Now, area of ABCD = a2 + ab + ab + b2 A bB
a+b
(a + b)2 = a2 + 2ab + b2

(ii) The product of (a – b) and (a – b)

Let’s multiply (a – b) by (a – b)

(a – b) u (a – b) = a (a – b) – b (a – b)

or, (a – b)2 = a2 – ab – ab + b2

Thus, = a2 – 2ab + b2 D a–b bC
(a – b)2 = a2 – 2ab + b2

Also, a2 + b2 = (a – b)2 + 2ab (a – b)2 b(a – b) a–b
a
When the length and breadth of the square ABCD is

decreased by b, b(a – b) b2 b
Area of ABCD = (a – b)2 + b (a – b) + b (a – b) + b2

or, a2 = (a – b)2 + ab – b2 + ab – b2 + b2 A a–b bB

or, a2 = (a – b)2 + 2ab – b2

or, (a – b2) = a2 – 2ab + b2

(iii) The product of (a + b), (a + b) and (a + b) (cube of binomials)

(a + b) (a + b) (a + b)= (a + b) (a + b)2

or, (a + b)3= (a + b) (a2 + 2ab + b2) a+b
b
= a(a2 + 2ab + b2) + b(a2 + 2ab + b2) a a a b b a+b
ab
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3 a a b b a+b

= a3 + 3a2b + 3ab2 + b3 a3 + 3ab2 + 3ba2 + b3 = (a + b)3

Thus, (a + b)3 = a3 + 3a2b + 3ab2 + b3

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Furthermore, (a + b)3 = a3 + 3a2b + 3ab2 + b3
or, (a + b)3= a3 + b3 + 3ab (a + b)
or, a3 + b3= (a + b)3 – 3ab (a + b)

(iv) The product of (a – b), (a – b) and (a – b)

(a – b) (a – b) (a – b) = (a – b) (a – b)2

(a – b)3 = (a – b) (a2 – 2ab + b2)

= a (a2 – 2ab + b2) – b (a2 – 2ab + b2)

= a3 – 2a2b + ab2 – a2b + 2ab2 – b3

= a3 – 3a2b + 3ab2 – b3

Thus, (a – b)3 = a3 – 3a2b + 3ab2 – b3

Furthermore, (a – b)3 = a3 – 3a2b + 3ab2 – b3

or, (a – b)3 = a3 – b3 – 3ab (a – b)

or, a3 – b3 = (a – b)3 + 3ab (a – b) (a – b)3

(v) The product of (a + b) and (a – b)

(a + b) (a – b) D a CD C Da b
b a2 – b2 C
= a (a – b) + b (a – b)
= a2 – ab + ab – b2 a b EF
a–b (a + b) (a – b)
a–b

A a+b B

= a2 – b2 A B AB Area of ABCD
Area of ABCD Area of ABCDEF a2 – b2 = (a + b) (a – b)

= a2 = a2 – b2

Thus,(a + b) (a – b) = a2 – b2

(vi) The product of (a + b) (a2 – ab + b2)

Here,(a + b) (a2 – ab + b2) = a (a2 – ab + b2) + b (a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3

= a3 + b3

Thus, a3 + b3 = (a + b) (a2 – ab + b2)

(vii) The product of (a – b) and (a2 + ab + b2)

Here,(a – b) (a2 + ab + b2) = a (a2 + ab + b2) – b (a2 + ab + b2)

= a3 + a2b + ab2 – a2b – ab2 – b3

= a3 – b3

Thus, a3 – b3 = (a – b) (a2 + ab + b2)

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Worked-out examples

Example 1: Find the squares of: a) 3x + 2y b) x– 1
Solution: x

a) Square of (3x + 2y)

= (3x + 2y)2 Let’s consider, 3x = a and 2y = b
= (3x)2 + 2.3x.2y + (2y)2 We know that (a + b)2 = a2 + 2ab + b2
= 9x2 + 12xy + 4y2 Then, (3x + 2y)2 = (3x)2 + 2.3x.2y + (2y)2

b) Square of x– 1
x
1
= x– x 2 Let’s consider, x = a and 1 =b
x

= (x)2 – 2.x. 1 + 12 We know that, (a – b)2 = a2 – 2ab + b2
x x 1 1 12
Then, x– x 2 = (x)2 – 2.x. x + x

= x2 – 2 + 1
x2
1
Example 2: Expand a) (y + 2z)3 b) 3a – a 3
Solution:

Let’s consider, y = a and 2z = b
a) (y + 2z)3 = y3 + 3y2.2z + 3y. (2z)2 + (2z)3 We know that

= y3 + 6y2z + 12yz2 + 8z3 (a + b)3 = a3 + 3a2b + 3ab2 + b3
Then,

(y + 2z)3 =y3 + 3.y2.2z + 3.y. (2z)2 + (2z)3

b) 3a – 1 3= (3a)3 – 3. (3a)2. 1 + 3.3a. 1 2– 1 3
a a a a
9 1
= 27a3 – 27a + a – a3

Example 3: Express 16x2 – 40xy + 25y2 as a perfect square.

Solution: = (4x)2 – 2.4x.5y + (5y)2 Let’s consider 4x = a and 5y = b.
16x2 – 40xy + 25y2 = (4x – 5y)2 Then, (4x)2 – 2.4x.5y + (5y)2 is in
the form a2 – 2ab +b2 and
a2 – 2ab + b2 = (a – b)2.

Example 4: Express 8a3 + 36a2b + 54ab2 + 27b3 as a perfect cube.

Solution: 8a3 + 36a2b + 54ab2 + 27b3
8a3 + 36a2b + 54ab2 + 27b3 = (2a)3 + .................... + (3b)3
= (2a)3 + 3.(2a)2.3b + 3.2a.(3b)2 + (3b)3 Consider, 2a = a and 3b = b, then
= (2a + 3b)3 write the expression in the form
a3 + 3a2b + 3ab2 + b3 = (a + b)3

Example 5: Find the product of a) (x – 3y) (x + 3y) b) (4a2 + 3b2) (4a2 – 3b2)
Using (a + b) (a – b) = a2 – b2
Solution:

a) (x – 3y) (x + 3y)= x2 – (3y)2
= x2 – 9y2

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Algebraic Expressions

b) (4a2 + 3b2) (4a2 – 3b2) = (4a2)2 – (3b2)2
= 16a4 – 9b4

Example 6: Find the product of a) (3x + y) (9x2 – 3xy + y2)

b) (2a – 5b) (4a2 + 10ab + 25b2)
Solution:

a) (3x + y) (9x2 – 3xy + y2) Consider 3x = a and y = b, then
= (3x + y) [(3x)2 – 3x.y + y2] (3x + y) (9x2 – 3xy + y2) is in the form
= (3x)3 + y3 (a + b) (a2 – ab + b2) = a3 + b3
= 27x3 + y3

b) (2a – 5b) (4a2 + 10ab + 25b2)
= (2a – 5b) [(2a)2 + 2a.5b + (5b)2]
= (2a)3 – (5b)3 = 8a3 – 125b3\

Example 7: If x + y = 7 and xy = 5, find the value of x2 + y2.

Solution:

We have, x2 + 2xy + y2 = (x + y)2

or, x2 + y2 = (x + y)2 – 2xy

1 = 72 – 2 u 5 = 49 – 10 = 39 1 1
a a2 a
Example 8: If a + = 6, find the values of (i) a2 + (ii) a – 2.

Solution:

Here, a + 1 = 6
a

(i) We have, a2 + b2 = (a + b)2 – 2ab

? a2 + 1 = a + 1 2 – 2.a. 1 = 62 – 2
a2 a a

= 36 – 2 = 34

(ii) We have (a – b)2 = a2 – 2ab + b2
1 = a2 – 2.a.a1 + 1
? a – a 2 a2

= a2 + 1 –2
a2

= 34 – 2 = 32

Example 9: Simplify (x + 3y)2 – (x – 3y)2

Solution: Alternative process

(x + 3y)2 – (x – 3y)2 (x + 3y)2 – (x – 3y)2

= x2 + 2.x.3y + (3y)2 – (x2 – 2.x.3y + (3y)2) = (x + 3y + x – 3y) [x + 3y – (x – 3y)]

= x2 + 6xy + 9y2 – x2 + 6xy – 9y2 = 2x(x + 3y – x + 3y)

= 12xy 2.7 u 2.7 – 1.3 u 1.3 = 2x × 6y = 12xy
2.7 – 1.3
Example 10: Simplify
Solution:

2.7 u 2.7 – 1.3 u 1.3 = (2.7)2 – (1.3)2 = (2.7 + 1.3) (2.7 – 1.3) = 2.7 – 1.3 = 4
2.7 + 1.3 2.7 + 1.3 2.7 + 1.3

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Algebraic Expressions

EXERCISE 8.2
General Section - Classwork

1. Let’s say and write the expanded forms of these expressions.

Expressions Expanded forms Expressions Expanded forms
a) (x + y)2 e) (x – y)2
b) (a + 1)2 f) (m – 1)2
c) (p + 2)2 g) (a – 2)2
d) (y + 3)2 h) (p – 3)2

2. Let’s say and write the following products in the forms of difference of two squared
terms.

a) (x + y) (x – y) = .......................... b) (x + 1) (x – 1) = ..........................

c) (p + 2) (p – 2) = .......................... d) (m + 3) (m – 3) = ..........................

e) (a + a1) (a – 1 ) = .......................... f) (x + 2x ) (x – 2 ) = ..........................
a x

g) ( 3 + p) ( 3 – p) = .......................... h) ( 4 + m) ( 4 – m) = ..........................
p p m m

3. Let’s say and write the following expressions in the forms of cube of sum or
difference of two terms.

a) x3 + 3x2y + 3xy2 + y3 = ..................... b) x3 – 3x2y + 3xy2 – y3 = .................

c) x3 + 3.x2.1 + 3.x.12 + 13 = ..................... d) a3 – 3.a2.2 + 3.a.22 – 23 = .................

e) p3+3.p2. 1 + 3.p. 1 + 1 = ..................... f) x3 – 3.x2. 1 + 3.x.x12 – 1 = .................
p p2 p3 x x3

4. Let’s say and write the following expressions in the forms of cube of sum or
difference of two cubed terms.

a) (x + y) (x2 – xy + y2) = ……….. b) (x – y) (x2 + xy + y2) = ………….

c) (a + 1) (a2 – a + 1) = …………. d) (p – 2) (p2 + 2p + 4) = …………..
e) (2 + 12) (22 – 1 + 41) = …………. f) (31 – 3) (91 + 1 + 9) = …………..

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Algebraic Expressions

Creative Section - A

5. Let’s look at these diagrams and write the areas in algebraic expression forms as

shown in the example.
Dx 1C

x x2 1.x x+1 Area of ABCD = x2 + 1.x + 1.x + 12
(x + 1)2 = x2 + 2x + 1

1 1.x 12

A x+1 B

D x 2C S x R Mx yE
x2 2.x
a) x+2 b) c) x2 x.y x+y

x x x

2 2.x y x.y

A x+2 B PQ P x+y R

Area of ABCD Area of PQRS Area of PREM

d) D x-1 C 1 e) H G f) S x – y R y

x–1 (x-1)2 1.x x (x-2)2 x x – y (x – y)2 x
EF
A 1.(x-1) B 12 P Q
1 x y

x x

Area of ABCD Area of EFGH Area of PQRS

6. Let’s find the area of each of these rectangles in algebraic expression forms.

a) D x 1C b) P x 2O c) H x yG

x2 – 12 x–1 x2 – 22 x–2 x2 – y2 x–y
x x x

A x+1 B M x+2 N E x+y F
12 1 y2 y
2

1 2 y

Area of ABCD Area of MNOP Area of EFGH

7. a) Find the squares of (i) 2a + 1 (ii) x – 3y (iii) a – 1 (iv) x + 1
a x

b) Find the cubes of (i) p + 2 (ii) 2a – b (iii) y + 1 (iv) x– 1
y (iv) (3x –3x31x )3
c) Expand (i) (3x – 1)2 (ii) (2y + 1 )2 (iii) (2p + q)3
2y
8. a) Express (i) x2 +2x + 1 (ii) a2 – 4ab + 4b2 (iii) 9p2 + 12pq + 4q2 as perfect

squares.

b) Express (i) x3 + 6x2 + 12x + 8 (ii) 8a3 – 36a2 + 54a – 27

(iii) p3 – 9p2q + 27pq2 – 27q3 as perfect cubes.

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Algebraic Expressions

9. Let’s find the expressions to be inserted to make these expressions perfect squares.

a) x2 + ……… + 16y2 (Hint. x2 + 2.x.4y + (4y)2, so the required terms is 8xy)

b) x2 + …....................…… + 4y2 c) x2 – …....................…… + 9y2

d) 4x2 + …....................…… + 25y2 e) 9a2 – …....................…… + 49b2

10. Let’s find the expressions for the blank space to make these expressions perfect cubes.

a) x3 + ……. + ……..+ 27 [Hint. x3 + 3. x2. 3 + 3. x. 32 + (3)3, so, the required

expression is 9x2 + 27x]

b) a3 + ….......….. + ….......….. + 1 c) x3 + ….......….. + ….......….. + 8

d) a3 – ….......….. + ….......….. – 27 e) x3 – ….......….. + ….......….. – 64y3

11. Let’s apply the appropriate formula to find the products.

a) (a + 3) (a – 3) b) (2x + 1) (2x – 1) c) (4 + 3p) (4 – 3p)

d) (2x – 3y) (2x + 3y) e) (x2 – y2) (x2 + y2) f) (2x2 + 5y2) (2x2 – 5y2)

g) (a + b + c) (a + b – c) h) (x – y + z) (x + y + z) i) (p – q – r) (p + q – r )

j) (x + y) (x – y) (x2 + y2) k) (x + 2) (x – 2) (x2 + 4) l) (2a + y) (2a – y) (4a2 + y2)

12. Let’s find the products using the formula (a + b) (a – b) = a2 – b2.

Hint. 99 × 101 = (100 – 1 ) × (100 + 1) = 1002 – 12 = 10000 – 1 = 9999

a) 19 × 21 b) 49 × 51 c) 78 × 82 d) 102 × 98

13. Let’s find the products by using formula a3 + b3 = (a + b) (a2 – ab + b2) or

a3 – b3 = (a – b) (a2 + ab + b2)

a) (x + 2) (x2 – 2x + 4) b) (x + 3) (x2 – 3x + 9) c) (x – 1) (x2 + x + 1)

d) (y – 4) (y2 + 4y + 16) e) (2x + 3y) (4x2 – 6xy + 9y2) f) (3a – 5b) (9a2 + 15ab + 25b2)

14. a) If (x + y) = 5 and xy = 3, find the value of x2 + y2.
b)
c) If (a – b) = 4 and ab = 2, find the value of a2 + b2.

d) If x + 1 = 3, find the value of x2 + 1 .
x x2
15. a)
If p – 1 = 7, find the value of p2 + 1 .
b) p p2

c) If a + 1 = 3, find the values of (i) a2 + 1 (ii) a – 1 2
a a2 a
d)
16. a) If x + 1 = 5, find the values of (i) x2 + 1 (ii) x – 1 2
x x2 x
b)
c) If p – 1 = 4, find the values of (i) p2 + 1 (ii) p + 1 2
p p2 p

If m2 – 1 = 6, find the values of (i) m2 + 1 (ii) m + 1 2
m m2 m

If (x + y) = 6 and xy = 2, find the value of x3 + y3.

If (x – y) = 5 and xy = 4, find the value of x3 – y3.

If a + 1 = 4, find the value of a3 + 1 .
a a3

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Algebraic Expressions

d) If m – 1 = 3, find the value of m3 – 1 .
m m3

17. a) If a + b = 2, find the value of a3 + b3 + 6ab.

b) If x + y = 5, find the value of x3 + y3 + 15xy

c) If x – y = 3, find the value of x3 – y3 – 9xy

d) If a – b = 7, find the value of a3 – b3 – 21xy
18. Simplify.

a) (a + b)2 + (a – b)2 b) (x + y)2 – (x – y)2 c) (2p – 3)2 + (2p + 3)2

d) (3x + y)2 – (3x – y)2 e) (x + 1 )2+ ( x – 1 )2 f) (2 – 1 )2 – (2 + 1 )2
x x a a

19. Simplify.

a) 2.1 u 2.1 – 0.9 u 0.9 b) 3.6 u 3.6 – 1.4 u 1.4
2.1 – 0.9 3.6 + 1.4

c) 2.5 u 2.5 u 2.5 – 1.4 u 1.4 u 1.4 d) 2.7 u 2.7 u 2.7 + 1.8 u 1.8 u 1.8
(2.5)2 + 2.5 u 1.4 + (1.4)2 (2.7)2 – 2.7 u 1.8 + (1.8)2

It’s your time - Project work

Paper folding:

20. a) Let’s take a few rectangular sheet of papers (photocopy paper). Then, fold them
to get square sheet of papers as shown in the diagrams.

b) Again, let’s fold each square sheet of paper as shown in the diagram and complete

(i) the sums. x1

x x2 x.y 1. = (x + 1)2 = x2 + 1.x + 1.x + 12

1 1.x 12 = x2 + 2x + 1

(ii) xy

x = ................ = ..........................
y x.y = ..........................

(iii) x xx x1
x2
x x2 x x2–12 x–1 x2–12 = ....x.2..–...1..2.... = ..........................
x

12 1 x+1
1
(iv) x x x xy

x x2 x2 x x2–y2 x–1 x2–y2 = ................ = ..........................
x x+y

y2 y
y

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Unit Laws of Indices

9

9.1 Laws of Indices - Review

Let’s consider an algebraic term 4x3.

Here, 4 is the coefficient, x is the base, and 3 is the exponent of the base. The exponent
of a base is also called its index (plural: indices).

Power is an expression that represents repeated multiplication of the same number (or
variable) whereas exponent refers to a quantity that represents the power to which the
number (or variable) is raised. Both terms are often used interchangeably in mathematical
operations.

There are certain rules for the operations of indices of different algebraic terms. These
rules are also called the Laws of indices.

(i) Product law of indices

Study the following illustrations and investigate the idea of the product law of
indices.

1 2 = 4 unit squares 123
3 4 2 = 21 × 21 = 21 +1 = 9 unit squares
22 32 4 5 6 3 = 31 × 31 = 31 +1

2 789

3

Similarly,

x2 x = x1 × x1 y2 y = y1 × y1
= x1 +1 = y1
+1

x y

Again,

3 = 8 unit cubes
2 = 27 unit cubes
14 = 31 × 31 × 31
2 = 21 × 21 × 21
23 7 8 = 21 + 1 + 1 3

56 2 33 = 31 + 1 + 1
2
3 3

x3 x = x1 × x1 × x1 y3 y = y1 × y1 × y1
= x1 + 1 + 1 = y1 + 1 + 1
y
xx y

Also, 22 × 2 = 22+1 = 23, 32 × 33 = 32 + 3 = 35, and so on.
Thus, if am and an are any two terms with the same base a and the powers m and n
respectively, then, am u an = am + n

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Laws of Indices

(ii) Quotient law of indices

Let’s study the following illustrations and investigate the idea of the quotient law

of indices. 22 2 × 2
2 2
22 ÷ 2 = = = 2 = 22 – 1 It’s easier!
In the case of division of
24 ÷ 22 = 24 = 2 × 2 × 2 × 2 = 22 = 24 – 2 the same bases, smaller
22 2 × 2 power is subtracted from
the greater power!!
23 ÷ 25 = 23 = 2 × 2 × 2 = 1 = 1
25 2 × 2 × 2 × 2 × 2 22 25 – 3

Thus, if am and an are any two terms with the same base a and powers m and n

respectively, then,

am ÷ an = am – n if m > n and am ÷ an = 1 m if m < n
an –

(iii) Power law of indices

Study the following illustrations and investigate the idea of the power law of

indices. = 22 u 2 I got it!
(22)2 = 22 u 22 = 24 (x2)3 = x2×3 = x6

(23)2 = 23 u 23 = 26 = 23 u 2 (y3)4 = y3 × 4 = y12

(24)3 = 24 u 24 u 24 = 212 = 24 u 3

Thus, if am is any term with the base a and the index m, (am)n = am u n

Furthermore, if a and b are any two terms,

(a u b)m = am u bm and a m = am
b bm

(iv)Law of negative index Verification

If a– m is a term with base a and power – m, then,

a– m = 1 or, 1 = a– m or am = 1 x3 = x × x × x × x × x = 1
am am a–m x5 x × x × x x2

For example, From the quotient law,
x3
2–2 = 1 , 1 = 2–4, 34 = 1 x5 = x3–5 = x–2

22 24 3–4 So, 1 = x –2
(v) Law of zero index x2

Study the following illustrations and investigate the idea of the law of zero index.

2q = 21 – 1 = 21 ÷ 21 = 21 =1
21
31
3q = 31 – 1 = 31 ÷ 31 = 31 =1

Thus, if aq is any term with base a and power 0, then aq = 1

(vi) Root law of indices

2 ) is the 2nd order of root, 3 is the 3rd order of root, and so on.

(or only

Therefore, n is the nth order of root.

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Laws of Indices

Again, the square root of 4 = 4= 22 or 2 22 2 =2

= 22

The square root of 16 = 16 = 24 or 2 24 4 = 22 = 4

= 22

The cube root of 27 = 33 = 3 33 3 =3

= 33

In this way, 2 22 = 2 , 2 24 4 3 33 3 and so on.

22 = 22, = 33...

In the similar way, n am = m

an.

m is a term with base a and power m , then, m = n am
n
Thus, if an an

Laws of indices at a glance

(i) Product law am u an = am + n
(ii) Quotient law am ÷ an = am – n when m > n
am ÷ an = 1 when m < n

an – m

(iii) Power law (am)n = am u n, (ab)m = ambm, am = am
b bm

(iv) Law of negative index a– m = 1 or am = 1
am a–m

(v) Law of zero index aq = 1, bq = 1, xq = 1 and so on

(vi) Root law of index m n am or n am = m

an = an

Worked-out examples

Example 1: Find the products in their exponential forms.

(i) 5 u 52 u 53 u 55 (ii) 32 u 92 u 272

(iii) (2a)4 u (2a)– 12 u (2a)2

Solution:

(i) 5 u 52 u 53 u 55 = 51 + 2 + 3 + 5 In product law, am × an = am + n

= 511

(ii) 32 u 92 u 272 = 32 u (32)2 u (33)2

= 32 u 32 u 2 u 33×2 In power law, (am)n = am × n

= 32 u 34 u 36 = 312

(iii) (2a)4 u (2a)–12 u (2a)2 = (2a)4 – 12 + 2
1
= (2a)– 6 = (2a)6 Law of negative index: a–m = 1
am

Example 2: Find the quotient in exponential forms.
(i) (73)5 ÷ 78 (ii) 164 ÷ 82 (iii) 272 ÷ 96

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Laws of Indices

Solution: = 715 ÷ 78 In quotient law of indices.
(i) (73)5 ÷ 78 = 715 – 8 = 77 am ÷ an = am – n if m > n

(ii) 164 ÷ 82 = (24)4 ÷ (23)2
= 216 ÷ 26
= 216 – 6 = 210

(iii) 272 ÷ 96 = (33)2 ÷ (32)6

= 36 ÷ 312
= 1 =1 In quotient law of indices,
1
312 – 6 36 am ÷ an = an – m if m < n

Example 3: Evaluate (i) 8 2 25 0.5 (iii) 125 1
Solution: 27 3 (ii) 36 3

64

(i) 8 2 = 23 2 = 2 3 × 2 22 = 4
3 3 3 3 =3 9

27 33

(ii) 25 0.5 = 25 1 = 52 1 = 5 2×1 = 5
36 36 2 2 2 6

62 6

(iii) 125 – 1 = 64 1 1
3 125 3
1 1 1 1
125 1 125– 3 1253 1
64 64 – 3 1 1× 64 3 64 3 64 3
1253 1
43 1 = 64– 1 = 1= = 1 = 125
3 3
= 1 1253
53 64 3

= 4 3 ×1 = 4
5 3 5

Example 4: Simplify (i) 3 125x6y3 , (ii) 3 64

Solution:

3 125x6y3 = 363 n m
3 53 x3 y3
(i) 53x6y3 = = 5x2y Root law of indices: am = an
3
(ii) 3 64 = 3 8 = 3 23 = 23 = 2

83 u 153
Example 5: Simplify (i) 64 u 103 (ii) (xa + b)a – b u (xb + c)b – c u (xc – a)c + a

Solution:

(i) 83 u 153 = (23)3 u (3 u 5)3
64 u 103 (2 u 3)4 u (2 u 5)3

= 29 u 33 u 53 = 29 u 33 u 53 = 29 – 7 × 53 – 3 = 22 × 50 = 4
24 u 34 u 23 u 53 27 u 34 u 53 34 – 3 3 3

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Laws of Indices

(ii) (xa + b)a – b u (xb + c)b – c u (xc – a)c + a = (x) (a + b) (a – b) u (x) (b + c) (b – c) u (x) (c – a) (c + a)
= xa2 – b2 × xb2 –c2 × xc2 – a2
= xa2 – b2 + b2 – c2 + c2 – a2

= x0 = 1

EXERCISE 9.1
General Section - Classwork

Let’s say and write the answers using the laws of indices as quickly as possible.

1. a) x2 × x = ............ b) a2 × a3 = ............ c) pa × pb = ............

d) y3 ÷ y = ............ e) z5 ÷ z2 = ............ f) xa ÷ xb = ............

g) (x2)3 = ............ h) (y3)2 = ............ i) (pa)b = ............

j) (x2y2)2 = ............ k) (ab3)2 = ............ l) (xpyq)m = ............

m) x° = ............ n) 5° = ............ o) (x + 2)° = ............

2. Let’s use the law of negative index and answer with positive index.

a) x – 2 = ............ b) p – 3 = ............ c) a – x = ............
= ............
d) 1 = ............ e) 1 = ............ f) 1
a–3 x –4 p–q = ............
= ............
3. Let’s use root law of index and write the answers.

1 = ............ 2 = ............ m

a) x2 b) 33 c) a n

d) 3 = e) 3 52 = ............ f) x ay

4. a) For what power of q is its value 1? ..............................

b) If 3x = 1, what is the value of x? ..............................

c) What is the value of 5x ×5–x ? ..............................

d) What is the value of 1a + b? ..............................

e) What should be the power of x so that its value will be 1 ? ..............................
x2

f) For what power of (a + 2) the value of a 3 2 is 3? ..............................
+

g) For what power of p will it be 3 p2 ? ..............................

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Laws of Indices

Creative Section - A

5. Let’s find the products or quotients in the exponential forms by using laws of

indices.

a) 22 × 25 × 2–1 b) 93 × 3– 3 × 272 c) (5x)3 × (25x2)2 × (5x)– 12

d) (xy)7 × (xy)– 5× (xy)– 6 e) 3ax2 × xa3 × ax4 × a2 f) 1 × 1 × 3 x6
x2 x –3

g) 38 ÷ 33 1 h) 256 ÷ 57 i) 82 ÷ 210
y –8 k) xm – 3 ÷ xm – 4 l) pa – b ÷p3a + b
j) (y2) – 3 ÷

6. Let’s apply the laws of indices and simplify.

a) (x2)3 b) (p2q3)2 c) (xy2)– 4 1 –3 xy –4
d) ab e) x2y2
j) 3 x6
f) x2 g) 3 y3 h) 3 x3y3 i) 3 x 9
y

7. Let’s evaluate.

a) (22)3 1 1 1 e) 22 – 1
32 2
b) (36)2 c) (52)2 d) (7–3)3

1 1 2 1 j) 8– 1
3
f) 42 g) 92 h) 83 i) 164

k) 81– 3 l) 16 1 m) 16 – 1 n) 625 – 0.5 o) 243 – 0.4
4 2 4 81 32

25 81

8. Let’s evaluate. –1

a) 22 b) 24 c) 3 53 d) 3 1 e) 3 29
h) 3 27–1 2–6 j) 5 32–2
f) 3 8 g) 4 16 m) 3 64
i) 4 1
k) 3 64 l) 3 43 81 o) 3 274
125
n) 4
256

Creative Section - B

9. Let’s simplify:

a) 22 × 23 × 24 b) 33 × 35 × 37 c) 42 × 24 × 82 d) 253 × 52 × 625
2 × 25 32 × 95 22 × 45 54 × 1253

e) 44 × 55 f) 84 × 95 g) 42 × 93 × 64 h) 35 × 255 × 225
252 × 162 162 × 273 82 × 274 93 × 1254

10. Let’s simplify:

a) xb – c × xc – a × xa – b b) (xa)b – c × (xb)c – a × (xc)a – b
d) (xa + b)a – b × (xb+ c)b – c × (xc + a)c – a
c) (xa – b)c × (xb – c)a × (xc – a)b

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Laws of Indices

e) xa + b + 1 × xa – b + 2 f) xp – q + 1 × xq – r + 1 × xr – p + 1
x2a + 3 x3

g) xa + b × xb + c × xc + a h) xa + b × xc + d
x2a × x2b × x2c xc + b xd + a

i) xa a + b × xb b + c × xc c + a j) xa a – b × xb b+c x –a c+a
xb xc xa x –b x –c
xc ×
k) (a2)x + y × (a2)y + z × (a2)z + x
(ax.ay.az)4 l) xa – b × xb – c × xc – a

11. Let’s simplify:

a) 2x + 1 + 2x b) 3y + 1 + 3y c) 5x + 2 – 5x d) 6x + 2 – 6x
3 × 2x 2 × 3y 6 × 5x 7 × 6x

e) 2x + 4 – 2x f) 6n + 2 – 6n g) 11p + 2 – 11p h) 2x + 3 + 2x + 2
5 × 2x 6n + 1 + 6n 11p + 1 + 11p 2x + 2 – 2x + 1

12. If a = 1, b = 2 and c = –3, find the value of:

a) 4a3b2 b) 3b2c2 c) ab + ba d) ac + cb – bc

e) abbcca f) ab+c bc+a ca + b g) ab–c bc–a ca–b

13. a) If x = 2, y = 3, m = 1 and n = 2, find the value of xm + n .
y4m – n

b) If x = 2, y = 4, m = –1 and n = 3, find the value of xm + n × yn – m
xm – n × yn + m

It’s your time - Project work

14. a) Let’s take any base number and index number greater than 1 and less
than 6. Then, verify the following laws of indices.

(i) product law (ii) Quotient law (iii) Power law
(iv) Law of negative index (v) Root law of indices (vi) Law of zero index

Example:

23 × 22 = (2 × 2 × 2) × (2 × 2) = 25 = 23 + 2

So, am × an = am + n

35 = 3 × 3 × 3 × 3 × 3 = 3 × 3 = 32 = 35 – 3 So, am = am – n
33 3 × 3 × 3 an

(22)3 = 22 × 22 × 22 = 22+2+2 = 26 = 22×3 So, (am)n = am×n

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Unit Factorisation, H.C.F. and L.C.M.

10

10.1 Factors and factorisation – Looking back

Classroom - Discussion

Let’s discuss the answers of the following questions.
In 3 × 4 = 12, what are the factors of 12 and what is the product of 3 and 4?
In 2 × 3 × 4 = 24, what are the factors of 24 and what is the product of 2, 3, and 4?
Similarly,
In 2 × x = 2x, what are the factors of 2x and what is the product of 2 and x?
In x × x2 = x3, what are the factors of x3 and what is the product of x and x2?
In 2y(y + 1) = 2y2 + 2y, what are the factors of 2y2 + 2y and what is the product of
2y and (y + 1)?
In (a + b) (a – b) = a2 – b2, what are the factors of a2 – b2 and what is the product of
(a + b) and (a – b)?

Thus, when two or more algebraic expressions are multiplied, the result is called the
product and each expression is called the factor of the product.
The process of finding out factors of an algebraic expression is known as factorisation.
It is also called resolution of the expression into its factors.
Now, let’s learn the process of factorisation of different types of expressions.

(i) Factorisation of expressions which have a common factor in each of its
term

Let’s take an expression, ax + ay. In this expression a is present in both terms. So, a is
common in both terms and it is called the common factor.

In such an expression, the common factor is taken out and each term of the expression
is divided by the common factor to get another factor.

Worked-out examples

Example 1: Factorise (i) 2ax +6bx (ii) 6x2y – 8xy2 + 10xy

Solution:

(i) 2ax + 6bx = 2x (a + 3b) 2x is the common factor in each term.

(ii) 6x2y – 8xy2 + 10xy 6x2y – 8xy2 + 10xy
= 2xy (3x – 4y + 5) = 3x × 2xy – 4y × 2xy + 5 × 2xy
So, 2xy is the common factor in each term.

Example 2: Resolve into factors 2x (a + b) – 3y (a + b)
Solution:
2x (a + b) – 3y (a + b) = (a + b) (2x – 3y) (a + b) is the common factor in each term.

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Factorisation, H.C.F. and L.C.M.

(ii) Factorisation of expressions having common factor in the
groups of terms

Let’s take an expression ax + bx – ay – by. In this expression, x is common in ax + bx
and y is common in –ay – by.

In such an expression, the terms are to be arranged in groups in such a way that each
group has a common factor.

Example 3: Factorise (i) a2 – ax + ab – bx (ii) a (x2 – y2) + x (y2 – a2)
Solution

i) a2 – ax + ab – bx Terms are arranged in groups.

= a (a – x) + b (a – x) a is common in a2 – ax and b is common in ab – bx.

= (a – x) (a + b) (a – x) is common in each term.

(ii) a (x2 – y2) + x (y2 – a2)= ax2 – ay2 + xy2 – a2x

= ax2 + xy2 – a2x – ay2

= x (ax + y2) – a (ax + y2)

= (ax + y2) (x – a)

Example 4: If (3x2 + 6x) sq. unit is the area of a rectangle, find the length and breadth
of the rectangle.

Solution 3x2 + 6x x + 2
Here, area of the rectangle = 3x2 + 6x

Now, factorising 3x2 + 6x = 3x(x + 2) 3x

? Length and breadth of the rectangle are 3x unit and (x + 2) unit or (x + 2) unit
and 3x unit.

EXERCISE 10.1
General Section

Let’s factorise, say and write the answers as quickly as possible.

1. a) ax + bx = ………………. b) ax – ay = ……………….

c) y2 – y = ………………. d) 2x3 + x2 = ……………….

e) x2 – x3 = ………………. f) x2y + xy2 = ……………….

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Factorisation, H.C.F. and L.C.M.

2. a) x(a + b) + y(a +b) = ………....…….. b) a(p + q) – (p + q) = ………….....…..
c) (y – 2) – y (y – 2) = ………....…….. d) x(a – b) + 2(a – b) = ……….....……..

Creative Section - A

3. Let’s factorise

a) 2ax + 4ay b) 6x2 – 9x c) 8px3 + 12px2

d) 4mx2 – 6mx + 2m2x e) 9b2y3 – 6b3y2 +15b2y2 f) –9a5 – 6a3x2 – 15a2

g) x(2x + 3) + 4(2x + 3) h) 2y(y – 4) – 3(y – 4) i) 7x (x + 2y) + y(x + 2y)

4. Let’s apply the process of factorisation and simplify.

a) 15 × 6 + 15 × 4 b) 3.4 × 103 – 2.6 × 103 c) 3.5 × 10–2 + 2.3 × 10–2

5. Let’s resolve into factors.

a) ax + ay + bx + by b) px + qx – py – qy
c) x2 + 2x + 3x + 6 d) x2 – 4x – 3x + 12
e) 6x2 – 2y2 + 4xy – 3xy f) ax2 + ay2 + bx2 + by2
g) x3 + x2 + x + x2y + xy + y h) x2 – x (y + z) + yz
i) a2 – a (2x – y) – 2xy j) x2 – (y – 3)x – 3y

k) x (a2 – b2) + a (b2 – x2) l) ab (c2 + 1) – c (a2 + b2)

Creative Section - B

6. Let’s find the length and breadth of the following rectangles of the given area.

a) b) c)

Area Area Area
= 3a2b + 6ab = x2 + 2x + xy + 2y = x2 + 3x + 4ax + 12a

7. a) If (4ax + 6ay) sq. unit is the area of a rectangle, find the length and breadth of the
rectangle.

b) Find the length and breadth of a rectangle of area (2x2 + 2xy + xy + y2) sq. unit.
Also, find the perimeter of the rectangle.

c) The area of a rectangular field is (x2 + 3xy + 2xy + 6y2) sq. unit. Find the
perimeter of the field.

(iii) Factorisation of expressions having the difference of two squared terms

Let’s find the product of the expressions (a + b) and (a – b).

(a + b) (a – b) = a(a – b) + b (a – b) = a2 – ab + ab – b2 = a2 – b2
Here, the expression a2 – b2 is the difference of two squared terms a2 and b2. From
the above illustration, its clear that (a + b) and (a – b) are the factors of a2 – b2.

Thus, to factorise an expression of the form a2 – b2, we should use the formula,

a2 – b2 = (a + b) (a – b)

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Factorisation, H.C.F. and L.C.M.

(iv) Factorisation of expressions having the sum or difference of two cubed
terms

Let’s find the product of the expressions (a + b) and (a2 – ab + b2)
(a + b) (a2 – ab + b2) = a(a2 – ab + b2) + b(a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3
= a3 + b3
From the above illustration, it is clear that (a + b) and (a2 – ab + b2) are the factors
of a3 + b3.

Thus, a3 + b3 = (a + b) (a2 – ab + b2)

Again, let’s multiply the expressions (a – b) and (a2 + ab + b2).
(a – b) (a2 + ab + b2) = a(a2 + ab + b2) – b(a2 + ab + b2)

= a3 + a2b + ab2 – a2b – ab2 – b3
= a3 – b3
From the above illustration, it is clear that (a – b) and (a2 + ab + b2) are the factors
of a3 – b3.

Thus, a3 – b3 = (a – b) (a2 + ab + b2)

Worked-out examples

Example 1: Factorise (i) x2 – 9 (ii) 4x2 – 25y2 (iii) a4 – 16
Solution:

(i) x2 – 9 = x2 – 32 It is in the form a2 – b2

= (x + 3) (x – 3) Using a2 – b2 = (a+ b) (a – b)

(ii) 4x2 – 25y2 Square root of 4x2 = 2x and square root of 25y2 = 5y
= (2x)2 – (5y)2

= (2x + 5y) (2x – 5y) Using a2 – b2 = (a + b) (a – b)
(iii) a4 – 16 Square root of a4 = a2 and square root of 16 = 4
Using a2 – b2 = (a + b) (a – b)
= (a2)2 – 42

= (a2 + 4) (a2 – 4)

= (a2 + 4) (a2 – 22) a2 – 4 = a2 – 22 is further factorised to (a + 2) (a – 2)
= (a2 + 4) (a + 2) (a – 2)

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Factorisation, H.C.F. and L.C.M.

Example 2: Factorise (i) (x2 + y2)2 – x2y2 (ii) 9 – (p – q)2

Solution:

(i) (x2 + y2)2 – x2y2 = (x2 + y2)2 – (xy)2 Consider, x2 + y2 = a and xy = b
= (x2 + y2 + xy) (x2 + y2 – xy) Then, a2 – b2 = (a + b) (a – b)
= (x2 + xy + y2) (x2 – xy + y2) (x2 + y2)2 – (xy)2

(ii) 9 – (p – q)2 = 32 – (p – q)2 = (x2 + y2 + xy) (x2 + y2 – xy)

= (3 + p – q) [3 – (p – q)]

= (3 + p – q) (3 – p + q)

Example 3: Resolve into factors x4 + x2y2 + y4

Solution:

x4 + x2y2 + y4 = (x2)2 + (y2)2 + x2y2

= (x2 + y2)2 – 2x2y2 + x2y2 Using x2 + y2 = (x+ y)2 – 2xy

= (x2 + y2)2 – x2y2

= (x2 + y2)2 – (xy)2

= (x2 + y2 + xy) (x2 + y2 – xy)

= (x2 + xy + y2) (x2 – xy + y2)

Example 4: Simplify by factorisation process (i) 452 – 252 (ii) 101×99
Solution:

(i) 452 – 252 = (45 + 25) (45 – 25) = 70 u 20 = 1400
(ii) 101 u 99 = (100 + 1) u (100 – 1)= 1002 – 12 = 10000 – 1 = 9999

Example 5: Factorise (i) x3 + 27 y3 (ii) 64a3 – 125b3

Solution:

(i) 8x3 + 27y3 = (2x)3 + (3y)3

= (2x + 3y) [(2x)2 – 2x . 3y + (3y)2]

= (2x + 3y) (4x2 – 6xy + 9y2)

(ii) 64a3 – 125b3 = (4a)3 – (5b)3

= (4a – 5b) [(4a)2 + 4a.5b + (5b)2]

= (4a – 5b) (16a2 + 20ab + 25b2)

EXERCISE 10.2

General Section

1. Let’s say and write the following expressions as the product of their factors.

Expressions Product of factors Expressions Product of factors
a) m2 – n2
b) m2 – 4 f) 4x2 – 1
c) x2 – y2
d) x2 – 9 g) 9y2 – 4

e) p2 – 16 h) p2 – 49

i) a2 – 1
4

j) b2 – 1
9

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2. Let’s say and write the differences of these squared numbers as quickly as possible.

a) 102 – 92 = ……………...................... b) 72 – 52 = ……………......................

c) 212 – 202 = ……………...................... d) 502 – 492 = ……………......................

e) 202 – 102 = ……………...................... f) 402 – 302 = ……………......................

Creative Section

3. Let’s resolve into factors.

a) x2 – 36 b) a2 – 49 c)25 – y2 d) 16 – 81p2

e) x3 – 16x f) 2a2 – 72 g) 5p3 – 80p h) 3y3 – 27y

i) 5a3 – 20ab2 j) 8x5y – 18x3y3 k) 25x2 – 1 l) 1 – 1
49 4x2 81

4. Let’s factorise: b) a4 – 81 c) x4 – y4 d) 16p4 – q4
a) x4 – 16 f) 32y4 – 162 g) a8 – b8 h) 256x8 – y8
e) 81x4 – 625

5. Let’s factorise: b) (x + y)2 – 25 c) (a2 + b2)2 – 4 d) (a2 – b2)2 – a2b2
a) (a – b)2 – 9 f) 49 – (a – b)2 g) (x + 3)2 – (y + 2)2 h) (a – 5)2 – (b – 4)2
e) 16 – (x + y)2

6. Let’s resolve into factors. b) x4 + x2 + 1 c) x4 + 7x2y2 + 16y4
a) x4 + x2y2 + y4 e) x4 – 3x2y2 + 9y4 f) a4 + 6a2b2 + 25b4
d) a4 – 7a2b2 + b4 h) 25x4 + 4x2y2 + 4y4 i) 4a4 – 13a2b2 + 9b4
g) 4x4 + 3x2y2 + 9y4

7. Let’s simplify by factorisation process.

a) 252 – 152 b) 642 – 442 c) 962 – 862 d) (101)2 – (100)2
h) 103 u 97
e) 51 u 49 f) 82 u 78 g) 101 u 99

8. Let’s resolve into factors.

a) x3 + 8 b) y3 + 27 c) 8a3 + 1 d) 8x3 + 27y3

e) 27a3 + 125b3 f) p3 – 8 g) 27m3 – n3 h) a3 – 1000b3

9. Let’s find the area of the shaded region using a2 – b2 = (a + b) (a – b).

a) 12 cm b) 18 cm c) 15 m

12 cm 5 cm 18 cm 15 m 6m
6m
3 cm

d) 3 cm e) 5 cm f)
24 m 30 ft

24 m 12 ft 30 ft 15 m 40 m

10 m10m 12 ft 15 m
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(v) Factorisation of trinomial expressions of the form x2 + px + q

Let’s take any two binomial expressions (x + a) and (x + b) and find their product.

The product of (x + a) and (x + b) = (x + a) (x + b)
= x (x + b) + a (x + b)
= x2 + bx + ax + ab
= x2 + (a + b)x + ab

Now, let’s compare the product x2 + (a + b)x + ab to the trinomial x2 + px + q.
Here, p = a + b and q = ab.
Thus, while factorising a trinomial expression of the form x2 + px + q, we should
find p = a + b and q = ab. From the known values of a + b and ab , we should find
the values of a and b. Then, the expression is factorised by grouping.

Worked-out examples

Example 1: If (a + b) = 5 and ab = 6, find the value of a and b.
Solution:
Here, ab = 6 = 2 u 3 which is a u b
Also, a + b = 2 + 3 = 5
? Either a = 2 and b = 3 or, a = 3 and b = 2

Example 2: If (a + b) = – 8 and ab = 12, find the value of a and b.
Solution:
Here, ab = 12 = 2 × 2 × 3 = 2 (2 × 3) = 2 × 6
Also, a + b = –8 = –2 – 6
? Either a = –2 and b = –6, or, a = –6 and b = –2

Example 3: Factorise x2 + 7x + 12

Solution: Product ab = 12
12 = 2 × 2 × 3 = 4 × 3
x2 + 7x + 12 = x2 + (3 + 4)x + 12 sum a + b = 7 = 4 + 3
= x2 + 3x + 4x + 12 So, x2 + 7x + 12 = x2 + (4 + 3)x + 12
= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

Example 4: Resolve into factors x2 – 10x + 21

Solution: In x2 – 10x + 21
x2 – 10x + 21= x2 – (3 + 7)x + 21 ab = 21 = (–3) × (–7)
a + b = (–3 – 7) = – 10
= x2 – 3x – 7x + 21

= x (x – 3) – 7 (x – 3)

= (x – 3) (x – 7)

Example 5: Factorise x2 + 9x – 36 Now, I can factorise!
In x2 + 9x – 36, I should express
Solution: 9 as the sum or difference of
any two factors of 36.
x2 + 9x – 36 = x2 + (12 – 3)x – 36 36 = 12 × 3
= x2 + 12x – 3x – 36 9 = 12 – 3
= x (x + 12) – 3 (x + 12)
= (x + 12) (x – 3)

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Example 6: Factorise x2 – 7x – 18. In x2 – 7x – 18
18 = 9 × 2 and
Solution:
7=9–2
x2 – 7x – 18 = x2 – (9 – 2)x – 18
= x2 – 9x + 2x – 18
= x (x – 9) + 2 (x – 9)
= (x – 9) (x + 2)

Example 7: Factorise (a + b)2 + 9(a + b) + 20

Solution:
Here, (a + b)2 + 9(a + b) + 20

Let (a + b) = x

Then, (a + b)2 + 9(a + b) + 20 = x2 + 9x + 20

= x2 + (5 + 4)x + 20 20 = 1 × 20, but 1 + 20 = 21
= x2 + 5x + 4x + 20 20 = 2 × 10, but 2 + 10 = 12
20 = 4 × 5 and 4 + 5 = 9
= x(x + 5) + 4(x + 5) So, x2 + 9x + 20 = x2 +(4 + 5) x + 20
= (x + 5) (x + 4)

Replacing the value of x, we get,

= (a + b + 5) (a + b + 4)

? (a + b)2 + 9(a + b) + 20 = (a + b + 5) (a + b + 4)

EXERCISE 10.3

General Section - Classwork

1. Let’s express the following expressions as the product of two factors.
a) a(a + 1) + 2(a + 1) = ……….....…… b) x (x – 3) – 2(x – 3) = ……….....……
c) y(5 + y) – 2(5 + y) = ……….....…… d) (x – 2) + x(x – 2) = ……….....……
e) p(p – 7) – (p –7) = ……….....…… f) xy (y + z) – 4(y + z) = ……….....……

2. Let’s study carefully, the following tricky ways of factorisation.

2+3 2×3 – 2–3 – 2×(–3)

x2 + 5x + 6 = (x + 2) (x + 3) x2 – 5x + 6 = (x – 2) (x – 3)

–2+3 –2×3 2–3 2×(–3)

x2 + x – 6 = (x – 2) (x + 3) x2 – x – 6 = (x + 2) (x – 3)

Now, let’s apply the tricky ways of factorization. Say and write the answers as
quickly as possible.

1+2 1×2 -1-2 -1×-2

a) x2 + 3x + 2 = ……..........……… b) x2 – 3x + 2 = ……..........………

2–1 2×(–1) -2+1 -2×1

c) x2 + x – 2 = ……..........……… d) x2 – x – 2 = ……..........………

e) x2 + 9x + 20 = ……..........……… f) x2 – 9x + 20 = ……..........………

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3. Let’s write the trinomial expressions from the product of the given factors.

1+2 1×2

a) (x + 1) (x + 2) = x2 + 3x + ................ (x + 1) (x + 2) = x2 + 3x + 2

b) (x + 2) (x + 3) = ……..........……… c) (x + 3) (x + 4) = ……..........………

d) (x + 3) (x – 1) = ……..........……… e) (x – 3) (x + 1) = ……..........………

Creative Section - A

4. Let’s add the area of squares and rectangles. Then, express the area as the product
of length and breadth.

a) D x 2C

x x2 1.x 1.x x+1 Area of ABCD = x2 + x + x + x + 1 + 1
1 (x + 2) (x + 1) = x2 + 3x + 2
A 1.x 1 1
B
b) S x+2

x 2R

x x2 Area of PQRS = .......................................................

1.x 1.x (x+2)

1.x 11 ........................... = ........................................................
2 1.x 11
Q
P x+2

c) H x 3G

x (x+1) Area of EFGH = .......................................................

1 x+3 F ........................... = .......................................................
E

d) P x 3O

x Area of MNOP = .......................................................

2 ........................... = .......................................................

MN

5. a) If a + b = 7 and ab = 10, find the value of a and b.
b) If a + b = – 9 and ab = 18, find the value of a and b.
c) If a + b = 4 and ab = – 32, find the value of a and b.
d) If a + b = – 1 and ab = – 30, find the value of a and b.

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6. Resolve into factors. b) x2 + 7x + 10 c) x2 + 7x + 12
e) x2 + 6x + 8 f) x2 + 9x + 20
a) x2 + 3x + 2 h) x2 + 11x + 30 i) x2 + 14x + 48

d) x2 + 8x + 15 b) x2 – 5x + 6 c) x2 – 6x + 5
g) x2 + 9x + 18 e) x2 – 10x + 21 f) x2 – 8x + 12
h) x2 – 13x + 30 i) x2 – 15x + 56
7. Factorise.
a) x2 – 3x + 2
d) x2 – 10x + 24
g) x2 – 13x + 36

8. Resolve into factors. b) x2 + 2x – 3 c) x2 + 2x – 8
a) x2 + x – 2 e) x2 + 5x – 14 f) x2 + 3x – 40
d) x2 + 3x – 18 h) x2 + 4x – 45 i) x2 + 4x – 60
g) x2 + 5x – 84
b) x2 – x – 6 c) x2 – 2x – 15
9. Resolve into factors. e) x2 – 4x – 12 f) x2 – 3x – 40
a) x2 – x – 2 h) x2 – 8x – 65 i) x2 – x – 72
d) x2 – 3x – 28
g) x2 – 7x – 44

10. Factorise. b) x3 + 17x2 + 42x c) x4 – 4x3 – 21x2
a) x3 + 15x2 + 56x e) x3 + 13x2 – 30x f) x4 + 2x3 – 8x2
d) x4 – 14x3 + 45x2 h) x5 + 11x4 – 102x3 i) x6 – 4x5 – 12x4
g) x5 – 18x4 + 72x3

11. Factorise. b) (a + b)2 + 14(a + b) + 33
a) (x + y)2 + 5(x + y) + 6 d) (x – y)2 – 9(x – y) – 22
c) (p – q)2 – 11(p – q) + 30

12. a) The area of a rectangle is a2 + 6a + 8 sq. units. Find its length and breadth. Also
find the perimeter of the rectangle.

b) The area of a rectangular ground is x2 + 5x – 36 sq. unit. If its length and breadth
are reduced by 2/2 unit, find the new area of the ground.

It’s your time - Project work

13. Let’s find the mistakes in these expressions. Then, correct the mistakes and
factorise the expressions.

a) x2 + (3 + 2)x + 7 b) x2 + (3 – 2)x + 6 c) x2 + (4 + 5) – 20 d) x2 – (5 – 4)x – 20

14. Let’s write any two trinomial expressions of your own in each of the given forms.
Then, factorise your trinomial expressions.

a) x2 + px + q b) x2 – px – q c) x2 + px – q d) x2 – px + q

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Factorisation, H.C.F. and L.C.M.

(vi) Factorisation of trinomial expressions of the form a2 + 2ab + b2:
Let’s find the product of (a + b) and (a + b) ab

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

= a2 + 2ab + b2 b ab b2 b

b

Thus, (a + b)2 are the factors of a2 + 2ab + b2. Similarly, (a – b)2 are the factors of
a2 – 2ab + b2. So, the expression of the form a2 + 2ab + b2 is factorised by making it

a perfect square.

(vii) Factorisation of trinomial expressions of the form px2 + qx + r:

In the trinomial expressions of the form px2 + qx + r, p and q are the numerical
coefficients of x2 and x respectively and r is any constant term.

To factorise such expressions, we need to find the two factors a and b of the product
of p and r such that a + b = q. Then, the expression is expanded to four terms and
factorisation is performed by grouping.

Worked-out examples

Example 1: Factorise a2 + 6a + 9

Solution: a2 + 6a + 9 is expressed in the form a2 + 2ab + b2
a2 + 6a + 9= a2 + 2.a.3 + 32

= (a + 3)2

Example 2: Resolve into factors 4x2 – 20x + 25.

Solution:
4x2 – 20x + 25 = (2x)2 – 2.2x.5+52 4x2 – 20x + 25 is expressed in the form a2 – 2ab + b2

= (2x – 5)2
Example 3: Resolve into factors 2x2 + 11x + 5.

Solution: 2 × 5 = 10
2x2 + 11x + 5 = 2x2 + (10 + 1)x + 5
In 2x2 + 11x + 5
= 2x2 + 10x + x + 5
= 2x (x + 5) + 1 (x + 5) 10 + 1
= (x + 5) (2x + 1) p × r = 2 × 5 = 10
q = 11 = ( 10 + 1)

Example 4: Factorise 6x2 + x – 2 6 × 2 = 12
Solution:
6x2 + x – 2 = 6x2 + (4 – 3)x – 2 In 6x2 + x – 2

= 6x2 + 4x – 3x – 2 4–3
= 2x (3x + 2) – 1 (3x + 2) p × r = 6 × 2 = 12
= (3x + 2) (2x – 1) q = 1 = ( 4 – 3)

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EXERCISE 10.4
General Section - Classwork

1. Let’s write the following expressions as the product of their factors.

a) 2x(x + 1) + 1(x + 1) = ....................... b) 2x(x – 1) – 1(x – 1) = .......................

c) 3x(2x + 1) + 2(2x + 1) = ....................... d) 3x(2x – 1) – 2(2x – 1) = .......................

e) 2x(3x – 2) + (3x – 2) = ....................... f) 2x(3x + 2) + (3x + 2) = .......................

2. Let’s rewrite these expressions in the forms of a2 + 2ab + b2 or a2 – 2ab + b2

as shown.

x2 + 2x + 1 = x2 + 2.x.1 + 12

9x2 – 30x + 25 = (3x)2 – 2.3x.5 + 52

a) x2 + 4x + 4 = ............................................................................

b) x2 – 8x + 16 = ............................................................................

c) 4x2 + 12x + 9 = ............................................................................

d) 9x2 – 12x + 4 = ............................................................................

Creative Section - A b) x2 + 8x + 16 c) x2 + 10x + 25
e) a2 + 14a + 49 f) a2 + 16a + 64
3. Resolve into factors. h) x2 – 2x +1 i) p2 – 18p + 81
a) x2 + 6x + 9 k) p2 – 24p + 144 l) x2 – 30x + 225
d) a2 + 12a + 36
g) x2 – 4x + 4
j) 2 – 20p + 100

4. Factorise. b) 4a2 + 20a + 25 c) 9a2 + 6a + 1
a) 4a2 + 4a + 1 e) 9a2 + 24a + 16 f) 16x2 + 24x + 9
d) 9x2 + 12x + 4 h) 25x2 – 80x + 64 i) 16a2 – 40a + 25
g) 9x2 – 24x + 16 k) 36 – 60p + 25p2 l) 144 – 120x + 25x2
j) 49p2 – 14p + 1

5. Resolve into factors. b) 2x2 + 7x + 5 c) 3x2 + 8x + 4
a) 2x2 + 5x + 3 e) 3x2 + 5x + 2 f) 3x2 + 10x + 3
d) 3x2 + 4x + 1 h) 4x2 + 11x + 7 i) 2x2 – 3x + 1
g) 4x2 + 12x + 5 k) 4x2 – 8x + 4 l) 8 – 22x + 9x2
j) 3x2 – 10x + 8 n) 6x2 – 19x + 14 o) 3a2 – 20a + 17
m) 13 – 17x + 4x2
b) 3x2 + x – 14 c) 4x2 + 5x – 6
6. Factorise. e) 6x2 + 11x – 2 f) 7x2 + 25x – 12
a) 2x2 + x – 3
d) 5x2 + 13x – 6

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g) 4x2 + 4x – 3 h) 6x2 + 11x – 10 i) 2a2 – 5a – 12
j) 3a2 – 5a – 8 k) 4a2 – 16a – 9 l) 2a2 – 7a – 15
m) 8x2 – 22x – 21 n) 6x2 – 11x – 7 o) 5x2 – 13x – 6

7. Resolve into factors. b) 3x2 + 5xy + 2y2 c) 2x2 – xy – 3y2
a) 2x2 + 5xy + 2y2 e) 15x2 – 17xy – 4y2 f) 24a2 – 38ab + 15b2
d) 8x2 + 2xy – 15y2 h) 14a2 – 38ab + 20b2 i) 8a2 – 45ab – 18b2
g) 15a2 – 28ab + 12b2

Creative Section B

8. Factorise: y2 b2 1 1
25 49 4x2 36p2
a) 25x2 + 2xy + b) 49a2 – 2ab + c) 4x2 + 2 + d) 9p2 – 1 +

9. Factorise:

a) a2 + b2 – c2 – d2 + 2ab – 2cd b) p2 + q2 – r2 – s2 – 2pq – 2rs

c) a2 – b2 – 2bc – c2 d) x2 – y2 + 2yz – z2

10. Resolve into factors: b) 3(a + b)2 – 10(a + b) + 8
d) 6(m – n)2 – 11(m – n) – 7
a) 2(x + y)2 – 9(x + y) + 10
c) 2(p – q)2 – 3(p – q) – 9

It’s your time - Project work

11. Let’s write two expressions of each of the following forms. Then, factorise the

expressions.

a) a2 + 2ab + b2 b) a2 – 2ab + b2

[Hint: Find (ax + b) (ax + b) or (cx – d) (cx – d) then factorise the expressions]

12. Let’s write an expression of each of the following forms. Then, factorise:

a) px2 + qx + r b) px2 – qx – r c) px2 + qx – r d) px2 – qx + r

[Hint: Find (ax + b) (cx – d) or, (ax – b) (cx + d) or, (ax – b) (cx – d), then factorise the expressions.]

10.2 Highest Common Factor (H.C.F)

Let’s take any two numbers 8 and 12.

Here, all possible factors of 8 are 1, 2, 4, 8 and all possible factors of 12 are 1, 2, 3, 4,
6, 12.

The common factors of 8 and 12 are 1, 2, 4 and the highest common factor is 4. Thus,
H. C. F. of 8 and 12 is 4.

Similarly, let’s consider any two algebraic terms x3 and x4.

All possible factors of x3 are x, x2, x3

All possible factors of x4 are x, x2, x3, and x4

The factors common to x3 and x4 are x, x2, and x3

Among these common factors, the highest common factor is x3.

So, H.C.F. of x3 and x4 is x3.

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Thus, to find the H.C.F. of monomial expressions at first, we should find the H.C.F. of

the numerical coefficient. Then, the common variable with the least power is taken as
the H.C.F. of the expressions.

10.3 H.C.F of polynomial expressions

To find the H.C.F. Of polynomial expression, the given polynomials are to be factorised
and a common factor or the product of common factors is obtained as their H.C.F. We
can also find the H.C.F. Of polynomial expression by division method.

Worked-out examples

Example 1: Find the H.C.F. of 8x4y3 and 12x3y5.

Solution:

Here, the first expression = 8x4y3 In x4y3 and x3y5 the least
= 2 × 2 × 2 × x4y3 power of x is x3, the least
power of y is y3.
The second expression = 12x3y5
= 2 × 2 × 3 × x3y5 So, H.C.F. is x3y3.

? H.C.F. = 2 × 2 × x3y3

= 4x3y3

Example 2: Find the H.C.F. of 2p3 + 6p2 pnd 2p3 – 18p.

Solution: = 2p3 + 6p2 = 2p2 (p + 3)
The first expression = 2p3 – 18p = 2p (p2 – 9)
The second expression = 2p (p2 – 32) = 2p (p + 3) (p – 3)
= 2p ( p + 3)
? H.C.F.

Example 3: Find the H.C.F. of 2a2 + 11a + 12, 6a2 + 7a – 3, 4a2 –9.

Solution:

The first expression = 2a2 + 11a + 12

= 2a2 + (8 + 3) a + 12

= 2a2 + 8a + 3a + 12

= 2a (a + 4) + 3 (a + 4)= (a + 4) (2a + 3)

The second expression = 6a2 + 7a – 3

= 6a2 + (9 – 2)a – 3

= 6a2 + 9a – 2a – 3

= 3a (2a + 3) – 1 (2a + 3) = (2a + 3) (3a – 1)

The third expression = 4a2 – 9

= (2a)2 – 32 = (2a + 3) (2a – 3)

? H.C.F. = (2a + 3)

EXERCISE 10.5
General Section - Classwork
Let’s say and write the H.C.F. of these expressions as quickly as possible.

1. a) H.C.F. of a and a2 is .............. b) H.C.F. of y2 and y3 is .................

c) H.C.F of a3 and a4 is .............. d) H.C.F. of 3y2 and 6y4 is ................

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Factorisation, H.C.F. and L.C.M.

e) H.C.F. of 4x5 and 12x3 is ............ f) H.C.F. of xy2 and x2y is .................
g) H.C.F. of p2q2 and pq3 is ........... h) H.C.F. of a3b2 and a2b3 is ...............

2. a) H.C.F. of (x + 1) (x + 3) and (x – 2) (x + 3) is ......................

b) H.C.F. of (a + 3) (a – 2) and (a – 3) (a – 2) is ......................

c) H.C.F. of (2p – q) (p + q) and (3p – q) (2p – q) is ...................

Creative Section - A

3. Let’s find the H.C.F. of the following monomial expressions:

a) 2x2y, 4xy2 b) 3a2b3, 6a3b2 c) 2pxy3, 9px3y

d) 8x4, 12x2, 16x5 e) 9x3y2, 6x2y3, 3x4y4 f) 5a2bc, 10ab2c, 15abc2

4. Let’s find the H.C.F. of the following polynomial expressions:

a) 2a + 14, 3a + 21 b) a2 + ab, a2x + abx

c) 3x + 9, x2 – 9 d) 2x2 + 4x, 2x2 – 8

e) x2 + 2xy, x3 – 4xy2 f) 6x2 – 9x, 4x2 – 9

g) (a + 2)2, a2 – 4 h) x3 – x, x2 – x – 2

i) a2 + 6a + 9, a2 – 9 j) 2x2 + x – 3, 4x2 – 9

k) x3 – 5x2, x2 – 6x + 5 l) 8x3 – 27, 4x2 – 12x + 9

m) a2 + 7a + 10, a2 + 6a + 8 n) a2 + 5a – 24, a2 – 5a + 6

o) 2x2 – x – 3, 2x2 – 11x + 12 p) 6x2 – 4x – 10, 3x2 – 20x + 25

q) x4 – 1, x3 – 1 r) x4 – 125x, x3 – 3x2 – 10x

Creative Section - B

5. Find H.C.F. of the following polynomials:

a) 2x + 4, x2 – 4, x2 – x – 6 b) 3x2 – 9x, x3 – 9x, x2 + x – 12

c) x2 – 4, x2 – 4x + 4, x2 + 5x – 14 d) x2 – 1, x2 + 2x – 3, x2 – 3x + 2

e) x2 + 6x + 8, x2 + 9x + 20, x2 + 7x + 10

f) 6x2 + x – 2, 8x2 – 14x + 5, 10x2 – 11x + 3

It’s your time - Project work

6. Let’s write a pair of trinomial expressions of your own in each of the following

forms. Then, find the H.C.F. of each pair.

a) x2 + bx + c b) x2 – bx – c c) x2 + bx – c d) x2 – bx + c

[Hint: Let’s get 1st expression as the product of (x + a) (x + b), a and b are integers.]

Let’s get 2nd expression as the product of (x + a) (x + c) or (x + c) (x + b) and so on.

10.4 Lowest Common Multiple (L.C.M)

Let’s consider any two algebraic terms a2 and a3.
A few multiples of a2 are a2, a3, a4, a5, a6 ...
A few multiples of a3 are a3, a4, a5, a6, a7, ...
The common multiples are a3, a4, a5, a6, ...

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Factorisation, H.C.F. and L.C.M.

Among these common multiples, a3 is the lowest one.

So, the Lowest Common Multiple (L.C.M.) of a2 and a3 is a3.
Thus, the common variable with the highest power of the given monomial expression
is the L.C.M. of the expressions.
In the case of polynomial expressions, their L.C.M. is obtained by the process of
factorisation. By this process, the product of common factors and the factors which are
not common is taken as the L.C.M. of the polynomials.

Worked-out examples

Example 1: Find the L.C.M. of 6x3y2 and 8x2y3.

Solution: In x3y2 and x2y3 the highest
Here, the first expression = 6x3y2 power of x is x3, the highest

The second expression = 2 × 3 × x3y2 power of y is y3.
= 8x2y3 So, L.C.M. is x3y3.
= 2 × 2 × 2 × x2y3

? L.C.M. = 2 × 2 × 2 × 3× x3y3 = 24x3y3

Example 2: Find the L.C.M. of x2 + 2xy and x3 – 4xy2.

Solution: = x2 + 2xy = x (x + 2y)
The first expression = x3 – 4xy2
The second expression = x (x2 – 4y2) = x [x2 –(2y)2] = x (x + 2y) (x – 2y)
= x (x + 2y) (x – 2y) = x (x2 – 4y2)
? L.C.M.

Example 3: Find the L.C.M. of 2x2 + 6x, 2x3 – 18x, 2x3 + 16x2 + 30x.

Solution:

The first expression = 2x2 + 6x = 2x (x + 3)

The second expression = 2x3 – 18x = 2x (x2 – 9) = 2x (x + 3) (x – 3)

The third expression = 2x3 + 16x2 + 30x

= 2x (x2 + 8x + 15)

= 2x (x2 + (5 + 3) x + 15)
= 2x (x2 + 5x + 3x + 15)

= 2x [x (x + 5) + 3 (x + 5)] = 2x (x + 5) ( x+ 3)

? L.C.M. = 2x (x + 3) (x – 3) (x + 5) = 2x (x2 – 9) (x + 5)

EXERCISE 10.6
General Section - Classwork

1. Let’s say and write the L.C.M. as quickly as possible.

a) L.C.M. of a and a2 is ............ b) L.C.M. of b2 and b3 is ............

c) L.C.M. of x3 and x4 is is ............ d) L.C.M. of 2x2 and 4x3 is ............

e) L.C.M. of ax3 and x5 is ............ f) L.C.M. of x2y and xy2 is ............

g) L.C.M. of x3y2 and x2y3 is ............ h) L.C.M. of axy and bx3y2 is ............

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Factorisation, H.C.F. and L.C.M.

2. a) L.C.M of (a + 1) (a – 1) and (a – 1) is .............................

b) L.C.M. of (x – 2) (x + 3) and (x – 2) (x – 3) is .............................

c) L.C.M. of a (x + y) (x – y) and b(x – y) is .............................

Creative Section - A

3. Find the L.C.M. of the following monomial expressions.

a) 2xy2, 4x2y b) 2a2b2, 3a3b3 c) 4x2y3, 6x3y2

d) 2x4, 3x2, 6xy4 e) 6x5y3,8x3y5,12xy f) 14xyz3, 21x2y3z, 28x3yz2

4. Find the L.C.M. of the following polynomial expressions.

a) ax2 + ax, a2x2 + a2x b) 2x2 + 4x, x3 + 2x2 c) 3a2b + 6ab2, 2a3 + 4a2b

d) a2x + abx, abx2 + b2x2 e) 3x2 + 6x, 2x3 + 4x2 f) 2a + 4, a2 – 4

g) 3a2 + 3a, 6a2 – 6 h) x2y – 5xy, x2 – 25 i) a4x2 – a2x4, 7a2x – 7ax2

j) x2 – xy, x3y – xy3 k) 4x2 – 2x, 8x3 – 2x l) x2 – 4, x2 + 5x + 6

m) x2 + x – 6, x2 – 9 n) 2x3 – 50x, 2x2 + 7x + 15

o) 4a3 – 9a, 2a2 + 3a – 9 p) x2 + 8x + 15, x2 + 7x + 12

q) a2 – 9a + 20, a2 – 2a – 15 r) a2 + 5a – 14, a2 – 8a + 12

s) 2a2 + 5a + 2, 2a2 – 3a – 2 t) 3x2 + 8x – 16, 3x2 – 16x + 16

u) 4x2 – x – 3, 3x2 – 2x – 1 v) 2x2 + 3x – 9, 4x2 – 12x + 9

Creative Section - B

5. Find the L.C.M. of the following polynomial expressions.

a) m2 + 2m, m2 – 4, m2 + 3m + 2 b) 4x – 20, x2 – 25, x2 – 3x – 10

c) (a + b)2, a2 – b2, 2a2 + ab – b2 d) x2 + 4x + 4, x2 – 4, x2 + 8x + 12

e) x2 – 6x + 9, x2 – 9, x2 – 10x + 21 f) x2 + 6x + 8, x2 + x – 12, x2 + 5x + 6

g) a2 + 4a – 5, a2 + 11a + 30, a2 – 5a – 6 h) 2a2 + 5a + 2, 3a2 – 7a + 2, 6a2 + 5a + 1

It’s your time - Project work

6. a) Let’s write any five pairs of monomial expressions with the same base but
different exponents. Then, find the H.C.F. and L.C.M. of each pair.

b) Now, find the product of each pair of expressions and the product of H.C.F. and
L.C.M. Verify: Product of two expressions = H.C.F. × L.C.M.

7. a) Let’s write any five pairs of monomial expressions with the same base but
different exponents. Then, find the H.C.F. of each pair of expressions.

b) Now, find the L.C.M. of each pair of expressions using the relation:
H.C.F. × L.C.M. = Product of two expressions.

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Unit Rational Expressions

11

11.1 Rational expressions 1 2 45, 79,
2 3
We have already learned about the rational numbers such as , , … and so on.
We p
can express a rational number in the form q where q z 0.

Similarly, algebraic expressions such as x2, 23a, x , a + b , … and so on are rational
y a – b
p
expressions. We can also express a rational expression in q form, where q z 0.

However, if the denominator of an expression is 0 (Zero), it is undefined and it cannot

be the rational expression. For example,

In 2 , if x = 0, then it is undefined in x + yy, if x = y, it is undefined.
x x –

11.2 Reduction of rational expressions to their lowest terms

We can reduce a monomial rational expression to its lowest terms by using the law of
indices. During the process, we simply subtract the smaller index of a variable from the
higher index of the same variable. For example,

4x4y3 = 2x4 – 2 = 2x 2
6x2y5 3y5 – 3 3y2

Alternatively, we can reduce the monomial rational expression to its lowest terms,
dividing the numerator and denominator by their H.C.F. For example,

In 4x4y3 , the H.C.F. of numerator and denominator is 2x2y3
6x2y5

Now, dividing the numerator and denominator by 2x2y3

4x4y3 ÷ 2x2y3 = 2x4 –2y3 – 3 = 2x2 y0 = 2x2
6x2y5 ÷ 2x2y3 3x2 –2y5 – 3 3x0 y2 3y2

In the case of polynomial numerator and denominator at first, we should factorise

them. Then the common factors from the numerator and denominator are cancelled. For

example, (x + 1) (x – 1) x–1
x2 –1 ax (x + 1) ax
ax2 +ax = =

Worked-out examples

Example 1: For what value of a is 2 undefined?
Solution: a–3

To be 2 undefined, a – 3 = 0, i.e. a = 3.
a–3

? The required value of a is 3.

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Rational Expressions

Example 2: Reduce 15a5b4 to its lowest terms.
10a4b5
Solution:

15a5 b4 = 3a5 – 4 = 3a am =am–n if m > n, am =an1–m if m < n
10a4b5 2b5 – 4 2b an an

Example 3: Reduce 4x2 – 25 to its lowest terms.
Solution: 2x2 – x – 10

4x2 – 25 = 2x2 (2x)2 – 52 – 10
2x2 – x – 10 – (5 – 4)x

= (2x + 5) (2x – 5) = x (2x + 5) (2x – 5) 5)= (2x + 5) (2x – 5) = 2x + 5
2x2 – 5x + 4x – 10 (2x – 5) + 2(2x – (2x – 5) (x +2) x+2

EXERCISE 11.1
General Section - Classwork

1. Let’s say and write the answers as quickly as possible.

a) For what value of x is 5 undefined ? .....................
x .....................

b) For what value of y is y 3 2 undefined ?
+

c) For what value of p, 5 becomes undefined? .....................
p–1

d) For what value of a, 2 –a becomes undefined? .....................
3+a

e) For what value of y is a+ b undefined? .....................
x– y .....................
p+q
f) For what value of x is 3x – 6 undefined ?

2. Let’s say and write the lowest terms of these rational expressions.

a) x2 = ................ b) 2x3 = ................ c) 6x4 = ................
x x 2x2
xy a2 – b2 a2 – b2
d) x2y2 = ................ e) a+b = ................ f) a–b = ................

Creative Section

3. Reduce the following rational expressions to their lowest terms.

a) 2x3 b) 6x4 c) 5x2y d) 8a4b3 e) 9ab3
4x2 8x5 10xy2 12a2b5 6a2b5

f) 4p5q5 g) 20p7q8 h) –14x3y2z i) 6a2b3c4 j) – 18p6q4r2
16p4q6 15p8q5 21xy2z2 – 18a3b2c3 – 24p4q6r4

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Rational Expressions

4. Reduce the following rational expressions to their lowest terms.

a) 3x + 6 b) 2y – 4 c) x2 + 4x
6x2 + 12x 4y2 – 8y x2 – 16

d) x2 + xy e) 3x + 9 f) a2 – b2
x3 – xy2 x2 – 9 a–b

g) 2x2 + 5x h) 10a2 – 15ab i) x – 2y
4x2 – 25 6ab – 9b2 x2 – 4y2

j) a2 – b2 k) x + 2 l) x2 – 3x
a4 – b4 x2 + 5x + 6 x2 – 7x + 12

m) a2 – 4 n) a3 – 36a o) 2a3 – 50a
a2 – 9a + 14 a2 + 3a – 18 a3 + 4a2 – 45a

p) (x – 7)2 q) (x + 6)2 r) x2 – 8x + 16
x2 – 49 x2 + 5x – 6 x2 – 2x – 8

s) x2 + 9x + 20 t) x2 – 5x – 14 u) 3x2 – 5x – 12
x2 + x – 12 2x3 – 13x2 – 7x 3x2 + x – 4

11.3 Multiplication of rational expressions
In the case of multiplication of monomial rational expressions, we should multiply the

numerators and denominators separately. Then, the product is reduced to the lowest

terms by using the laws of indices. For example,

24x3y2 × 39a3b2 = 3x3y2a3b2 = 3ax
6a2b3 2x2y3 a2b3x2y3 by
3

In the case of multiplication of polynomial rational expressions, the numerators and

denominators are factorised. Then, they are reduced to their lowest terms. For example,

2x2 + 6x × x2 – 1 = 2x (x + 3) × (x + 1) (x – 1)
x3 – x2 x2 – 9 x2 (x – 1) (x + 3 (x –3)

= 2 (x + 1)
x (x – 3)

11.4 Division of rational expressions

In the case of division of rational expressions, we should multiply the dividend by the
reciprocal of the divisor as like the process of multiplication of rational expressions.
For example,

8x5y3 6x4y2 48x5y3 515a3b 20x5y3a3b 20xy
9a4b2 ÷ 15a3b = 9a4b2 × 6x4y2 = 9a4b2x4y2 = 9ab
33

a2 + 5a + 6 ÷ a2 – 9 = a2 + 3a + 2a + 6 × a2 – 3a + a –3
a2 – 1 – 2a – (a + 1) (a – 1) (a + 3) (a – 3)
a2 3

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Rational Expressions

=a (a + 3) + 2 (a + 3) ×a (a – 3) + 1 (a – 3)
(a + 1) (a – 1) (a + 3) (a – 3)

= (a + 3) ( a + 2) × (a – 3) ( a +1)
(a + 1) (a – 1) (a + 3) (a – 3)

= a+2
a–1

EXERCISE 11.2
General Section - Classwork

1. Let’s simplify mentally, say and write the answers as quickly as possible.

a) x × y2 = .................... b) a2 × b = ....................
y x2 b2 a

c) 2x3 × 9y3 = .................... d) p2 ÷ p = ....................
3y2 4x2 q2 q

e) x3 ÷ x = .................... f) 10x5 ÷ 5x2 = ....................
y3 y 6y4 3y

Creative Section - A
2. Find the products or quotients.

a) 4x2y3 × 9x4 b) 12a2b3 × 5x3y5 c) 28a3x3y2 27p3b2
3x3y2 8y3 25x2y4 6a3b4 45p2b3 × 7a4x2y3

d) 6a4 ÷ 8a3 e) 8a5b3 ÷ 6a4b2 f) 32x7y6z5 ÷ 8x5y4z3
9b3 3b2 9x4y2 15x3y 27p4q3r2 9p3q2r

3. Simplify:

a) x3c2 ÷ x2y3 × xy4 b) 18x3y4 × 25x2yz ÷ 5x4y5
a4y2 a3b2 c2z2 15a5b3 6a2bc2 3a6b4c

c) 7x2y3 × 12ab ÷ 3a2b2 d) 4a4b2 ÷ 12a3 2 × 3a2x2yc
8a5b4 21xy 4x3y2 6x3y4 9a4y5z 2b2y2z

Creative Section - B

4. Simplify:

a) x2 – a2 × 2a b) 2x + 6 3x2 + 9x c) a2 – ba2b2÷ a2 b – ab2
ax + a2 x–a x2 – 9 ÷ 2x2 – 6x a2b + a2b2

d) x2 – xy ÷ x2 (x – y) e) 4a2 – 9b2 × x2y + xy2 f) 25x2 – 16y2 × x – 2
x2 + xy x3 + x2y x2 – y2 4a – 6b x2 – 4 5x – 4y

g) 4x2 – 81y2 × a – 2a2 h) a2 + 5a + 6 ÷ a2 – 9 i) x2 +5x + 6 ÷ x2 – 9
1– 4a2 2x – 9y a2 – 1 a2 – 2a – 3 a2 + 3a – 4 a2 – 16

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Rational Expressions

j) 16x2y2 – z4 × (a – 2)2 ÷ a2 – 4 k) x2 – 4y2 × x2 – xy ÷ y2 – xy
a2 – 4 4xy + z2 (a +2)2 2x2 – 5xy + 2y2 x+y 2x – y

l) a2 – a – 20 × a2 – a – 2 ÷ a + 1 m) a2 – 8a – 9 × a2 – 25 ÷ a2 +4a – 5
a2 – 25 a2 – 2a – 8 a2 + 5a a2 – 17a + 72 a2 – 1 a2 – 9a + 8

11.5 Addition and subtraction of rational expressions

In the case of addition and subtraction of rational expressions with a common denomi-
nator, we should simply add or subtract the numerators. Then, the sum is reduced to its
lowest terms. For example,

x2 – y2 = x2 – y2 = (x + y) (x – y) =x + y
x–y x–y x – y x–y

In the case of addition or subtraction of the rational expression with different
denominator, follow the process given below.

(i) Find the L.C.M. of the denominators.

(ii) Divide the L.C.M. by each denominator and multiply the quotient so obtained by
the corresponding numerator.

(iii) Carry out the operation of addition or subtraction in the numerator.

(iv) Reduce the sum to its lowest terms.

Worked-out examples

Example 1: Simplify (i) x + 2x (ii) a2 + b2 – 2ab
Solution: 9a 9a a2 – b2 a2 – b2 a2 – b2

(i) 9xa+ 2x = x+2x = 3x = x
9a 9a 9a 3a

(ii) a2 + b2 – 2ab = a2 + b2 – 2ab = (a – b)2 = (a – a) (a – b) = a–b
a2 – b2 – a2 – b2 a2 – b2 + b) (a – (a + b) (a – b) a+b
a2 b2 (a b)

Example 2: Simplify (i) x 1 1 + 1 (ii) x+2 – x–2
+ x–1 x–2 x+2
Solution:
1 1 x– 1+x +1 2x
(i) x + 1 + x–1 = (x + 1) (x – 1) = x2 – 1 L.C.M. of (x+1) and (x–1) is (x+1) (x–1)

(ii) x+2 – x–2 = (x + 2)2 – (x – 2)2 L.C.M. of (x–2) and (x+2) is (x–2) (x+2)
x–2 x+2 (x – 2) (x + 2)

= x2 + 2.x.2 + 22 – (x2 – 2.x.2 + 22)
(x – 2) (x + 2)

= x2 + 4x + 4 – x2 + 4x – 4 = 8x
x2 – 4 x2 – 4

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Rational Expressions

Example 3: Simplify (i) p–2 – p–2 (ii) a 2 1 + 2a – a2 + 3
p+2 p2 – 4 + a–1 a2 – 1
Solution:
p–2 p–2 p–2 p–2
(i) p+2 – p2 – 4 = p+2 – p2 – 22

= p–2 (p – 2)
p+2 – (p + 2) (p – 2)

= p–2 – 1
p+2 p+2

= p–2– 1 p–3
(p + 2) = p+2

(ii) 2 + 2a a2 + 3 2 2a a2 + 3
a+1 a–1 – a2 – 1 = a + 1 + a – 1 – (a + 1) (a – 1)

= 2(a – 1) + 2a (a + 1) – (a2 + 3) L.C.M. of (a+1) (a–1) and (a+1) (a–1) is (a+1) (a–1)
(a + 1) (a – 1)

2a – 2+ 2a2 + 2a – a2 – 3
= (a + 1) (a – 1)

= a2 + 4a – 5 = a2 + 5a – a – 5
(a + 1) (a – 1) (a + 1) (a – 1)

= a(a + 5) –1 (a + 5) = (a+ 5) (a – 1) = a +5
(a + 1) (a – 1) (a + 1) (a – 1) a+1

Example 4: Simplify 2+ 5x – x+2
Solution: x2 + 3x + 2 x2 – x – 6 x2 – 2x – 3

2 + 5x – x+2
x2 + 3x + 2 x2 – x – 6 x2 – 2x – 3

= 2 5x – x+2
x + 2x + x+ 2 + x2 – 3x + 2x – 6 x2 – 3x + x – 3

= x(x + 2 x+ 2) + 5x – x+2
2)+1( x(x – 3)+2( x – 3) x(x – 3)+1( x – 3)

= 2 + 5x – x+2
(x + 2) ( x+ 1) (x – 3) ( x+ 2) (x – 3) ( x+ 1)

= 2(x – 3) + 5x (x + 1) – (x + 2)2 L.C.M. of (x+2) (x+1), (x–3) (x+2) and (x–3) (x+1)
(x + 1) ( x+ 2) (x – 3) is (x + 1) (x + 2) (x – 3)

= 2x – 6 + 5x2+ 5x – (x2 + 4x + 4)
(x + 1) (x + 2) ( x – 3)

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Rational Expressions

5x2 + 7x – 6 – x2 – 4x – 4
= (x + 1) (x + 2) ( x – 3)

4x2 + 3x – 10
= (x + 1) (x + 2) ( x – 3)

4x2 + 8x – 5x – 10
= (x + 1) (x + 2) ( x – 3)

4x(x + 2) – 5 (x + 2) (x + 2) (4x – 5) 4x – 5
= (x + 1) (x + 2) ( x – 3) = (x + 1) (x + 2) ( x – 3) = (x + 1) ( x – 3)

EXERCISE 11.3
General Section - Classwork

1. Let’s simplify, say and write the answers as quickly as possible.

a) a + 3a = .................... b) 4x + 2x = ....................
x x 3a 3a

c) 12y – 2y = .................... d) 7xy – xy = ....................
5x 5x 3ab 3ab

e) a x b + a 3x b = .................... f) 2x2 + a x2 b = ....................
+ + a–b –

g) x x y – x y y = .................... h) a a2 b – a b2 b = ....................
– – + +

Creative Section - A

2. Let’s simplify:

a) x x y + x x y b) 3p – 3q c) a+b + a–b
+ + p–q p–q 2ab 2ab

d) p2 q – p q2 q e) x y2 – y y2 f) a – b
p– – x2– x2– a2–b2 a2–b2
3
g) x2 4 + 4 4 – 4x 4 h) a + a2 + 8a + 15
x2 – x2 – x2 – a2 + 8a + 15

i) x2 + 4x j) a2 – a2 – 4a + 6 + 3
x2 + 6x + 8 x2 + 6x + 8 5a + 6 a2 – 5a a2 – 5a + 6

3. Let’s simplify:

a) 1 + 1 b) p 1 2 – 1 c) 2a 1 1 + 1
x+1 x–1 – p–1 + 2a – 1

d) a2b – ab2 e) 1 – 1 f) x 1 2 + 4
a–b a+b y2 (x – y) x2 (x – y) – 4 – x2

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Rational Expressions

g) 1 – 1 h) 2x2 – 2x i) 1 + 1
x2 – 1 x–1 x2 – 25 x–5 x (x + a) x (x – a)

j) x 2 2 – 8 k) x + y l) 9x2 + y2 – 3x – y
– x2 – 4 (x + y)2 x2 – y2 9x2 – y2 3x + y

m) 2a 1 3b – 4a – 6b n) (a – 1 – a) – 1
+ 4a2 – 9b2 b) (c (a – c) (b – c)

o) x+2 2 + 3 p) a–2 – a+1
x2 + x – x2 – 1 a2 + 4a + 4 a2 – 4

q) a–2 – a+1 r) 2 – 1
a2 – 1 a2 – 2a + 1 x2 + x x2 + 3x + 2

Creative Section - B

4. Let’s simplify:

a) 1 – 2 + 1 b) 1 – 1 + 3
x x+1 x+2 x+1 x–1 x2 – 1

c) x + x 2 – 4x d) x 2 1 + 2x – x2 + 3
x+2 x– x2 – 4 + x–1 x2 – 1

e) 3 + 4 + a+2 f) x + x – 9
a–1 a+1 a2 – 1 2x + 3 2x – 3 4x2 – 9

g) 2x + 3y – 18y2 h) x2 + y2 – x2 – y2
2x + 3y 2x – 3y 4x2 – 9y2 xy y(x + y) x(x + y)

i) 2xy – x–y + x+y j) x+y – x–y – 4xy
x2 – y2 x+y x–y x–y x+y x2 + y2

5. a) (x – 1 – 4) + (x – 1 – 5) + (x – 1 – 3)
3) (x 4) (x 5) (x

b) (x – 1 + 2) – 3 + 2
3) (x (x + 2) (x – 4) (x – 4) (x – 3)

c) (x – 2 – 3) – 2 + (x – 1 – 2)
2) (x (x – 1) (x – 3) 1) (x

d) a + b + (c – c – b)
(a – b) (a – c) (b – c) (b – a) a) (c

e) 2 – 2 + 1
x2 – 5x + 6 x2 – 4x + 3 x2 – 3x + 2

f) x–1 + x–2 + x–5
x2 – 3x + 2 x2 – 5x + 6 x2 – 8x + 15

g) y+3 + 6 – 16y
y2 + 6y + 9 y2 – 9 8y2 – 24y

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Unit Equation, Inequality and Graph

12

12.1 Equation and inequality - Looking back
Classroom - Exercise

1. Let’s tell and write the equations and find the values of the variables.

a) The sum of x and 3 is 7, equation is ......................... and x = ..................

b) The difference of x and 2 is 5, equation is ......................... and x = ..................

c) The product of 4 and y is 12, equation is ......................... and y = ..................

d) The quotient of p divided by 2 is 3, equation is ......................... and p = ..................

2. Let’s tell and write the values of variables as quickly as possible.

a) x + 3 = 9, x = ............... b) y – 1 = 5, y = ..................

c) a = 2, a = .................. d) 3p = 15, p = ...................
5

3. Let’s tell and write the possible solution sets.

a) 3 < x < 7, solution set = {..................} b) 1 < x < 5, solution set = {..................}

c) 2≥x≥0 , solution set = {..................} d) 0≤x≤3, solution set = {..................}

The sum of x and 3 is 7, i.e. x + 3 = 7 is a mathematical statement.

Here, unless x is replaced by any number, we cannot decide whether
x + 3 = 7 is a true or false statement. For example,

If x is replaced by 2, then 2 + 3 = 5, which is false.

If x is replaced by 4, then 4 + 3 = 7, which is true.

Such mathematical statements which cannot be predicted as true or false statements
until the variable is replaced by any number are known as open statements.

The open statement containing ‘equal to’ (=) sign is called an equation. An equation can
be true only for a fixed value of variable.

On the other hand, the open statement containing trichotomy signs (<, >, ≤, ≥) is
known as inequality. For example,

x<4 (x is less than 4)
p>–5 (p is greater than – 5)
y ≤4 (y is less than and equal to 4)
q ≥3 (q is greater than and equal to – 3), etc. are some inequalities.

An inequality can be true for more than one values of variable

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Equation, Inequality and Graph

12.2 Properties of inequality

To solve an inequality, we apply the following properties of inequality.

(i) If x > y, then If the same number is added to or subtracted from both sides of
x+a>y+a inequality, the inequality relationship remains the same.
x–a>y–a

(ii) If x > y, then

ax > ay and If both sides of an inequality are multiplied or divided by the same
x y positive numbers, the inequality relationship remains the same.
a > a

Where a z 0.

(iii) If x > y, then

– ax < – ay and If both sides of an inequality are multiplied or divided by the same
x y negative numbers, the inequality relationship is reversed.
–a < –a

12.3 Graphs of linear inequalities

We can represent an inequality on a number line. In the graphical representation of
an inequality, an empty circle ( ) means the circled value is not included and a filled
circle ( ) means the circled value is included in the inequality.

Study the following illustrations.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x > 2 (It does not include 2.) x t 2 (It includes 3.)

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x < -2 (It does not include –2.) x d –2 (It includes –2.)

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

–5 < x < 5 (It does not include –5 and 5.) –5 d x d 5 (It includes –5 and 5.)

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

–7 d x < 6 –4 < x d 8
(It includes – 7 but does not include 6.) (It includes 8 but does not include – 4.)

12.4 Solution of inequalities in one variable

Let’s consider an inequality x < 5.

Here, when x is replaced by the numbers less than 5, the inequality is true. So, the set of
solution to this inequality is {... – 1, 0, 1, 2, 3, 4}. Thus, we get the solution to an inequality
by solving it using the properties of inequality.

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Equation, Inequality and Graph

Worked-out examples

Example 1: Solve and find the solution set of x + 3 < 7. Draw the graph of the

solution set.

Solution:

Here, x+3 <7

or, x + 3 – 3 < 7 – 3 (subtracting 2 from both sides)

or, x < 4

? The required solution set is {... – 1, 0, 1, 2, 3}

Now, the graph of the solution set is

x <4

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

Example 2: Solve and find the solution set of 3x – 1 ≥ x + 5. Draw the graph of the

solution set.

Solution:

Here, 3x – 1 t x + 5

or, 3x – 1 + 1 t x + 5 + 1 (Adding 1 to both sides)

or, 3x t x + 6

or, 3x – x t x + 6 – x (Subtracting x from both sides)

or, 2x t 6
2x 6
or, 2 t 2 (Dividing both sides by 2)

or, x t 3 xt3

? The required solution set is {3, 4, 5, ...}

Now, the graph of the solution set is -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

Example 3: Solve – 7 ≤ 3x – 1 ≤ 8 and draw the graph of its solution set.

Solution:

Here, – 7 d 3x – 1 d8

or, – 7 + 1 d 3x – 1 + 1 d 8 + 1

or, – 6 d 3x d 9 –2 d x d 3
–6 3x 9
or, 3 d 3 d 3

or, – 2 d x d 3 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

? The solution set is {– 2, – 1, 0, 1, 2, 3}

Example 4: Solve – 3 ≤ 2 – 5x ≤ 22 and draw the graph of its solution set.

Solution: 2 – 5x d 22 1 t x t –4
Here, – 3 d

or, – 3 – 2 d 2 – 2 – 5x d 22 – 2 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

or, – 5 d – 5x d 20 Both sides are divided by negative numbers (– 5),
–5 t ––55x 20 so, the signs of inequality are reversed.
or, –5 t –5

or, 1 t x t – 4

? The solution set is {– 4, – 3, – 2, – 1, 0, 1}

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Equation, Inequality and Graph

Example 5: If x ≤ – 2, find the value of y in y = 2x – 3.
Solution:
Here, x d –2
Now, y = 2x – 3 = 2 u (– 2) – 3 = – 4 – 3 = –7
? y d – 7.

Example 6: The cost of a pen is Rs 20 and that of pencil is Rs 7. If Sunayana has
Solution: Rs 60 and she wants to buy 2 pens and a few numbers of pencils, find
the maximum numbers of pencils that she can buy.

Let the required number of pencil be x.
The cost of 2 pens = 2 u Rs 20 = Rs 40
The cost of x numbers of pencils = Rs 7x

Now, 7x + 40 d 60

or, 7x + 40 – 40 d 60 – 40

or, 7x d 20

or, 7x d 20
7 7
or, x d 276

Since x is a whole number, the required maximum number of pencils is 2.

EXERCISE 12.1

General Section - Classwork

1. Let’s say and write the solution sets as quickly as possible.

a) x + 1 > 3, solution set is {...............................................}

b) x – 1 < 3, solution set is {...............................................}

c) 2x t – 4, solution set is {...............................................}
{...............................................}
d) x d 2, solution set is {...............................................}
3

e) – 1 < x < 3, solution set is

f) – 5 d x d 2, solution set is {...............................................}

2. Let’s say and write down the inequalities represented by each of the following
graphs:

a) b)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

.............................................................. ..............................................................

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c) d)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

.............................................................. ..............................................................

e) f)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

.............................................................. ..............................................................

g) h)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

.............................................................. ..............................................................

Creative Section

3. Lets draw graphs of the following inequalities.

a) x > 3 b) x < – 2 c) x t – 5 d) x d 4
h) – 6 d x < 9
e) 6 < x < 10 f) – 3 d x d 3 g) – 5 < x d 7

4. Let’s find the solution sets of the following inequalities and represent them

graphically.

a) x + 2 < 8 b) x – 3 > 1 c) x + 3 d 7 d) x – 2 t 3

e) 2x + 1 < 9 f) 3x – 1 d 8 g) 4x + 5 > 9 h) 5x – 3 t 7

i) x > – 3 j) 2x d – 2 k) 2 (x + 5) < 4 + 5x l) 5 + 2 (2x – 3) t 3x + 4
2 3

m) x+2 <2 n) x–4 d–5 o) x –2 > 3x – 1 p) x+2 – x tx–2
3 2 3 4 6 3

5. Let’s solve the following inequalities and draw graphs to represent the solutions.

a) 3 < x + 1 < 7 b) – 2 < x – 3 < 2 c) – 5 d x + 1 d 8

d) 1 d 2x + 3 d 9 e) – 3 < 3x – 9 < 3 f) 7 < 5x – 3 d 27
g) 2 < 3 – x < 10 h) – 6 d 4 – 2x d 6 1 3
i) – 2 d x + 1 d 2

j) – 5 < x+2 d 4 k) – 1 d x + 2 < 1 l) 5 d x + 1 d 6 21
3 3 3 3 3 2 2

6. a) If x d – 3, find the value of y in y = 3x – 2.

b) If x > – 1, find the value of y in y = 3 – 2x.

c) If x d 5, find the value of y in 2x + 3y + 5 = 0.

7. a) When 5 is subtracted from three times a number, the result is greater than 6.
Find the least whole number that satisfies the statement.

b) When 2 is added to five times a number, the result is less than 13. Find the
greatest whole number that satisfies the statement.

8. a) The cost of a pen is Rs 15 and the cost of a box is Rs 24. Shashwat has Rs 90 and
he wants to buy a box and a few numbers of pens. Find the maximum number
of pens that he can buy.

b) A toffee costs Rs 6 and a packet of biscuit costs Rs 8. If Bishwant has Rs 50 and
he has already bought two toffees, find the maximum number of biscuits that he
can buy.

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Equation, Inequality and Graph

12.5 Linear equations with one variable

Let’s take an equation x + 5 = 9
It is an equation with only one variable which is x and the power of the variable x
is 1. Such an equation is called a linear equation with one variable. Therefore, a first
degree equation in one variable is always a linear equation.

12.6 Solution of linear equation

Let’s consider an equation, x + 3 = 8.
This equation can be true only for a fixed value of x which is 5. So, 5 is called the
solution (or root) to the equation. The process of getting a solution to an equation is
called solving equation.

Worked-out examples

Example 1: Solve a) 3 (x – 4) = 2 (x – 1) b) y 1 1 + 2 = 3
– y+1 y
Solution:
b) y 1 + 2 = 3
a) 3 (x – 4) = 2 (x – 1) –1 y+1 y
or, 3x – 12 = 2x – 2
or, 3x – 2x = – 2 + 12 or, y + 1 + 2y – 2 = 3
or, x = 10 (y – 1) (y + 1) y

or, 3y – 1 = 3
y2 – 1 y

or, 3y2 – y = 3y2 – 3
or, y = 3

Example 2 : If the sum of three consecutive odd numbers is 27. Find the numbers.
Solution :

Let the first odd number be x.

Then, the second consecutive odd number = x + 2

And, the third consecutive odd number = x + 4

Now, x + (x + 2) + (x + 4) = 27

or, 3x + 6 = 27

or, 3x = 27 – 6

or, x = 21
3
=7

The first odd number, x = 7

The second odd number = x + 2 = 7 + 2 = 9

The third odd number = x + 4 = 7 + 4 = 11

So, the required consecutive odd numbers are 7, 9, and 11.

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Example 3: A sum of Rs 80 is divided into two parts. If two times the greater part is

10 less three times the smaller part, find the parts of the sum.

Solution :

Let the greater part of the sum be Rs x.

Then, the smaller part of the sum = Rs (80 – x)

Now, 2x = 3 (80 – x) – 10

or, 2x = 240 – 3x – 10

or, 5x = 230

or, x = 46

The greater part of the sum = Rs x = Rs 46

The smaller part of the sum = Rs (80 – x) = Rs (80 – 46) = Rs 34.

So, the required parts of the sum are Rs 46 and Rs 34.

Example 4: The length of a rectangular ground is three times its breadth and its

perimeter is 160 m. Find its length and breadth.

Solution:

Let the breadth of the ground be b m, Answer checking
Then, its length = 3bm l = 60 m and b = 20 m
Perimeter = 2(l +b)
Now, perimeter of the ground = 160m
or, 2 (l + b) = 160 m =2(60 + 20)
or, 2 (3b + b) = 160 m = 2 × 80

or. 8b = 160 m = 160 m which is given

b = 20 m in the question.

The breadth of the ground = b = 20 m

The length of the ground = 3b = 3 × 20 m = 60 m

So, the required length and breadth of the ground are 60 m and 20 m respectively.

Example 5: Four years ago, father was eight times as old as his son was. Six years
hence he will be three times as old as his son will be. Find their present
age.

Solution :

Four years ago, let the age of the son was x years.

Then, four years ago, the age of the father was 8x years.

The present age of the son = (x + 4) years.

The present age of the father = (8x + 4) years.

Six years hence, the age of the son = x + 4 + 6 = (x + 10) years.

Six years hence, the age of the father = 8x + 4 + 6 = (8x + 10) years.

According to question, Answer checking
8x + 10 = 3 (x + 10) Present age of son is 8 years and father is 36 years.
4 years ago son was 4 years old
or, 8x + 10 = 3x + 30 4 years ago father was 32 years
or, 8x – 3x = 30 – 10 32 years = 8 × 4 years
or, 5x = 20 6 years hence son will be 14 years and father will be 42 years
or, x = 4 42 years = 3 × 14 years

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