NCERT Textbook: Algebraic Expressions and Identities | NCERT Textbooks (Class 6 to Class 12) - CTET & State TET PDF Download

 Page 1


ALGEBRAIC EXPRESSIONS AND IDENTITIES  93
8.1  Addition and Subtraction of Algebraic
Expressions
In earlier classes, we have already become familiar with what algebraic expressions
(or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x
2
, 4xy + 7 etc.
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add  7x
2
 – 4x + 5 and 9x – 10, we do
7x
2
 – 4x + 5
+ 9x – 10
7x
2
 + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
more examples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx        – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy  +  9yz +3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
in the second expression and 5x in the third expression, are carried over as they are,
since they have no like terms in the other expressions.
Algebraic Expressions
and Identities
CHAPTER
8
Reprint 2024-25
Page 2


ALGEBRAIC EXPRESSIONS AND IDENTITIES  93
8.1  Addition and Subtraction of Algebraic
Expressions
In earlier classes, we have already become familiar with what algebraic expressions
(or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x
2
, 4xy + 7 etc.
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add  7x
2
 – 4x + 5 and 9x – 10, we do
7x
2
 – 4x + 5
+ 9x – 10
7x
2
 + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
more examples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx        – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy  +  9yz +3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
in the second expression and 5x in the third expression, are carried over as they are,
since they have no like terms in the other expressions.
Algebraic Expressions
and Identities
CHAPTER
8
Reprint 2024-25
94  MATHEMATICS
Example 2: Subtract 5x
2
 – 4y
2
 + 6y – 3 from 7x
2
 – 4xy + 8y
2
 + 5x – 3y.
Solution:
7x
2
 – 4xy + 8y
2
 + 5x – 3y
5x
2
– 4y
2
+ 6y – 3
(–)                (+)            (–)    (+)
2x
2
 – 4xy + 12y
2
 + 5x – 9y + 3
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting –3 is the same as adding +3. Similarly , subtracting 6y is the same as
adding – 6y; subtracting – 4y
2
 is the same as adding 4y
2 
and so on. The signs in the
third row written below each term in the second row help us in knowing which
operation has to be performed.
EXERCISE 8.1
1. Add the following.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p
2
q
2
 – 3pq + 4, 5 + 7pq – 3p
2
q
2
(iv) l
2
 + m
2
, m
2
 + n
2
, n
2
 + l
2
,
2lm + 2mn + 2nl
2. (a) Subtract 4a – 7ab + 3b + 12  from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p
2
q – 3pq + 5pq
2
 – 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq
2
 + 5p
2
q
8.2 Multiplication of Algebraic Expressions:
Introduction
(i) Look at the following patterns of dots.
Pattern of dots Total number of dots
4 × 9
5 × 7
Reprint 2024-25
Page 3


ALGEBRAIC EXPRESSIONS AND IDENTITIES  93
8.1  Addition and Subtraction of Algebraic
Expressions
In earlier classes, we have already become familiar with what algebraic expressions
(or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x
2
, 4xy + 7 etc.
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add  7x
2
 – 4x + 5 and 9x – 10, we do
7x
2
 – 4x + 5
+ 9x – 10
7x
2
 + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
more examples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx        – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy  +  9yz +3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
in the second expression and 5x in the third expression, are carried over as they are,
since they have no like terms in the other expressions.
Algebraic Expressions
and Identities
CHAPTER
8
Reprint 2024-25
94  MATHEMATICS
Example 2: Subtract 5x
2
 – 4y
2
 + 6y – 3 from 7x
2
 – 4xy + 8y
2
 + 5x – 3y.
Solution:
7x
2
 – 4xy + 8y
2
 + 5x – 3y
5x
2
– 4y
2
+ 6y – 3
(–)                (+)            (–)    (+)
2x
2
 – 4xy + 12y
2
 + 5x – 9y + 3
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting –3 is the same as adding +3. Similarly , subtracting 6y is the same as
adding – 6y; subtracting – 4y
2
 is the same as adding 4y
2 
and so on. The signs in the
third row written below each term in the second row help us in knowing which
operation has to be performed.
EXERCISE 8.1
1. Add the following.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p
2
q
2
 – 3pq + 4, 5 + 7pq – 3p
2
q
2
(iv) l
2
 + m
2
, m
2
 + n
2
, n
2
 + l
2
,
2lm + 2mn + 2nl
2. (a) Subtract 4a – 7ab + 3b + 12  from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p
2
q – 3pq + 5pq
2
 – 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq
2
 + 5p
2
q
8.2 Multiplication of Algebraic Expressions:
Introduction
(i) Look at the following patterns of dots.
Pattern of dots Total number of dots
4 × 9
5 × 7
Reprint 2024-25
ALGEBRAIC EXPRESSIONS AND IDENTITIES  95
m × n
(m + 2) × (n + 3)
(ii) Can you now think of similar other situations in which
two algebraic expressions have to be multiplied?
Ameena gets up. She says, “W e can think of area of
a rectangle.” The area of a rectangle is l × b, where l
is the length, and b is breadth. If the length of the
rectangle is increased by 5 units, i.e., (l + 5) and
breadth is decreased by 3 units , i.e., (b – 3) units,
the area of the new rectangle will be (l + 5) × (b – 3).
(iii) Can you think about volume? (The volume of a
rectangular box is given by the product of its length,
breadth and height).
(iv) Sarita points out that when we buy things, we have to
carry out multiplication. For example, if
price of bananas per dozen = ` p
and for the school picnic bananas needed = z dozens,
then we have to pay = ` p × z
Suppose, the price per dozen was less by ` 2 and the bananas needed were less by
4 dozens.
Then, price of bananas per dozen = ` (p – 2)
and bananas needed = (z – 4) dozens,
Therefore, we would have to pay = ` (p – 2) × (z – 4)
To find the area of a rectangle, we
have to multiply algebraic
expressions like l × b or
(l + 5) × (b – 3).
Here the number of rows
is increased by
2, i.e., m + 2 and number
of columns increased by
3, i.e., n + 3.
To find the number of
dots we have to multiply
the expression for the
number of rows by the
expression for the
number of columns.
Reprint 2024-25
Page 4


ALGEBRAIC EXPRESSIONS AND IDENTITIES  93
8.1  Addition and Subtraction of Algebraic
Expressions
In earlier classes, we have already become familiar with what algebraic expressions
(or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x
2
, 4xy + 7 etc.
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add  7x
2
 – 4x + 5 and 9x – 10, we do
7x
2
 – 4x + 5
+ 9x – 10
7x
2
 + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
more examples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx        – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy  +  9yz +3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
in the second expression and 5x in the third expression, are carried over as they are,
since they have no like terms in the other expressions.
Algebraic Expressions
and Identities
CHAPTER
8
Reprint 2024-25
94  MATHEMATICS
Example 2: Subtract 5x
2
 – 4y
2
 + 6y – 3 from 7x
2
 – 4xy + 8y
2
 + 5x – 3y.
Solution:
7x
2
 – 4xy + 8y
2
 + 5x – 3y
5x
2
– 4y
2
+ 6y – 3
(–)                (+)            (–)    (+)
2x
2
 – 4xy + 12y
2
 + 5x – 9y + 3
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting –3 is the same as adding +3. Similarly , subtracting 6y is the same as
adding – 6y; subtracting – 4y
2
 is the same as adding 4y
2 
and so on. The signs in the
third row written below each term in the second row help us in knowing which
operation has to be performed.
EXERCISE 8.1
1. Add the following.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p
2
q
2
 – 3pq + 4, 5 + 7pq – 3p
2
q
2
(iv) l
2
 + m
2
, m
2
 + n
2
, n
2
 + l
2
,
2lm + 2mn + 2nl
2. (a) Subtract 4a – 7ab + 3b + 12  from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p
2
q – 3pq + 5pq
2
 – 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq
2
 + 5p
2
q
8.2 Multiplication of Algebraic Expressions:
Introduction
(i) Look at the following patterns of dots.
Pattern of dots Total number of dots
4 × 9
5 × 7
Reprint 2024-25
ALGEBRAIC EXPRESSIONS AND IDENTITIES  95
m × n
(m + 2) × (n + 3)
(ii) Can you now think of similar other situations in which
two algebraic expressions have to be multiplied?
Ameena gets up. She says, “W e can think of area of
a rectangle.” The area of a rectangle is l × b, where l
is the length, and b is breadth. If the length of the
rectangle is increased by 5 units, i.e., (l + 5) and
breadth is decreased by 3 units , i.e., (b – 3) units,
the area of the new rectangle will be (l + 5) × (b – 3).
(iii) Can you think about volume? (The volume of a
rectangular box is given by the product of its length,
breadth and height).
(iv) Sarita points out that when we buy things, we have to
carry out multiplication. For example, if
price of bananas per dozen = ` p
and for the school picnic bananas needed = z dozens,
then we have to pay = ` p × z
Suppose, the price per dozen was less by ` 2 and the bananas needed were less by
4 dozens.
Then, price of bananas per dozen = ` (p – 2)
and bananas needed = (z – 4) dozens,
Therefore, we would have to pay = ` (p – 2) × (z – 4)
To find the area of a rectangle, we
have to multiply algebraic
expressions like l × b or
(l + 5) × (b – 3).
Here the number of rows
is increased by
2, i.e., m + 2 and number
of columns increased by
3, i.e., n + 3.
To find the number of
dots we have to multiply
the expression for the
number of rows by the
expression for the
number of columns.
Reprint 2024-25
96  MATHEMATICS
Notice that all the three
products of monomials, 3xy,
15xy, –15xy, are also
monomials.
TRY THESE
Can you think of two more such situations, where we may need to multiply algebraic
expressions?
[Hint: • Think of speed and time;
• Think of interest to be paid, the principal and the rate of simple interest; etc.]
In all the above examples, we had to carry out multiplication of two or more quantities. If
the quantities are given by algebraic expressions, we need to find their product. This
means that we should know how to obtain this product. Let us do this systematically . T o
begin with we shall look at the multiplication of two monomials.
8.3  Multiplying a Monomial by a Monomial
Expression that contains only one term is called a monomial.
8.3.1  Multiplying two monomials
W e begin with
4 × x = x + x + x + x = 4x  as seen earlier.
Similarly , 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Now, observe the following products.
(i) x × 3y = x × 3 × y = 3 × x × y = 3xy
(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y  = 15xy
(iii) 5x × (–3y) = 5 × x × (–3) × y
= 5 × (–3) × x × y  = –15xy
Note that 5 × 4 = 20
i.e., coefficient of product = coefficient of
first monomial × coefficient of second
monomial;
and x × x
2
 = x
3
i.e.,  algebraic factor of product
= algebraic factor of first monomial
× algebraic factor of second monomial.
Some more useful examples follow .
(iv) 5x × 4x
2
 = (5 × 4) × (x × x
2
)
= 20 × x
3
 = 20x
3
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x
2
yz
Observe how we collect the powers of different variables
in the algebraic parts of the two monomials. While doing
so, we use the rules of exponents and powers.
8.3.2  Multiplying three or more monomials
Observe the following examples.
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x
2
y
2
 × 6x
3
y
3
 = (4xy × 5x
2
y
2
) × 6x
3
y
3
 = 20x
3
y
3
 × 6x
3
y
3
 = 120x
3
y
3
 × x
3
y
3
= 120 (x
3
 × x
3
) × (y
3
 × y
3
) = 120x
6
 × y
6
 = 120x
6
y
6
It is clear that we first multiply the first two monomials and then multiply the resulting
monomial by the third monomial. This method can be extended to the product of any
number of monomials.
Reprint 2024-25
Page 5


ALGEBRAIC EXPRESSIONS AND IDENTITIES  93
8.1  Addition and Subtraction of Algebraic
Expressions
In earlier classes, we have already become familiar with what algebraic expressions
(or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x
2
, 4xy + 7 etc.
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add  7x
2
 – 4x + 5 and 9x – 10, we do
7x
2
 – 4x + 5
+ 9x – 10
7x
2
 + 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
more examples.
Example 1: Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx        – 4y
+ –2xy – 3zx + 5x (Note xz is same as zx)
5xy  +  9yz +3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
in the second expression and 5x in the third expression, are carried over as they are,
since they have no like terms in the other expressions.
Algebraic Expressions
and Identities
CHAPTER
8
Reprint 2024-25
94  MATHEMATICS
Example 2: Subtract 5x
2
 – 4y
2
 + 6y – 3 from 7x
2
 – 4xy + 8y
2
 + 5x – 3y.
Solution:
7x
2
 – 4xy + 8y
2
 + 5x – 3y
5x
2
– 4y
2
+ 6y – 3
(–)                (+)            (–)    (+)
2x
2
 – 4xy + 12y
2
 + 5x – 9y + 3
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting –3 is the same as adding +3. Similarly , subtracting 6y is the same as
adding – 6y; subtracting – 4y
2
 is the same as adding 4y
2 
and so on. The signs in the
third row written below each term in the second row help us in knowing which
operation has to be performed.
EXERCISE 8.1
1. Add the following.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p
2
q
2
 – 3pq + 4, 5 + 7pq – 3p
2
q
2
(iv) l
2
 + m
2
, m
2
 + n
2
, n
2
 + l
2
,
2lm + 2mn + 2nl
2. (a) Subtract 4a – 7ab + 3b + 12  from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p
2
q – 3pq + 5pq
2
 – 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq
2
 + 5p
2
q
8.2 Multiplication of Algebraic Expressions:
Introduction
(i) Look at the following patterns of dots.
Pattern of dots Total number of dots
4 × 9
5 × 7
Reprint 2024-25
ALGEBRAIC EXPRESSIONS AND IDENTITIES  95
m × n
(m + 2) × (n + 3)
(ii) Can you now think of similar other situations in which
two algebraic expressions have to be multiplied?
Ameena gets up. She says, “W e can think of area of
a rectangle.” The area of a rectangle is l × b, where l
is the length, and b is breadth. If the length of the
rectangle is increased by 5 units, i.e., (l + 5) and
breadth is decreased by 3 units , i.e., (b – 3) units,
the area of the new rectangle will be (l + 5) × (b – 3).
(iii) Can you think about volume? (The volume of a
rectangular box is given by the product of its length,
breadth and height).
(iv) Sarita points out that when we buy things, we have to
carry out multiplication. For example, if
price of bananas per dozen = ` p
and for the school picnic bananas needed = z dozens,
then we have to pay = ` p × z
Suppose, the price per dozen was less by ` 2 and the bananas needed were less by
4 dozens.
Then, price of bananas per dozen = ` (p – 2)
and bananas needed = (z – 4) dozens,
Therefore, we would have to pay = ` (p – 2) × (z – 4)
To find the area of a rectangle, we
have to multiply algebraic
expressions like l × b or
(l + 5) × (b – 3).
Here the number of rows
is increased by
2, i.e., m + 2 and number
of columns increased by
3, i.e., n + 3.
To find the number of
dots we have to multiply
the expression for the
number of rows by the
expression for the
number of columns.
Reprint 2024-25
96  MATHEMATICS
Notice that all the three
products of monomials, 3xy,
15xy, –15xy, are also
monomials.
TRY THESE
Can you think of two more such situations, where we may need to multiply algebraic
expressions?
[Hint: • Think of speed and time;
• Think of interest to be paid, the principal and the rate of simple interest; etc.]
In all the above examples, we had to carry out multiplication of two or more quantities. If
the quantities are given by algebraic expressions, we need to find their product. This
means that we should know how to obtain this product. Let us do this systematically . T o
begin with we shall look at the multiplication of two monomials.
8.3  Multiplying a Monomial by a Monomial
Expression that contains only one term is called a monomial.
8.3.1  Multiplying two monomials
W e begin with
4 × x = x + x + x + x = 4x  as seen earlier.
Similarly , 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Now, observe the following products.
(i) x × 3y = x × 3 × y = 3 × x × y = 3xy
(ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y  = 15xy
(iii) 5x × (–3y) = 5 × x × (–3) × y
= 5 × (–3) × x × y  = –15xy
Note that 5 × 4 = 20
i.e., coefficient of product = coefficient of
first monomial × coefficient of second
monomial;
and x × x
2
 = x
3
i.e.,  algebraic factor of product
= algebraic factor of first monomial
× algebraic factor of second monomial.
Some more useful examples follow .
(iv) 5x × 4x
2
 = (5 × 4) × (x × x
2
)
= 20 × x
3
 = 20x
3
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x
2
yz
Observe how we collect the powers of different variables
in the algebraic parts of the two monomials. While doing
so, we use the rules of exponents and powers.
8.3.2  Multiplying three or more monomials
Observe the following examples.
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x
2
y
2
 × 6x
3
y
3
 = (4xy × 5x
2
y
2
) × 6x
3
y
3
 = 20x
3
y
3
 × 6x
3
y
3
 = 120x
3
y
3
 × x
3
y
3
= 120 (x
3
 × x
3
) × (y
3
 × y
3
) = 120x
6
 × y
6
 = 120x
6
y
6
It is clear that we first multiply the first two monomials and then multiply the resulting
monomial by the third monomial. This method can be extended to the product of any
number of monomials.
Reprint 2024-25
ALGEBRAIC EXPRESSIONS AND IDENTITIES  97
TRY THESE
Find 4x × 5y × 7z
First find 4x × 5y and multiply it by 7z;
or first find 5y × 7z and multiply it by 4x.
Is the result the same? What do you observe?
Does the order in which you carry out the multiplication matter?
Example 3: Complete the table for area of a rectangle with given length and breadth.
Solution: length breadth area
3x 5y 3x × 5y = 15xy
9y 4y
2
..............
4ab 5bc ..............
2l
2
m 3lm
2
..............
Example 4: Find the volume of each rectangular box with given length, breadth
and height.
length breadth height
(i) 2ax 3by 5cz
(ii) m
2
n n
2
p p
2
m
(iii) 2q 4q
2
8q
3
Solution: V olume = length × breadth × height
Hence, for (i) volume = (2ax) × (3by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz
         for (ii) volume = m
2
n × n
2
p × p
2
m
= (m
2
 × m) × (n × n
2
) × (p × p
2
) = m
3
n
3
p
3
         for (iii) volume = 2q × 4q
2
 × 8q
3
= 2 × 4 × 8 × q × q
2
 × q
3 
 = 64q
6
EXERCISE 8.2
1. Find the product of the following pairs of monomials.
(i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) 4p
3
, – 3p
(v) 4p, 0
2. Find the areas of rectangles with the following pairs of monomials as their lengths and
breadths respectively .
(p, q); (10m, 5n); (20x
2
, 5y
2
); (4x, 3x
2
); (3mn, 4np)
We can find the product in other way also.
4xy × 5x
2
y
2
 × 6x
3
 y
3
= (4 × 5 × 6) ×  (x × x
2
 × x
3
) × (y × y
2
 × y
3
)
= 120 x
6
y
6
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