Prime Mathematics 8

Prime Mathematics Book - 8

Combined mean:

Let mean of first n1, observation is X1 and mean of next n2 observation is X2 then ,
sum of first n1 observations (∑fx)1 = n1X1

sum of next n2 observations (∑fx)2 = n2X2

total number of observations = N = N1 + N2

sum of all the observations ∑fx = (∑fx)1 + (∑fx)2 = n1X1 + n2X2

Combined Mean (X) = ∑fx
N

=n1xn11 ++nn22x 2

In the same way, combined mean X = n1X1 + n2X2 + n3X3 + .....

n1 +n2 + n3 + .....

Example 9: The average height of 15 boys is 5.3 feet and that of 10 girls is 4.9 feet.

Find the average height of all the students.

Solution: Here

Number boys (n1) = 15, mean height of boys X = 5.3 feet
1

Number of girls (n2) = 10, mean height of girls X = 4.9 feet.
2
∴ Combined Mean (X) = n1Xn11 ++nn22X 2

= 15 × 5.3 + 10 × 4.9
15 + 10

79.5 + 49 128.5
= 25 = 25 = 5.14

∴ Mean height of the students is 5.14 feet.

Exercise 1.2.1

1. Find the arithmetic mean of the following data:
a) 89, 92, 85, 88, 61, 62, 60, 53, 59, 55
b) 70, 110, 120, 80, 40, 210
c) 20, 15, 12, 5, 8
d) 34, 11, 23, 79, 47, 67, 58, 81

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Prime Mathematics Book - 8

2. Find the mean of the following data.
a) x 15 16 17 18 19 20 21

f 3 9 12 24 30 10 3

b) Wages (Rs) 250 275 300 325 350 400
No. of workers
10 14 15 8 2 1

c) Daily wages(Rs) 450 500 550 600 650 700 750
No. of workers
12 13 14 13 12 11 5

d) Marks obtained 5 10 15 20 25

No. of students 6 9 12 7 6

3. Calculate the mean of the following data.

a) Classes 10 - 20 20 -30 30 - 40 40 - 50 50 - 60
9 4
Frequency 11 11 15

b) Classes 1-5 6 - 10 11 - 15 16 - 20 21 - 25
Frequency 2 7 8 6 2

c) Marks obtained 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100
No. of students 9 12 14 15 10

d) Ages (years) 18 - 24 24 - 30 30 - 36 36 - 42 42 - 48 48 - 54

No. of workers 12 16 24 22 18 10

4. a) Find the mean of prime numbers between 20 and 50.
b) If the arithmetic mean of 8, 7, 9, x, 5, is 7, find the value of x.
c) If the mean height of the data given below is 62.4 kg, find the missing frequency.

Height(inches) 60 61 62 63 64

No. of children 2 3 8 P 5

d) If the mean of the following data is 5.6, find m 10
4
x 2 4 6 8
f 7 4m5

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Prime Mathematics Book - 8

5. a) Mean weights of 30 girls and 20 boys of grade VIII are respectively 39 kg and 46 kg.
Find the mean weight of the students of grade VIII.

b) The mean of 7 numbers is 9. If the mean of the first 4 numbers is 12, find the mean
of the remaining 3 numbers.

c) Average marks of 40 students of a class is 75.5. If average marks of 22 girls is 80,
find the average marks of remaining 18 boys.

d) The mean of the marks secured by 26 students of section A of class VIII is 80, that
of 30 students of section B is 78 and that of 24 students of section C is 72. Find the
mean of the students of three sections of class VIII.

6. a) If X = 17.5, ∑fx=455, find ∑f.
b) If ∑fx= 480 and N=40, find X.

c) If the mean marks of 32 students is 42, find total marks of all the students.

d) If N= 70+a, X= 50 and ∑fx= 2000+55a, find the value of a.

1.2.2 Median

Abhijit, Ajay, Santosh, Rejina, Kabita, Kristi and Sunita are standing in a row according

to their heights. Rajina is standing in the middle. We say Rajina's height is the median

height. When the given observations are arranged in ascending or descending order,

the value of the middle term is the median.

The median divides the arranged data exactly into two equal parts.
If N is the number of observation, Median = N + 1 item

( )2

(i) Median of raw (individual) data:
- Arrange the given data in ascending or descending order. Let the number of

observations in the data is N.

Abhijit Ajay Santosh Rejina Kabita Kristi Sunita

1. If N is odd: ( )= Value ofN+1th
Median (Md.) 2
2. If N is even: item.

Median (Md.) = Average of two middle observations

[( ) ( ) ]1 n th n
2
=2 2
observation + + 1 th

observation

248 11 Statistics

Prime Mathematics Book - 8

Example 10: Find the median of: 16, 42, 22, 25, 33, 32, 26, 36, 27, 34, 45.

Solution: Arranging the data in ascending order

16, 22, 25, 26, 27, 32, 33, 34, 36, 42, 45

Where number of observations (N) = 11

N being odd,

∴ Median = The value of n+1 th term

( )2

( ) 11+1 th
= The value of
2 term = The value of 6th term. = 32.

∴ Median = 32

Example 11: Find the median of: 4, 6, 63, 41, 7, 19, 30, 39, 80, 24, 53, 40, 69, 75

Solution: Arranging the given data in ascending order:

4, 6, 7, 19, 24, 30, 39, 40, 41, 53, 63, 63, 75, 80.

Where number of terms (n) = 14 which is even.
( ) ( ) n th
∴ Median = Average of n th term and 2 + 1 term
2
[( ) ( ) ] 14
1 14 th 2 + 1 th
=2 2
observation + observation

1 [ 7th term + 8th term ] = 1 [ 39 + 40 ] 79 = 39.5
=2 2 =2

∴ Median = 39.5

Example 12: The following data have been arranged in ascending order:

25, 38, 46, 48, 52, m - 3, m + 5, 60, 72, 81, 84, 87

If the median of the data is 55, find the value of m.

Solution:
Given data: 25, 38, 46, 48, 52, m -3, m + 5, 60, 72, 81, 84, 87

Where n = 12 which is even
( ) ( )= Average of n th
n th term and 2 + 1
∴ Median 2 term

1 12
2
[( ) ( ) ]=2
12 th + + 1 th

2 item item

=12 [ 6th term + 7th term ]

=12 [m-3 + m+5] =21 [2m+2] = m+1

or, 55 = m+1 ∴ m = 54
(ii) Median of un-grouped (desecrate) data:
- Data should be ordered (arranged in ascending/descending order).

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Prime Mathematics Book - 8

- Prepare commutative frequency distribution table.

( ) n+1 th
2
- item.
Median =
( ) n+1 , the c.f. value just greater than n+12 . n+1
- Median = value corresponding to c.f. value or if there is not c.f value
( ) ( )2 2

Example 13: Calculate the median of the data.

Marks 12 15 20 35 36 38 42 48
12 11 8 6 4 3
No. of students 2 4
Commutative frequency
Solution: Frequency (cf)
Marks (f) 2
6
(X) 18
29
12 2 37
15 4 43
20 12 47
35 11 50
36 8
38 6
42 4
48 3

N=50

( ) ( )
n+1 th 50+1 th
2 2 item = 25.5th item
Here, N = 50 ∴ median = item =

In the c.f. column, there is not c.f. value 25.5 and the c.f. value just greeter

than 25.5 is 29. Corresponding to it, median = 35

Example 14: Find the median of the data.

3, 18, 3, 12, 12, 15, 7, 7, 10, 7, 10, 12, 10, 15, 10, 15, 12

Solution: Arranging the data in ascending order

3, 3, 7, 7, 7, 10, 10, 10, 10 , 12, 12, 12, 12, 15, 15, 15, 18


Here,

Number of observation N = 17
( ) ( ) n+1 17+1 th
Now, median = 2 th 2 item = 9th item

item =

In the above arranged data, 9th term is 10.

Thus median = 10.

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Prime Mathematics Book - 8

Alternative method:
Arranging and presenting the data in commutative frequency distribution table.

Values Frequency Commutative frequency
(X) (f) (cf)


3 2 2
7 3 5
4 9
10 13
16
12 4 17
15 3
18 1

N=17

Here, Number of observation (N) = 17.

( ) ( ) 17+1
Now, median = n+1 th item= th

22 item = 9th item

The Corresponding value in the column of x to c.f. value 9 is 10.

∴ Median = 10.
Note:

F In case of less number of observations, desecrate data can also be simply
arranged and locate the median just as in the case of individual data.

( ) F It there is c.f. value = n +1 , median is the observation corresponding
2 n +1
( )to it. If there is not c.f. value = 2 , median is the observation
corresponding to the c.f value just greater than it .

Exercise 1.2.3

1. Find the median of the data:
a) 75, 92, 105, 124, 140, 142, 150, 158, 160
b) 40cm, 35cm, 45cm, 30cm, 44cm, 48cm
c) 28, 32, 55, 80, 42, 66, 73, 92, 60, 87
d) Rs.10, Rs.7, Rs.25, Rs.16, Rs.28, Rs.32, Rs.36

2. Find the median of the data.
a) x 4 9 12 14 20 21

f 234431

b) Wage per day (Rs.) 250 300 350 400 450 500
Number of worker 6 9 12 10 9 4

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Prime Mathematics Book - 8

Marks Obtained 35 42 59 67 72 88 94
c) Number of students 2 4 6 8 10 7 4

3. a) The data 90, 100, 5x + 10 , 115, 120, are in ascending order. If median of the data
3

is 110, find the value of x.
b) x + 1, 2 x + 2, 3x + 3 and 4x + 4 are in ascending order. If the median of the data

is 10, find the value of x.
c) 9 - x, 12 - x, 13 + x, 15 + x are in ascending order. Find the median of the data.
d) 29, 32, 41, m, 60, 67, 72, 8, are in ascending order and median of the data is

58. Find the values of m.

1.2.4 Mode

In the data, the value which occurs maximum times is called mode and is denoted by Mo.
- Individual series has no mode or all are modes.
- In a discreet series, the value with the greatest frequency is the mode.
- A data can have more than one mode. If there exist two modes (two values with same

highest frequencies) the data is called bimodel data. If there exist more than two
modes, the data is called multi- model data.

Example 15: Find the mode of the data: 11, 4, 16, 11, 7, 14, 7, 11, 7, 14, 16, 11.
Solution: Arranging the data in frequency distribution table.

X 4 7 11 14 16
F 1 3 422

Here the highest frequency is 4 that of 11.
∴ Mode (Mo) = 11.
Example 16: The given data shows the record of sales of 100 pair of shoes.

Size of shoe 5 6 789 10
Number sold 7 15 25 30 13 10

Find the model size of the shoes.

Solutions: Here, the maximum frequency is 30 that of size 8.

∴ The mode = 8
∴ Model size = 8

Exercise 1.2.4

1. Find the mode of the data:

a) 2, 3, 2, 3, 4, 4, 5, 6, 6, 4, 7, 5

b) a, b, b, e, c, a, e, c, e, a, d, b, e, e, b.

c) Rs3, Rs20, Rs15, Rs8, Rs15, Rs5, Rs20, Rs10, Rs18, Rs20

252 11 Statistics

Prime Mathematics Book - 8

d) 5, 7, 9, 1, 5, 3, 9, 7, 1, 7, 9, 7

2. Find the mode of the data.

a) X 125 130 135 140 145 150
f 9 11 12 9 6 3

b) Size of shoes 456789

No. of shoes sold 12 18 20 25 15 10

1.3 Measure of dispersion

Marks obtained by two groups of students are given below.

A: 20, 30, 40, 50, 60, 70, 80, 90, 100

B: 40, 45, 50, 55, 60, 65, 70, 75, 80
Which group has better performance?
→ One way of comparing two groups is by observing their average marks. But average

marks of both the group is 60. Other way you may also use the highest and lowest
marks but statistics do not concern with individuals.

→ In group A values are scattered much away from the mean value where as in
group B, the values are scattered less from the data. Dispersion is the measure
of scatterness of the data from the mean (average). Dispersion is measured by
(i) Range (ii) Quartile deviation (iii) Mean deviation (iv) standard deviation. As an
introduction of dispersion, we will discuss about range.

Range:

Range is the difference of the largest and the smallest value. It is the

simplest and crude measure of dispersion (secureness). If the largest value is L

and smallest value is S then Range = L - S

Range can also be expressed relativity as decimal coefficient as

Coefficient of rtahnegrea=ngLLe+-SaSnd its coefficient from the data.
Example 17: Calculate

12 years, 25 years, 37 years, 18 years, 42 years, 50 years, 54 years

Solution: Given data: 12 years, 25years, 37years, 18years, 42years, 50years, 54years

Here, the largest value (L) = 54years and

the lowest value (S) = 12years

Range = L-S = 54years - 12years = 42years.

And, = L-S = 54yrs - 12yrs = 42 = 0.64
Coefficient of range L+S 54yrs + 12yrs 66

A. Range of grouped data: In case of grouped data,

The largest value (L) = Upper boundary of the highest class.

The smallest value (s) = Lower boundary of the lowest class.

Statistics 253

Prime Mathematics Book - 8

Example 18: Find the range of the data.

Class - interval 5-10 10-15 15-20 20-25
Frequency 6 10 5 4

Solution: In the given data, the highest class is 20-25 where its upper limit is 25.

∴ L=25 Now, can you tell,
And the lowest class is 5-10 which group of
where its lower limit is 5 students we
∴S=5 discussed in the
Now, beginning of the
topics is better?

Range = L - S = 25 - 5 = 20

Note:
F Range has unit but coefficient is simply a number.

F While comparing the marks of two groups, data with smaller range or coefficient of range is better.

Exercise 21.5

1. Find the range and coefficient of range of the following:
a) 18, 22, 24, 25, 12, 15, 20, 16
b) 191cm, 202cm, 150cm, 300cm, 195cm, 198cm, 205cm
c) 150c, 180c, 170c, 210c, 250c, 300c, 270c, 290c
d) 5225m, 3450m, 8848m, 7820m, 4360m, 8470m, 6955m

2. Calculate the range of the data.

a) Marks 10 11 12 13 14 15 16 17 18 19 20
b) 663 2
c) Frequency 2 4 7 8 9 8 8
40 41
Ages (years) 25 30 34 38 87
No. of persons 11 9 10 12
25
Income(Rs.1000) 6 8 12 15 20 2

No. of persons 8 10 15 16 4

3. Find the range of the data. Also calculate the coefficient of range.

a) Marks 5-10 10-15 15-20 20-25 25-30 30-35
No. of students 1 4 10 16 12 8

b) Wages (Rs) 100-150 150-200 200-250 250-300 300-350
No. of workers 10 15 20 18 16

254 11 Statistics

Prime Mathematics Book - 8

Area Revision Test (Statistics)

1. Monthly average temperature of a certain locality are given below. Draw a line
graph to represent the data.

Months Baisakh Jestha Asad Shrawan Bhadra Aswin
31° 34° 28° 31° 20°
Temperature (°c) 25°
AB O
2. Draw pie chart of the data. 9 30

Blood group A B
No. of Persons 18 15

3. Calculate the mean of the data.

Marks Obtained 20-30 30-40 40-50 50-60 60-70 70-80
No. of Students
4 10 14 12 8 2

4. Find the median of the data. 34, 11, 23, 79, 47, 67, 58, 81
5. (a) If mean x = 175, Sum of the terms ∑fx = 455, find the number of term ∑f.

(b) If the median of the data 29, 32, 41, n, 60, 67, 72, 8 arranged in ascending is
58, find the value of m.

Answers

(Statistics) 18
17
Exercise - 1.1(Precession of data) 16
15
25 14production (in 1000)

1. 20 2. 13

15No. of workers 12
10 11
5
0 400 500 600 700 800 900 1000 10

Wages (Rs.) 0 2009 2010 2011 2012 2013

Year

3. 400 Temperature 30 Profit (Lakhs Rs. )
4.
300 Baisakh Jestha Ashadh ShrawanBhadra Aswin
200 20
100
10
0
0
2013
2012
2011
2010
2009
2008
2007
2006

Years

Statistics / Answers 255

Applied Mathematics Book - 8

Number of students of a school Index Expenditure of a family Index

5. II = Class I 700 1000 = Food
I 500 600 = Clothing
= Class II 6. = Education
III V = Class III 800 = Rent
IV = Class IV = Misc
= Class V

Number of teachers of ABC school Index

= Pre-primary 8. Income of a family Index
= Primary
7. = L. Secondary = House rent
= Secondary = Business
1000 600 420 720 = Pension
= Agriculture
480 = Salary

1200 800 900 1080

9. Index 10. Oil
720
= Bricks 147.60
540 Hydro electricity
540 = Cement Nuclear
360 540 = Steel 43.20 21.60
900 = Timber 75.60
= Labour 720
Natural gas
= Misc Coal

11. (i) Pie Chart (ii) Area of oceans of the world (iii) Pacific, Arctic (iv) 18.61%
(v) Pacific 183.7 million sq. km, Atlanta 106.68 million sq. km, Indian 73.7 million sq. km Antarctic 19.8

million sq. km, Arctic 12.12 million sq. km.
12. (i) Food : Rs. 5760, Rent : Rs. 1240.64, Clothing : Rs. 1440, Transport : Rs. 1359.36, Education : Rs. 3600
(ii) Expenditure of a family.

Topics Food Rent Clothing Transport Education

Expenditure (Rs.) 5760 2240.64 1440 1359.39 3600

Exercise - 1.2.1 (Mean)

1. (a) 70.4 (b) 105 (c) 12 (d) 50 2. (a) 18.22 (b) 291 (c) 5832.125 (d) 14.75

3. (a) 31.8 (b) 12.8 (c) 75.83 (d) 35.82 4. (a) 35.86 (b) 6 (c) 7 (d) 5

5. (a) 41.8 (b) 5 (c) 70 (d) 76.85 6. (a) 26 (b) 12 (c) 1344 (d) 300

Exercise - 1.2.3 (Median)
1. (a) 140 (b) 42 (c) 61 (d) 25 2. (a) 12 (b) 72 (c) 20 (d) 350 3. (a) 64 (b) 3 (c) 12.5 (d) 56

1. (a) 4 (b) e (c) 20 (d) 7 Exercise - 1.2.4 (Mode)
2. (a) 135 (b) 7 (c) 72.5 (d) 48.18

Exercise - 1.3 (Range)

1. (a) 13, 0.35 (b) 150, 0.33 (c) 15o, 0.33 (d) 5398m, 0.439

2. (a) 10 (b) 26 (c) 19 3. (a) 30, 0.75 (b) Rs. 250, 0.56

256 11 Statistics / Answers

Estimated periods - 20

At the end of this unit the student will be able to Objectives:

● know about the factorization of algebraic expressions.
● factorize algebraic expressions of different forms.
● use of a2 - b2=(a + b) (a - b).
● know about the concept of (a ± b)3.
● find the H.C.F and L.C.M of algebraic expressions.
● know about the addition, substraction, multiplication and division of rational algebraic expressions.

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

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

Teaching Materials:

Chart of algebraic expressions, tiles, local materials, etc.

Activities:

It is better to
● discuss to factorize the given algebraic expressions.
● discuss about the use of a2 - b2 = (a + b) (a - b).
● discuss about the use of (a ± b)2 = a2 ± 2ab + b2.
● discuss about the factorization of trinomial expressions of different forms.
● discuss about the concept of (a ± b)3 and its uses.
● discuss to find the H.C.F and L.C. M of algebraic expressions.
● discuss the about the addition, substraction, multiplication and division of rational algebraic expressions.

Prime Mathematics Book - 8

Unit - 1 Algebraic Expressions Estimated periods : 20

Review:

Algebra is a branch of mathematics in which we deal symbols and letters to represent
the numbers. The symbols or letters which we use are called variables or constants. The
symbols which have only one (fixed) value are called constants and having different
values are variables. The mathematical operational sign +, -, ×, ÷ are used between
the variable and constant, then the expression is formed which is called an algebraic
expression. For examples, 2x, 3x2 + 4y, x2 - 5x + 7, etc.

The naming of algebraic expression are given according to terms which the expression
contains. The expression that contains only one term is called monomial expression. For
example, 5, 2x, 7x2y etc. The expression that contains two terms is called binomial
expression. For example, x + 2y, 2a - 7b2 etc.

The expression that contains three terms is called trinomial expression.
For example, a + 2b - 7, 4x3 + 9x - y etc.

The expression contains more than three terms is called multi-nomial expression.
For example, 7x2 + 5xy - 4x - 7.

The algebraic expression containing the variables with non-negative integral power is
called a polynomial. For example, x + 3, 4x2 - 5x + 7 etc.

We have already studied about the addition, subtraction, multiplication, division of
the algebraic expression in grade VI and VII. Now, we will discuss about the factorization
of algebraic expression in this grade.

1.1 Factorization

We have 5 × 7 = 35 i.e the product of 5 and 7 is 35. So 5 and 7 are the factors of 35.
Similarly, m(x + y)= mx + my i.e. the product of m and (x + y) is mx + my. So, m and
(x + y) are the factors of mx + my. When m and (a + b) are given, we multiply and get
the product am + bm. Conversely, when the product am + bm is given, we factorize
and get the factors m and (a + b).Therefore, the factorization is the reverse of
expansion. For example,

Factorization
3a2 + 2ab = a(3a + 2b)

Expansion

The given algebraic expression can be expressed in the product form of prime

factors by a process known as the factorization. For example: 7a + a3 = a(7+a2)
because a is common in both terms.

258 12 Algebraic Expressions

Prime Mathematics Book - 8

Types of factorization:

Type - I Factorization by taking common:

If each term of an algebraic expression includes a common factor. Then the
expression can be factorized by taking out that common factor. For example, 4ax2 - 2xy
is an algebraic expression. 1st term = 4ax2 = 2 × 2 × a × x × x and 2nd trem = 2xy = 2 × x
× y . In both terms, 2 and x are common factors. So, 2x is taken out common from each
term of the given expression.

∴ 4ax2 − 2xy = 2x(2ax − y)
In this way, we can factorize an algebraic expression by taking out its common factor(s).

Example 1: Factorize the following expressions:

a) 3x − ax + byx b) 3x2y − 27xy2
Solution:

a) Here, 3x - ax + byx

1st term =3x =3 × x , 2nd term =ax =a × x and 3rd term = byx = b × y × x ,

In each term, x is a common factor,

∴ 3x − ax + byx = x(3 − a + by)

b) Here,

3x2y − 27xy2

= 3 × x × x × y − 3 × 3 × 3 × x × y × y

= 3 × x × y (x − 3 × 3 × y)

∴ 3x2y − 27xy2 = 3xy(x − 9y)

Exercise 1.1(a)

Resolve into factors: 2. ax2 − bxy 3. 4x2y − 2axy 4. 5ab−10bc
1. mx − my

5. − m2 + 2mn2 6. 12p2 − 6q2 7. − 4mn2 + 2mn 8. x + x3

9. 12x2 + xy + xz 10. 2x3 − 2x2 + 8x 11. 60x4 − 4x2y + 12x

12. (a + b)(x + y) − (a + b)(x − y) 13. x(a + b) − y(a + b) + z(a + b)

14. x(2x + y) − 3(2x + y) + 4y(2x + y)

Algebraic Expressions 259

Prime Mathematics Book - 8

Type - II. Factorization by grouping:

Sometimes,nothing is common to all the terms of the given algebraic expression.
But the terms of the expression can be arranged in suitable groups which have a common
factor. After taking common from among the group, the expression can be factorized.

For example: a − b + b2 − ab

Arrange the terms for taking common "a" from terms and taking common "b" from

next terms.

So, a − ab − b + b2 = a(1 − b) − b(1 − b)

= (1 − b)(a − b) [ (1 − b) is common to both groups]
∴
Example 2: Factorise the following expressions: ∴

a) a2 + bc + ab + ac b) x2y + 4xy − xy2 − 4y2

Solution: a) Here,

a2 + bc + ab + ac = a2 + ab + ac + bc = a(a + b) + c(a + b)

= (a + b)(a + c) [ (a + b) is common to both groups]

b) Here,
x2y + 4xy − xy2 − 4y2 = x2y − xy2 + 4xy − 4y2 = xy(x − y) + 4y(x − y)
= (x − y)(xy + 4y) = (x − y)y(x + 4) = y(x - y)(x + 4)

Example 3: Factorize: (x2 − y2)z + (y2 − z2)x

Solution: Here,

(x2 − y2)z + (y2 − z2)x [Nothing is common]

= x2z − y2z + y2x − z2x [Expand the expression]

= x2z − z2x + y2x − y2z [Grouping the terms]

= xz(x − z)+y2(x − z) [Taking common with in the group]

= (x − z)(xz + y2) [Taking common(x − z)from the terms]

Resolve into factors: Exercise 1.1(b) 3. ax + by + ay + bx
1. mx + xy − my − y2 6. pq + 1 − p − q
4. m2 − 15n − 5m + 3mn 2. mx + 2ny − 2m − nxy 9. a2 − (b + 3)a + 3b
7. 2a2 + 5a − 6a − 15 5. x − y + x2 − xy 12. x3 − x2 − xy + x + y − 1
10. (a2 − b2)c + (b2 − c2)a 8. x2y − xy + 2x2y − 2xy 14. 4x - y + 16x2 - y2
13. 2x2 + 3ax + 2ax + 3a2 11. mn2 + (1 + m2)n + m
14. 2ab + a2b − 2b − ab
260 12 Algebraic Expressions

Prime Mathematics Book - 8

Type - III Factorization of difference between two squares:

a a a b
x y a-b


x

a a

b bz y ba

b b a+b
fig: 1 fig: 2 fig: 3

First draw a square ABCD of a side "a" units as shown in figure 1, then draw a square
of a side "b" unit in the square ABCD as shown in figure 1 and shade parts except square
having side "b" unit of the square ABCD. Divide the square ABCD in three parts x,y and
z as shown in figure 2 and separate all three parts by a paper cutter. Arrange the parts
x and y as shown in figure 3. Then figure 3 is a rectangular form in which length and
breadth are (a + b) and (a − b) respectively. The area of whole square ABCD is a2 and
area of the unshaded part of the square ABCD is b2 in the figure 1. So, the area of the
shaded part of the figure 1 is a2 - b2.

Area of the figure 3 = area of the rectangle = length × breadth = (a + b) × (a − b)

Area of the shaded part of the figure:-I and whole area of the figure 3 are equal.

∴ a2 - b2 = (a + b) (a − b)

This suggests that if an expression is a difference of two squares, then it can be
factorized as a product of sum and difference.

Example 4: Factorize the following expressions:

a) x2 − 9y2 b) m2 − 1 c) a2 + 2ab − c2 + b2
Solution: 4n2

a) Here,

x2 − 9y2 = (x)2 − (3y)2

Using a2 − b2 = (a + b) (a − b) = (x + 3y) (x − 3y)

b) Here,

( ) 12
1 2n
m2 − 4n2 = (m)2 −

Using a2 − b2 = (a + b) (a − b)

( ) ( ) + 1 m - 1
=m 2n 2n

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c) Here,

a2 + 2ab − c2 + b2 = a2 + 2ab + b2 − c2 = (a + b)2 − c2

= (a + b + c) (a + b − c)

Example 5: Find the value by using a2 − b2 = (a + b) (a − b) of 295×295−180 × 180 .
295 −180
Solution: Here,
=295 × 295 − 180 × 180
295 −180


=(29259)52 − (180)2
−180


=(295 +(219850)−(219850)- 180)




=(295 + 180)(295 − 180)
(295 − 180)
= 295 + 180b = 475

Exercise 1.1(c)

1. Factorize the following expressions:

a) a2 − 16 b) 9 − c2 c) 25a2 − b2 d) a2 − 16b2 e) 9x2 − 49y2

f) p2 − 1 g) x2 − 1 h) 5x2 − 20y2 i) 13m2 − 11 j) ab3−9a3b
q2 9y2

k) 25 − 1 l) (2x − 3y)2 − (2x + 3y)2 m) zx2 − zy2 n) (m + 2)2 − 4
9a2
81p2
o) 1 − 121q2 p) m2 + n2 − 2mn − 9 q) a2 + 6ab − 9 + 9b2

2. Find the value of the following by using a2 − b2 = (a + b) (a − b):

a) 35 × 35 − 21 × 21 b) 189 × 189 − 11 × 11 c) 105 × 95

d) 139 × 121 e) 207 × 193 f) 219 × 219 − 181 × 181
219 − 181

g) 352 × 352 − 248 × 248 x cm
352 − 248

3. Find the area of the shaded portion in the adjoining figure on x cm
the right side. If x = 8cm, find the actual area of the shaded
portion. 2cm
2cm
262 12 Algebraic Expressions

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4. Find the area of the shaded portion in the adjoining

figure. If the area of a circle is given by πr2, where

π= 22 and r = 1.4 cm, R = 7 cm, find the actual area of R
7
the shaded portion.

5. A circular well of radius 0.35 meter is situated at the
middle of a square field having the length of a side x meter. Find the area of the field
without the well.

6. a) Show that a2 - 4 = (a + 2) (a - 2) in a diagram.
b) Show that (a + b)2 = a2 + 2ab + b2 in a diagram.
c) Show that (a - b)2 = a2 - 2ab + b2 in a diagram.
d) Show that a2 - 9 = (a + 3) (a - 3) in a diagram.
e) Show that x2 + 3x + 2 = (x + 1)(x + 2) in a diagram.

Type - IV Factorization of the trinomial expressions of the perfect square:

What are the expanded form of (a + b)2 and (a − b)2 ?
We know that,

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

We can apply above mathematical expression for factorization of the trinomial
expressions of the perfect square.

Example 6: What should be filled in the blanks to make the expressions a perfect square ?

a) x2+ ------------ + 49 b) y2 − -------------- +36

Solution: a) Here,
x2 + -------- + 49 = (x)2 + -------- + (7)2
Now, comparing it with a2 + 2ab + b2 , we get

a = x and b = 7

Therefore, 2ab = 2 × x × 7 = 14x,

So, x2 + 14x + 49 = (x + 7)2

∴ 14x should be filled in the black of x2 +-------- + 49
to make the expression a perfect square.

b) Here,

y2 − -------------- + 36 = (y)2 − -------- + (6)2

Now, comparing it with a2 − 2ab + b2 , we get

a = y and b = 6

Therefore, 2ab = 2 × y × 6 = 12y,

So, y2 − 12y + 36 = (y − 6)2

∴ 12y should be filled in the black of y2 − -------- + 36
to make the expression a perfect square.

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Example 7: Factorize: x2 + 18x + 81
Solution: Here,
x2 + 18x + 81
= (x)2 + 2 × x × 9 + (9)2 = (x + 9)2 = (x + 9)(x + 9)

Example 8: Resolve into Factors: x4 + y4 + x2y2
Solution: Here,
x4 + y4 + x2y2

= (x2)2 + 2x2y2 + (y2)2 − x2y2 [ 2 × x2 × y2 = 2x2y2 is needed to make a perfect ∴
square. So, we write x2y2 = 2x2y2 − x2y2]

= (x2 + y2)2 − (xy)2

= (x2 + y2 + xy)(x2 + y2 − xy)

= (x2 + xy + y2) (x2 − xy + y2)

Exercise 1.1(d)

1. Fill the suitable term in the blanks to make the expression a perfect square:

a) x2 +---------- + 25 b) x2+----------+ 16 c) 4m2 +----------+ y2

d) p2 − ---------- + 9 e) 9x2−----------+ 25y2 f) p2 + ---------- + 4
g) m2 + 8mn + --------- h) 225m2 − ---------- + 64y2 p2

i) 1 − ---------- + 36y2

2. Resolve into factors: b) a2 − 6a + 9 c) p2 − 4p + 4
a) x2 + 12x + 36
d) y2 − 14xy + 49x2 e) 49p2 + 14p + 1 f) 3x2 − 6x + 3
g) 49q2 − 70q + 25
j) 4a2 − 20a + 25 h) a2 + 18ax + 81x2 i) 9a2 + 42ab + 49b2

k) 25y2 − 60y + 36

3. Resolve into factors:

a) a4 + a2b2 + b4 b) a4 + 1+ a2 c) p2 + 6pq + 9q2 − 25 d) x2 − y2 + 2yz − z2

e) 9x2 + 2ab + 4y2 − a2 − b2 −12xy f) a2 - 2a + 1 − b2 g) 4x2 + 4xy + y2 − m2

h) (1 - x2) (1 - y2) + 4xy

264 12 Algebraic Expressions

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Type - V Factorization of trinomial expression of the form:

x2 + bx + c and x2 − bx + c

Let us suppose that b = p + q and c = p × q = pq

Then,

x2 + bx + c = x2 + (p + q)x + pq

= x2 + px + qx + pq

= x(x + p) + q(x + p)

= (x + p)(x + q)

Similarly,

x2−bx+c = x2-(p + q)x + pq

= x2-px - qx + pq

= x(x − p) − q(x − p)

= (x − p)(x − q)

From the above,(x + p) and (x + q) are the factors of x2 + bx + c when b = p + q and
c = p × q, where p and q are two numbers. In such trinomial expression, the coefficient
of x is the sum of two numbers and the constant term is the product of those two
numbers.

From the above discussion, it is clear that when we factorize a trinomial expression,
first we split the middle term into two parts in such a way that the sum of there two
parts is equal to the middle term and their product is equal to the product of 1st term
and 3rd term. Then we factorize it by the suitable grouping method.

For example, let us take a trinomial expression a2 + 7a + 12. Compare a2 + 7a + 12 with
x2 + bx + c Then, b = 7 and c = 12. We want p and q Such that b = p + q and c = p × q = pq.

∴ p + q = 7 and p × q = 12

We want two numbers whose sum is 7 and their product is 12. For this we first find the
factors of 12. The factors of 12 are 1, 2, 3, 4, 6 and 12. We choose two factors whose product
is 12 and sum is 7. The possible numbers are 3 and 4 because 3 + 4 = 7 and 3 × 4 = 12.

Now,

a2 + 7a + 12 = a2 + (3 + 4)a + 12

= a2 + 3a + 4a +12

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

= (a + 3)(a + 4)

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Example 9: Find two numbers whose sum is 21 and the product is 80.
Solution: First we have to take a number 80 as a product of two numbers.
So, 80= 1 × 80, 80 = 2 × 40, 80 = 4 × 20, 80 = 5 × 16, 80 = 8 × 10, Since the sum of

the two required numbers is 21, we choose that pair of factors whose sum is 21.
So, 21 = 5 + 16.
∴ The required numbers are 5 and 16.

Example 10: Factorise the following expressions: a) x2 +11x + 24 b) m2 − 13m + 40

Solution: a) x2 +11x + 24

Here,

x2 + 11x + 24

First we have to find two numbers whose sum is 11 and the product is 24.

So, 24 = 1 × 24,

24 = 2 × 12, 24 = 3 × 8, 24 = 4 × 6. But 3 + 8 = 11,

∴ Two numbers are 3 and 8.

Now,

x2 + 11x + 24 = x2 + (3 + 8)x + 24 = x2 + 3x + 8x + 24 = x(x + 3) + 8(x + 3)

= (x + 3)(x + 8)

b) m2 − 13m + 40

Here,

m2 − 13m + 40

First we have to find two numbers whose sum is 13 and the product is 40.

So, 40 = 1 × 40, 40 = 2 × 20, 40 = 4 × 10, 40 = 5 × 8. But 5 + 8 = 13,

∴ Two required numbers are 5 and 8.

Now,

m2 − 13m + 40 = m2 − (5 + 8)x + 40 = m2 - 5m - 8m + 40 = m(m − 5) − 8(m − 5)

= (m − 5)(m − 8)

Example 11: Factorise : (m+n)2−10(m+n)+21

Solution: Here,
(m + n)2 − 10(m + n) + 21
Put m + n = x,then the given expression is x2 − 10x + 21

Now, x2 − 10x + 21 = x2 − (7 + 3)x + 21 [∴ 7 + 3 = 10 and 7 × 3 = 21 ]

= x2 - 7x - 3x + 21

= x(x − 7) − 3(x − 7)

= (x − 7)(x − 3)

Substituting the value of x, we get

(m + n)2 − 10(m + n) + 21 = (m + n − 7)(m + n - 3).

266 12 Algebraic Expressions

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Exercise 1.1(e)

1. Find two numbers whose product is p and the sum is s:

a) p = 6, s = 5 b) p = 72, s = 17 c) p = 48, s = 19
f) p = 154, s = 25
d) p = 98, s = 21 e) p = 140, s = 27

2. Resolve into factors:

a) x2 + 5x + 6 b) x2 + 14x + 45 c) x2 − 17x + 72 d) m2 + 12m + 32
g) x3 + 12x2 + 11x h) a4 + 12a3 + 32a2
e) a2 + 14a + 48 f) p2 + 17p + 66 k) a2 + 22ab + 105b2 l) x2 − 13xy + 40y2
o) m2 − 16m + 39 p) x2 y2 + 8 - 6xy
i) x2 − 6x + 8 j) m2 + 4m + 3
m) x2 − 10x - 39 n) y2 + 12y + 35

3. Resolve into factors:

a) x2 − 26x + 168 b) (a + b)2 − 21(a + b) + 98 c) (p − q)2 − 18(p − q) + 45

d) (m + n)2 + 23(m + n) + 76 e) x2y2 − 12xyz + 27z2

Type-VI Factorization of Trinomial Expression of the form:

x2 + px − q and x2 − px − q

Let us suppose that, p = a − b and q = a × b = ab

Then,

x2 + px − q = x2 + (a − b)x − ab

= x2 + ax - bx - ab

= x(x + a) − b(x + a)

= (x + a)(x − b)

Similarly,

x2 − px − q = x2 − (a − b)x − ab

= x2 − ax + bx − ab

= x(x − a) + b(x − a)

= (x − a) (x + b)

From the above discussion, it is clear that when the constant −q is negative, then
the coefficient of x (i.e. middle term) is the difference of two numbers.

So, for factorization of trinomial expression as like above, we split the middle term
into two parts as the difference and then apply the suitable grouping method.

For example, let us take a trinomial expression x2 + 3x − 28. Compare x2 + 3x − 28 with
x2 + px − q. Then, p = 3 and q = 28. We want a and b.

Algebraic Expressions 267

Prime Mathematics Book - 8

Such that p = a − b and q = ab

∴ a − b = 3 and ab = 28

We want two numbers whose difference is 3 and their product is 28. For this we find
the factors of 28. The factors of 28 are 1, 2, 4, 7, 14, 28. We choose two factors whose
product is 28 and difference is 3. The possible numbers are 7 and 4 because 7 − 4 = 3 and
7 × 4 = 28.

Now,
x2 + 3x - 28 = x2 - (7 - 4)x - 28
= x2 - 7x + 4x - 28
= x(x - 7) + 4(x - 1)
= (x - 7)(x + 4)

Example 12: Find two numbers whose product is 91 and difference is 6.
Solution: First we have to take a number 91 as a product of two numbers
So, 91 = 1 × 91, 91 = 7 × 13
Since the difference of two numbers is 6, we choose that pair of factors whose

difference is 6. So, 13 - 7 = 6

∴ The required two numbers are 13 and 7.

Example 13: Factorize the following expressions: a) x2 + 4x − 45 b) y2 − 7yz − 60z2

Solution: a) x2+4x−45

Here,

x2 + 4x − 45

First we have to find two numbers whose product is 45 and difference is 4.

So, 45 = 1 × 45, 45 = 3 × 15, 45 = 5 × 9, 4 ≠ 45 − 1, 4 ≠ 15 − 3, 4 = 9 − 5

∴ The required two numbers are 9 and 5.

Now,

x2 + 4x − 45 = x2+(9 − 5)x − 45 = x2 + 9x − 5x − 45 = x(x + 9) − 5(x + 9)

= (x + 9)(x − 5)

b) y2 − 7yz − 60z2
Here,
y2 − 7yz − 60z2
By the same above process,
Now,
60 = 12 × 5 and 7 = 12 − 5
y2 − 7yz − 60z2 = y2− (12 − 5)yz − 60z2 = y2 − 12yz + 5yz − 60z2
= y(y − 12z) + 5z(y − 12z)
= (y − 12z)(y + 5z).

268 12 Algebraic Expressions

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Exercise 1.1(f)

1. Find two numbers whose product is p and the difference is d:

a) p = 5, d = 4 b) p = 30, d = 7 c) p = 12, d = 4
d) p = 105, d = 8 e) p = 78, d = 7 f) p =112, d = 9

2. Resolve into factors: b) x2 − 11x − 26 c) y2 + 2y − 15
a) x2 + 7x − 78 e) 18 − b2 − 7b f) m2 + 8m − 105
d) a2 − 13a − 48 h) x2 + x − 56 i) x2 − 5x − 36
g) p2 − 60 − 7p

3. Resolve into factors: b) 4x3 − 8x2 − 12x c) p2 − 8p − 33
e) (a + b)2 − 4(a + b) − 21 f) (p − 2q)2 + 3(p − 2q) − 70
a) x3 − 7x2 − 8x
d) x2 y2 − 4xy − 117
g) m3 − 5m2n − 36mn2

Type - VII Factorization of Trinomial Expression of the form:

ax2 + bx + c and ax2 − bx + c

To factorize the trinomial expressions like ax2 + bx + c or ax2 − bx + c, first we find
two numbers whose product is equal to ac and their sum is equal to b. Then we proceed
the same as in type V. Study the following examples carefully.

Example 14: Factorize: 3x2 + 22x + 7.

Solution: The given expression is 3x2 + 22x + 7

Here, 3 × 7 = 21. So, we have to find two numbers whose product is 21 and the
sum is 22. Such two numbers are 21 and 1.

21 × 1 = 21 and 21 + 1 = 22

Now,

3x2 + 22x + 7 = 3x2 + (21 + 1)x + 7

= 3x2 + 21x + 1x + 7

= 3x(x + 7) + 1(x + 7)

= (x + 7)(3x + 1)

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Example 15: Factorize: 12a2 − 32a + 5

Solution: Here,

12a2 − 32a + 5

12 × 5 = 60

So, we have to find two numbers whose product is 60 and the sum is 32. That
two numbers are 30 and 2,because 30 × 2 = 60 and 30 + 2 = 32.

Now,

12a2 − 32a + 5 = 12a2 − (30 + 2)a + 5

= 12a2 − 30a − 2a + 5

= 6a(2a − 5) − 1(2a − 5)

= (2a − 5)(6a − 1)

Example 16: Factorize: 6(a + 2b)2 + 17(a + 2b) + 7

Solution: The given expression is 6(a + 2b)2 + 17(a + 2b) + 7.

If we put a+2b=x,then the expression is 6x2 + 17x + 7.

Now,

6x2 + 17x + 7 = 6x2 + (14 + 3)x + 7 ∴[ 6 × 7 = 42 and sum = 17

∴ 14 + 3 = 17, 14 × 3 = 42]
= 6x2 + 14x + 3x + 7

= 2x(3x + 7) + 1(3x + 7)

= (3x + 7)(2x + 1)

Replacing the value of x, we get;

{3(a + 2b) + 7} {2(a + 2b) + 1}

= (3a + 6b + 7) (2a + 4b + 1)

Exercise 1.1(g)

1. Resolve into factors: b) 3x2+5x+2 c) 3m2 + 10m+ 8 d) 7p2−30p+8
a) 2x2−9x+9 f) 6y2−17y+7 g) 2p2−23p+56 h) 15p2−13p+2

e) 3a2−7a+4

i) 6x2+17x+12

2. Resolve into factors: b) 4m2 − 8mn + 3n2 c) 21x2 + 25xy + 4y2
a) 3a2 + 10b2 − 13ab

d) 18m2 − 51m2n + 21mn2 e) 2(a − b)2 − 21(a − b) + 49 f) 6(2a + b)2 + 19(2a + b) +15

g) 7(m − 2n)2 − 30(m − 2n) + 8 h) 5(2x − y)2 − 14(2x − y) + 8

270 12 Algebraic Expressions

Prime Mathematics Book - 8

Type VIII Factorization of Trinomial Expression of the form:

ax2 + bx − c and ax2 − bx − c.

To factorize the trinomial expressions like ax2 + bx − c or ax2 − bx − c, first we find
two numbers whose product is equal to ac but negative and their difference which is
equal to b. Then we split the middle term bx into two parts to get those numbers and
proceed the same as in Type V, Study the following examples carefully .

Example 17: Factorize: 12x2 + 28x − 5

Solution: The given expression is 12x2 + 28x − 5

Here, 12 × -5= - 60.

So, we have to find two numbers whose product is 60 and difference is 28.
That two numbers are 30 and 2 because 30 × 2 = 60 and 30 − 2 = 28.

Now,

12x2 + 28x − 5 = 12x2 + (30 − 2)x − 5

= 12x2 + 30x − 2x − 5 = 6x(2x + 5) − 1(2x + 5)

= (2x + 5)(6x − 1)

Example 18: Factorize: 3m2 − 11m − 20

Solution: The given expression is 3m2 − 11m − 20

Here, 3 × − 20 = −60

So, we have to find two numbers whose product is 60 and difference is 11.
That two numbers are 15 and 4,because 15×4=60 and 15−4=11.

Now,

3m2 − 11m − 20 = 3m2 − (15 − 4)m − 20

= 3m2 − 15m + 4m − 20 = 3m(m − 5) + 4(m − 5)

= (m − 5)(3m + 4)

Example 19: Factorize: 3(a − 2b)2 − 13(a − 2b) − 30

Solution: The given expression is 3(a − 2b)2 − 13(a − 2b) − 30

If we put a − 2b=x,then the expression is 3x2 − 13x − 30.

Now,

3x2 − 13x − 30 =3x2 − (18 − 5)x − 30 [ 3 × − 30 = − 90∴
∴18 × − 5 = − 90 and 18 − 5 = 13]

= 3x2 − 18x + 5x − 30

= 3x(x − 6) + 5(x − 6)

= (x − 6)(3x + 5)

Replacing the value of x, we get;

{(a − 2b) − 6} {3(a − 2b) + 5} = (a − 2b − 6) (3a − 6b + 5).

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Exercise 1.1(h)

1. Resolve into factors:

a) 2x2 + 5x − 7 b) 3m2 + 6m − 9 c) 5x2 − 14x − 3 d) 10p2 − 3p − 1
g) 3x2 − 13x − 30 h) 4y2 + 13yz − 12z2
e) 6a2 − 4ba − 10b2 f) 3m2 − mn − 10n2

2. Factorize: b) 12b2 − b − 1 c) 10x2 + 19xy − 15y2
a) 3a2 + 13a − 16
d) (x + 1)2 − 6(x + 1) − 16 e) 28 + 27(y − 1) − (y − 1)2 f) 3(p + 2q)2 − (p + 2q) − 10
g) 2(x − 2y)2 + 5(x − 2y) − 12
h) 5(2x + 3y)2 − 9(2x + 3y) − 14

23.4 Geometrical Concept of (a ± b)3

a. Geometrical Concept of (a+b)3

a2b a3

a ab2 a
b b3
ab2 b a2b a
a
a unseen piece
a2b

a b ab2

→ Take a wooden cube of side a + b and cut into pieces as shown.
→ Separate the pieces. You will get following pieces.

b3 b a2b a2b a2b
b b

b a a a
bb a a
a
a3

a ab2 ab2 ab2
bbb

ab b b
aa a a

Find the volume of each pieces separately. Out of that 8 pieces, 2 pieces are cube and
other 6 pieces are cuboid.

272 12 Algebraic Expressions

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Now,

Volume of the whole cube = sum of the volume of all 8 pieces

or, (a + b)3 = a3 + b3 + a2b + a2b + a2b + ab2 + ab2 + ab2 [ volume of cube = l 3 ∴

volume of cuborid = l × b × h]

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

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

∴ (a + b)3 = a3 + 3a2b + 3ab2 + b3

= a3 + 3ab(a + b) + b3

b. Concept of (a+b)3 by Algebraic multiplication:

Let (a + b) be a binomial.
The cube of (a + b) is (a + b)3.

Now, Again,
(a + b)3 = a3 + 3ab(a + b) + b3
(a + b)3 = (a + b) (a + b) (a + b) or, (a + b)3 − 3ab(a + b) = a3 + b3
or, (a + b) [(a + b)2 − 3ab] = a3 + b3
= (a2 + ab + ab + b2)(a + b) or, (a + b) (a2 + 2ab + b2 − 3ab) = a3 + b3
or, (a + b) (a2 − ab + b2) = a3 + b3
= (a2 + 2ab + b2)(a + b) ∴ a3 + b3 = (a + b)3 − 3ab(a + b)
= (a + b) (a2 − ab + b2)
= a2(a + b) + 2ab(a + b) + b2 (a + b)

= a3 + a2b + 2a2b + 2ab2 + ab2 + b3

= a3 + 3a2b + 3ab2 + b3

∴(a+b)3 = a3 + 3a2b + 3ab2 + b3

= a3 + 3ab(a + b) + b3

c. Geometrical Concept of (a − b)3 :

(a - b)2b

(a - b)b2 (a - b) a (a - b)b2 (a - b)3
(a - b) b a-b
a b b3
(a - b)2b (a - b)2b
(a - b)
a b (a - b)b2 a-b a-b

unseen piece

→ Take a wooden cube of side a units.
→ Reduce each side by b units and cut as shown.
→ Separate the pieces. You will get following pieces.

Algebraic Expressions 273

Prime Mathematics Book - 8 (a - b)2b (a - b)2b (a - b)2b
b3
b bb
b
b (a - b) (a - b)
(a - b)
b
(a - b)3(a - b) (a - b) (a - b) (a - b)

(a - b)b2 (a - b)b2 (a - b)b2
b b b
(a - b) b b b
(a - b)
(a - b) (a - b) (a - b)

Out of these 8 pieces, 2 pieces are cube and other 6 pieces are cuboids.
Now,
Volume of the whole cube = sum of the volume of all 8 pieces
or, a3 = (a−b)3 + b3 + (a −b)2. b + (a−b)2 . b + (a−b)2. b + (a−b). b2 + (a−b). b2 + (a−b). b2
or, a3 = (a − b) 3 +b3 + 3(a − b)2 . b + 3(a − b). b2
or, a3 = (a − b)3 + b3 + 3(a2 − 2ab + b2)b + 3(a − b). b2
or, a3 = (a − b)3 + b3 + 3a2b − 6ab2 + 3b3 + 3ab2 −3b3
or, a3 = (a − b)3 + b3 + 3a2b − 3ab2
or, (a − b)3 = a3 − 3a2b + 3ab2 − b3
∴ (a − b)3 = a3 − 3a2b + 3ab2 − b3 = a3 − 3ab(a − b) − b3.

d. Concept of (a − b)3 by Algebraic multiplication:

Let (a − b) be a binomial.

The cube of (a − b) is (a − b)3.

Now, Again,

(a − b)3 = (a − b) (a − b) (a − b) (a − b)3 = a3 − 3ab(a − b) − b3

= (a2 − ab − ab + b2) (a − b) or, (a − b)3 + 3ab(a − b) = a3 − b3

= (a2 − 2ab + b2) (a − b) or, (a − b) [(a − b)2 + 3ab] = a3 − b3

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

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

= a3 − 3a2b + 3ab2 − b3 ∴ a3 − b3 = (a − b)3 + 3ab(a − b)

∴ (a − b)3 = a3 − 3a2b + 3ab2 − b3 = (a − b) (a2 + ab + b2)

= a3 − 3ab(a − b) − b3

274 12 Algebraic Expressions

Note: Requir e d formulae:- Prime Mathematics Book - 8
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)3 = a3 + 3ab(a + b) + b3 a3 + b3 = (a + b)3 − 3ab(a + b)
(a − b)3 = a3 − 3a2b + 3ab2 − b3 a3 + b3 = (a + b) (a2 − ab + b2)
(a − b)3 = a3 − 3ab(a − b) − b3 a3 − b3 = (a − b)3 + 3ab(a − b)
a3 − b3 = (a − b) (a2 + ab + b2)

Example 20: Find the cube of the following by using formula:

a) 2x + y b) x − 3 c) 2x − y + z

Solution: a) Here, cube of 2x + y = (2x + y)3

= (2x)3 + 3(2x)2y + 3(2x)y2 + y3

= 8x3 + 3 × 4x2y + 6xy2 + y3

= 8x3 + 12x2y + 6xy2 + y3

b) Here, cube of x − 3 = (x − 3)3

= (x)3 − 3(x)2(3) + 3(x)(3)2 − (3)3

= x3 − 9x2 + 27x − 27

c) Here, cube of 2x − y + z = (2x − y + z)3 = [(2x − y) + z]3

put 2x−y = a and z = b, then

(a+b)3 = a3`+ 3a2b + 3ab2 + b3, we get

(2x − y + z)3 = (2x − y)3 + 3(2x − y)2z + 3(2x − y)z2 + z3

= (2x)3−3(2x)2y + 3(2x)y2 − y3 + 3[(2x)2 − 2.2x.y + y2]z
+ 6xz2 − 3yz2 + z3

=8x3−12x2y+6xy2−y3+12x2z−12xyz+3y2z+6xz2−3yz2+z3

Example 21: If a + b = 5 and ab = 3, find the value of a3 + b3.

Solution: Here, a + b = 5 and ab = 3
Now,
a3 + b3 = (a + b)3 − 3ab(a + b) = (5)3 − 3 × 3(5) = 125 − 45 = 80.

Example 22: If x− 1 = 4, find the value of x3 − 1
x x3

Solution: Here, x− 1 = 4 and x3 − 1 =?
x x3

( ) ( ) ( ) 1 1 3 − 1 1 1
Now, x3 − x3 = (x)3 − x x x 3+ 3 . x. x x − x
=

= (4)3 + 3(4) = 64 + 12 = 76

Algebraic Expressions 275

Prime Mathematics Book - 8

Example 23: Simplify: (2x + y)3 − (2x − y)3
Solution: Here,
(2x + y)3 − (2x − y)3
= [(2x)3 + 3(2x)2y + 3(2x)y2 + y3] − [(2x)3 − 3(2x)2y + 3(2x)y2 − y3]
= 8x3 + 12x2y + 6xy2 + y3 − 8x3 + 12x2y − 6xy2 + y3
= 12x2y + 12x2y + y3 + y3
= 24x2y + 2y3

Exercise 1.1(i)

1. Find the cube of: b) m + 2 c) a − 5 d) 2x − 3
a) x + 3
e) 4 - 3a f) 2x + 3y g) 3m − 4n h) 1 + 3y

i) 3x2 + y2 j) m − 1 k) x + 1 l) q − 1
m) x + 2y + a m x 2q

n) 3x − 2y − z

2. Expand the following cube:

a) (3x + 2y)3 b) (x2 − y)3 c) (4a − b)3
d) (2a + 5)3
3a2( )e)− 1 2 ( )-f) m n3
3a2 n m

3. Write the following expressions in the form of (a + b)3 or(a − b)3:

a) 27a3 + 27a2b + 18ab2 + 8b3 b) 8m3 − 18m2n + 27mn2 − 27n3

c) 64x3 + 240x2y + 300xy2 + 125y3

4. Find the value of x3 + y3 if, b) x + y = 5 and xy = 3
a) x + y = 6 and xy = 4

5. Find the value of x3 − y3 if b) x − y = 5 and xy = 14
a) x − y = 4 and xy = 6

6. Find the value of x3 −a 3 if x − a = 6 and xa = 10.

7. Factorize: b) 8x3 + 27y3 c) 64p3 − 216 d) 27m3 + 125n3
a) x3 − 27

8. Find t he value of:

a) x3 + 1 1 b) y3 1 1
x3 if x + x = 7 y3 if y − y = 9

11 d) n3− 1 1
c) m3 + m3 if m + m = 12 n3 if n − n = x

276 12 Algebraic Expressions

9. Find the value of: Prime Mathematics Book - 8
a) b3 + c3 + 18bc if b + c = 6
c) p3 + q3 − 9pq if p + q = 3 b) m3 − 8n3 − 24mn if m − 2n = 4
d) x3 − y3 − 18xy if x − y = 6

10. Simplify: b) (x + y)3 − (x − y)3 c) (x + z)3 − (x − z)3
a) y3 + z3 − (y + z)3 e) (2x − 3y)3 + (x − y)3 f) (x + y)3 − 3xy(x + y)
d) (x − y)3 + 3xy(x + y)
(7.1)3−(3.1)3
g) (4.9)3+(2.1)3 h) (7.1)2+7.1×3.1+(3.1)2
(4.9)2−4.9×2.1+(2.1)2

1.2 Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of
Algebraic Expressions.

1.2(a) Highest Common Factor (HCF): Let us take two numbers 12 and 18.
Write the factors of 12 and 18.
Factors of 12 = 1, 2, 3, 4, 6, 12 Factors of 18 = 1, 2, 3, 6, 9, 18
What are the common factors of 12 and 18 ?

The common factors of 12 and 18 are 1, 2, 3, 6. Among them the highest common
factor is 6.

∴ HCF of 12 and 18 is 6.

Similarly, let us take the algebraic expressions 3x3y2 and 6x2y.
The factors of 3x3y2 = 3, x, x2, y, y2, xy, x2y, xy2, x3y2
The factors of 6x2y = 2, 3, x, x2, y, xy, x2y
The common factors of both expressions are 3, x, x2, y, xy and x2y. Out of

these common factors, the highest factor is 3x2y. It divides both expressions
exactly. Therefore, 3x2y is HCF of the given expressions. The highest factor
which divides all the given expression exactly is known as the highest common
factor (HCF).

In case of polynomials, we first factorize the given expressions into factors
and find their HCF in the same process as above.


Note: If nothing is common to all the given expressions, then 1 is common to the
given expressions. In such case, HCF of the given expression is 1.

Example 24: Find the HCF of ab2c, a2bc3 and 2a3b2c2

Solution: Here,

1st expression = ab2c = a × b × b × c

2nd expression = a2bc3 = a × a × b × c × c × c

3rd expression = 2a3b2c2 = 2× a × a × a × b × b × c ×c

∴ HCF = a × b × c = abc

Algebraic Expressions 277

Prime Mathematics Book - 8

Example 25: Find the HCF of x2 + xy, x2 − y2 and xy + y2.
Solution: Here,
1st expression = x2 + xy = x(x + y)
2nd expression = x2 − y2 = (x + y)(x − y)

3rd expression = xy + y2 = y(x + y)

In the above three expressions, (x+y) is only common factor.
Therefore, HCF of the given expression is (x+y).

Example 26: Find the HCF of m3 − mn2 and x2y − xy2.
Solution: Here,
1st expression = m3 − mn2 = m(m2 − n2) = m(m + n)(m − n)

2nd expression = x2y− x y2 = xy(x − y)

None of the factors are common to the above two expressions.

∴ HCF of the given expression = 1.

Example 27: Find the HCF of x2 + 5x + 6, x2 − 9 and x2 + x − 6.

Solution: Here,
1st expression = x2 + 5x + 6

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

= (x + 3)(x + 2)

2nd expression = x2 − 9 = (x)2 − (3)2 = (x + 3)(x − 3)
3rd expression = x2 + x − 6

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

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

∴ HCF = common factors of all three expressions = (x + 3)

Exercise 1.2(a)

1. Find the HCF of:

a) a2b and ab3 b) 4a2b and ab3 c) 9x2y3 and 15xy2

d) ab2c, a2bc3 and a2bc2 e) 6a2x, 8ax2y and 12x3y2a2

f) 4m2n2p, 12m3np2 and 16m2n4p5 g) 2a − 4 and a − 2 h) x2 − 4 and 3x + 6

i) p2q − pq2 and 2p2 − 2pq j) a2 − b2 and ab − b2 k) x2 + 2xy + y2 and x2 − y2

l) x2 − 25 and x2 + 4x − 5 m) x3 − x and x2 − x + 2 n) x2 + 5x + 6 and x2 + x − 6

o) m2 − m2n2 and n2 − n4

278 12 Algebraic Expressions

Prime Mathematics Book - 8

2. Find the HCF of:

a) x2 − 11x + 30 and x2 − 36 b) x3 − 9x and x2 + x − 6
c) x2 − y2, x2 − xy and x2y − xy2 d) a3 − ab2, a2+ab and a2b + ab2
e) a2 + 2a − 3, a2 − 1 and a2 − 3a + 2 f) x2 − 6x + 8, x2 + 4x + 4 and x2 − 4
g) 2x2 − 6x, x2 − x − 6 and x3−9x h) x2 − 4, x2 − 2x and x2 − 7x + 10
i) m3 − 3m + 2, 3m2 − 2m − 8 and 2m2 − 9m + 10 j) y3 − 5y + 6, y2 − y − 2 and y2 + 5y + 4

k) b3 + 6b2 − 4b − 24, b2 + 5b + 6 and b2 − 4 l) 3x2 + 5x − 12, 2x2+7x+3 and x2 + 4x + 3

1.2(b) Lowest Common Multiple (LCM):

Let us take two numbers 10 and 14. Write the multiples of these two numbers.
Multiples of 10 (M10 ) = {10, 20, 30, 40, 50, 60, 70, 80, ............}
Multiples of 14 (M14 ) = {14, 28, 42, 56, 70, 84, ...............}

What are the common multiples of 10 and 14 ?
The common multiple of 10 and 14 is 70. 70 is lowest common multiple of 10 and 14.
The common multiples of 10 and 14 are {70, 140, 210, 280, 350, ........}.
The lowest common multiple is 70. Therefore, the LCM of 10 and 14 is 70.

Similarly, let us take the algebraic expressions x2y, xy2 and x3y4.
Multiple of x2y = x2y, x3y, x2y2, x3y2, x3y3, x3y4, ...........
Multiple of xy2 = xy2, x2y2, x3y2, x4y2, xy3, xy4, x2y3, x3y4, ...........
Multiple of x3y4 = x3y4, x4y4, x3y5, x4y5, .................

Here, we see that a common multiple is x3y4. No multiple smaller than this is common
to all. Therefore, x3y4 is the LCM of the given expressions. The lowest multiple of all
the given algebraic expressions which is exactly divisible by all the given expressions is
known as the lowest common multiple of the given algebraic expressions. It is written
as LCM in short form.

The lowest common multiple of the given algebraic expressions is also the product of
common factors and remaining non common factors of the given expressions.

In case of polynomials, we first factorize the given expression into factors and find their
LCM in the same process as above. Study the following example.

Let us take the algebraic expressions: x2 − x, x2 − 1 and x3 − 1. First we find the factors of them.
1st expression = x2−x = x(x − 1)
2nd expression = x2−1 = (x)2 − (1)2 = (x + 1)(x − 1)
3rd expression = x3−1 = (x)3 − (1)3 = (x − 1)(x2 + x × 1 + 12) = (x − 1)(x2 + x + 1)

The common factor to all three expressions is (x − 1). There is no factor common to
any two given expressions. Therefore, the product of common factor and non-common
factors is x (x − 1)(x + 1)(x2 + x + 1). These product is exactly divisible by all three
expressions. Therefore, x (x − 1)(x + 1)(x2 + x + 1) is the LCM of the given expressions.

∴ LCM of polynomials = common factors × non-common factors.

Note: In case of three or more expressions, the terms even common in two

expressions are taken as the common factors.

Algebraic Expressions 279

Prime Mathematics Book - 8

Example 28: Find the LCM of 3xy, 6xy2 and 12x2y2.

Solutions: Here, = 3× x × y
1st expression 3xy

2nd expression 6xy2 =2×3×x×y×y

3rd expression 12x2y2 =2×2×3×x×x×y×y

Common factors in all three expressions = 3 × x × y

Common factors in 2nd and 3rd expressions = 2× y

Factors left in 3rd expressions = 2× x

∴ LCM = 3 × x × y × 2× y × 2× x = 12x2y2

Example 29: Find the LCM of x3 − 4x and x2 + 7x + 10.

Solution: Here, = x(x2 − 4)
1st expression = x3 − 4x

= x{(x)2 − (2)2} = x(x + 2)(x − 2)
2nd expression = x2 + 7x + 10 = x2 + (5 + 2)x + 10

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

= (x + 2)(x + 5)

Common factor in both expression is = (x + 2)

Factors left in 1st expression is = x(x − 2)

Factors left in 2nd expression is = (x + 5)

∴ LCM = (x + 2) × x(x − 2) × (x + 5) = x(x2 − 4)(x + 5).

Example 30: Find the LCM of 6x3 + 5x2 − 6x, 2x4 + x3 − 3x2 and 3x3 − 5x2 + 2x.

Solution: Here, 1st expression = 6x3 + 5x2 − 6x

= x(6x2 + 5x − 6)

= x{6x2 + (9 − 4)x − 6}

= x(6 x2 + 9x − 4x −6)

= x{3x (2x + 3)−2(2x + 3)}

= x(2x + 3)(3x − 2)

2nd expression = 2x4 + x3 − 3x2

= x2(2x2 + x − 3)

= x2{2x2 + (3 − 2)x − 3}

= x2(2x2 + 3x − 2x − 3)

= x2{x(2x + 3) − 1(2x + 3)}

= x × x (2x + 3)(x − 1)

3rd expression = 3x3 − 5x2 + 2x

= x(3x2 − 5x + 2)

= x{3x2 − (3 + 2)x + 2}

= x(3x2 − 3x − 2x + 2)

= x{3x(x − 1) − 2(x − 1)}

= x(x − 1) (3x − 2)

280 12 Algebraic Expressions

Prime Mathematics Book - 8

Common factor in all three expression is = x
Common factor in 1st and 2nd expression is = (2x + 3)
Common factor in 1st and 3rd expression is = (3x − 2)
Common factor in 2nd and 3rd expression is = (x − 1)
Factors left in 2nd expression is =x

∴ LCM = Common factors in all three expression × Common factors in any
two expression × Remaining factors in any expression = x × (2x + 3)
× (3x − 2) × (x − 1) × x

= x2(2x + 3)(3x − 2)(x − 1).

Exercise 1.2(b)

1. Find the LCM of: b) 4a and 6ab c) x3y2 and 3xy3 d) 3mn and 6mn2
a) 3x and 6x2y f) 2x and 2x−4 g) 5xy and 10y2 h) 8x2y and 10x3y2
e) 6p2q and 6pq2

2. Find the LCM of:

a) 3x2 - 3 and x2 - 1 b) 4b2 - 9 and 4b + 6 c) x + y and x2 + xy

d) 2(a2 - 1) and a2 - 2ab + 1 e) 5x + 20 and x2 - 16 f) x2 - xy and xy - y2

g) x2 -4 and x2 - 4x + 3 h) 3y2 + 7y + 2 and 2y2 + 3y - 2 i) m2 - 4mn + 4n2 and m2 - 4n2

3. Find the LCM of:

a) 4a2b, 6ab and 8ab2 b) 3xy2, 10x2y2 and 15x2y

c) x2 - 2x, x - 2 and x + 2 d) a2 - b2, a2 + 2ab + b2 and a2 + b2

e) x2 - xy, x2 - y2 and xy - y2 f) x2 + x, x2 - 1 and 2x + 2

g) m2 - 1, m2 + m - 2 and m2 - 2m + 1 h) 4 - x2, 4 + 2x and 2x + x2

i) a2 - 4, a2 + 4a + 4 and a2 + 3a + 2 j) x2 - 10x + 25, x2 - x - 20 and x2 - 25

k) 2x2-8, 4x+8 and x2-4x-5 l) x3 - x2 - 42x, x2 - 5x - 14 and x4 + 4x2 - 12x2

m) 3m2 - 7m - 20, 6m2 + 13m + 5 and 2m2 - 9m + 4

n) a4 + a2 + 1 and a3 + a2 + a

1.3 Rational Algebraic Expressions

a. Rational Expressions:

Let us discuss about the following numbers : 3, − 2, 3 , 3x , a
5 5y b

Tfohremnuofmabefrrasc3t,io−n2qpan,dwh53earereqr≠atoioannadl numbers. Because these numbers are in the
p and q both are integers.

Algebraic Expressions 281

Prime Mathematics Book - 8

Similarly, 3x and a both are also rational but the numerator and denominator both
5y b

contain algebraic expression. Such type of expression is called rational algebraic

expression. For examples : x2−x1, 3x , a2 − 1 , x4 − 7 etc.
y b x+2

A rational expression is defined only for those values which do not make the
denominator is 0. If the denominator of the rational expression is 0, then the rational
expression is undefined.

For example : if x = 2 in x 5 2 , then 5 is undefined. Similarly, the rational
- x-2

expressions 4 3a , x y2 and 4p2 are undefined if x = 4 in 4 3a , x = a in x y2 and
-x -a 2-p -x -a

p = 2 in 4p2 receptively.
2-p

b. Reduction of Rational Expressions into their lowest Terms:

To reduce a rational expressions into their lowest term, we first factorise the
numerator and denominator separated and then cancel the common factors from

both numerator and denominator.

For example: let the given rational expression be 5x2y .
15xy2

5x2y 5 × x × x × y x
Then, 15xy2 = 3 × 5 × x × y × y = 3y

Example 31: For what value of x is x 3 undefined ?
-11

Solution: Since x 3 is undefined if the denominator x - 11 is 0
- 11

∴ x - 11 = 0 or, x = 11

So, 3 is undefined for x = 11.
x−11

282 12 Algebraic Expressions

Prime Mathematics Book - 8

Example 32: Reduce to the lowest terms:

(a) 8my2 (b) a2+6a+8
12mpy a2−16

Solution: (a) 8my2 (b) a2 + 6a + 8
12mpy a2 − 16

= 2×2×2×m×y×y = a2 + (4 + 2)a + 8
2×2×3×m×p×y (a)2 − (4)2

= 32py = a(2a+ 4a + 2a + 8
+ 4) (a − 4)

= a(a + 4) + 2(a + 4)
(a + 4) (a − 4)

= (a+4) (a+2) = (a + 2)
(a+4) (a−4) (a − 4)

Exercise 1.3(a)

1. For what value of x, each of the following rational expressions are undefined ?

a) x 5 3 b) 5q2 c) 2ab d) x3 e) x+2
− p−x x−5 4 − x2 x+9

2. For what value of a, each of the following rational expression are not defined ?

a) 2x b) 5x3 c) x 7 e) 3m2 −2mn
3−a 4−a y−a d) a−1 a2−16

3. Reduce to the lowest terms:

a) 4x2 b) 6x2y 4x2ab3 4x - 6y e) a2 + ab
5x4 15xy2 c) 14xa2b d) 8x - 12y a2 - ab

f) m(m-n) g) x3+3x2 h) (x-5)2 i) 3mn
n(n-m) x3+x2 2x-10 3n2 - 12n

4. Simplify:

a) a2 + 2ab + b2 b) 5x2 - 5y2 c) 2m - 6n d) m2 + 6m + 9 e) 5y3 - 45y
a2 - b2 7x + 7y m2 - 9n2 m2 - 9 4y2 - 12y

f) q3 - 1 g) x2 - 5x + 6 h) m2 + m- 4-m12 i) y2 + y - 12 j) x2 - 3x + 2
1 - q2 x2 - 4 12 y2 - y - 6 x2 - 4

k) p2 + 9p + 1250 l) y2 - 1 5
p2 + 2p - y2 + 6y +

Algebraic Expressions 283

Prime Mathematics Book - 8

c. Addition and subtraction of Rational Expressions having same Denominator:
While adding or subtracting rational expressions having the same denominator, the

numerator are added or subtracted and the sum or difference of the numerators of
rational expressions is written over the common denominator. Then the obtaining

expression is reduced into its lowest term if possible.

4x 3x 2a2 6a
Example 33: Simplify : a) x - 1 + x - 1 b) a - 3 − a - 3
Solution: a) Here, b) Here,

4x 3x 2a2 6a
= x - 1 + x - 1 = a - 3 − a - 3

4x + 3x = 7x = 2a2 - 6a = 2a(a - 3) = 2a
= x - 1 x-1 a-3 (a - 3)

Example 34: Simplify : a) x2 + 3 5x+3 x2 3x
x+3 + x+3 b) x2-4x+3 - x2-4x+3

Solution : a) Here, b) Here,

= x2+3 + 5x+3 x2 3x
x+3 x+3 = x2-4x+3 - x2-4x+3

= x2 + 3x ++ 53x + 3 = x2-3x
x2-4x+3
x2 + 5x + 6
= x + 3 = x(x-3)
x2-(3+1)x+3

= x2 + 3x + 2x + 6 = x(x-3)
x+3 x2 - 3x - 1x + 3
x(x-3)
= x(x+3()x++32)(x+3) = x(x-3)-1(x-3)

= (x+3) (x+2) = x(x-3) =x+2 = x
(x+3) (x-3) (x-1) x-1

Exercise 1.3(b)

1. Simplify:

a) 7 + 3 b) 3x + 2x c) 7a - 4a d) x 5 2 + x 4 2
x x 5 5 3 3 + +

e) x+y - y f) x + 1 + 2x + 5 g) 2m - 2n h) 8 - 2x
m+n m+n 3 3 m-n m-n x-4 x-4

i) 3x + x 3 1 j) m n + n n k) x +y - x-y l) ab - ab
x+1 + m- -m x -y x-y a+b a+b

284 12 Algebraic Expressions

Prime Mathematics Book - 8

2. Simplify:

a) x+3 + x - 3 b) 5x +1 - x+1 c) 5x - 5y d) x - 11 + 14
x-5 x - 5 x2 -3 x2 - 3 x2 - y2 x2 - y2 x2 - 9 x2 - 9

e) 3x - x 9 3 f) x2 + 2x + 1 g) a3 - 4a
x-3 - x+1 x+1 a-2 a-2

h) m2 + 6 + 2m + 6 i) 5m2 - 35m - 60
m2 + 5m m2 + 5m 4-m 4-m

j) 3p2 + 6pq + 3q2 k) x2 - 4xy + 4y2 l) 2x2 + 24xy + 16y2
p+q p + q x-2y x-2y x-2y 3x + 4y 3x + 4y 3x + 4y

d. Reduce the Rational Expressions to their lowest common denominators (LCD):
To reduce the given rational expressions having the same denominators, the following
steps are used:

- Find the LCM of all the denominators.
- Divide the LCM by each of the denominator.
- Multiply each of the numerators and denominators by its quatiant.

To understand the process of reduction into LCD, study the following examples.

Example 35: Reduce 3a , 3b and 5 into LCD.
x2yz xyz2 xy2z

Solution: The given rational expressions are, 3a , 3b and 5
x2yz xyz2 xy2z

LCM of x2yz, xyz2 and xy2z = x2y2z2

Now, x2y2z2 ÷ x2yz = yz x2y2z2 ÷ xyz2 = xy x2y2z2 ÷ xy2z = xz

∴ 3a × yz , 3b × xy and 5 × xz
x2yz yz xyz2 xy xy2z xz

= 3ayz , 3bxy and 5xz
x2y2z2 x2y2z2 x2y2z2

Example 36: Reduce 2 , a2 - 3 + 12 and a2 - 4 - 8 into LCD.
a2 - a - 6 7a 2a

Solution: The given rational expressions are

2 , a2 - 3 + 12 and 4
a2 - a - 6 7a a2 - 2a - 8

a2 - a - 6 = a2 - (3 - 2)a - 6 = a2 - 3a + 2a - 6
= a(a - 3) + 2 (a - 3)
= (a - 3)(a + 2)

Algebraic Expressions 285

Prime Mathematics Book - 8

a2 - 7a + 12 = a2 - (4 + 3)a + 12 = a2 - 4a - 3a + 12

= a(a - 4) - 3(a - 4)

= (a - 4)(a - 3)

a2 - 2a - 8 = a2 - (4 - 2)a - 8 = a2 - 4a + 2a - 8

= a(a - 4) + 2(a - 4)

= (a - 4)(a + 2)



∴ 2 , a2 - 3 + 12 and a2 - 4 - 8
a2 - a - 6 7a 2a

= (a - 2 + 2) , (a - 3 - 3) and (a - 4 + 2)
3)(a 4)(a 4)(a

LCM of the denominators of the given rational expressions = (a - 3) (a + 2) (a - 4).
Now,
(a - 3) (a + 2) (a - 4) ÷ (a - 3)(a + 2) = a - 4

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

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

∴ (a - 2 + 2) , (a - 3 - 3) and (a - 4 + 2)
3)(a 4)(a 4)(a

= (a - 2(a - 4) -4) , (a - 3(a + 2) + 2) and (a - 4(a - 3) - 3)
3)(a + 2)(a 4)(a - 3) (a 4)(a + 2) (a

Exercise 1.3(c)

1. Reduce to the lowest common denominator (LCD):

a) 4 , 5a b) x y z c) 2 , 3x , 5y d) 2x , 3y e) a 5 b, a 2 b
2x 3y a, b, c a2bc ab2c abc2 3a 2b + -

f) 3x , 2y , 5z g) a 3 b, a3 2 b3, a2 + 5 + b2 h) 1 x , 1 2a , 1 3-b2a
x-y x+y x2 - y2 - - ab - 4a2 + 2a

i) x2 + a + 2, x2 + b + 6 , x2 + 2c + 3 i) x-1 , x-2 , x-3
3x 5x 4x x2 - 5x + 6 x2 - 4x + 3 x2 - 3x + 2

286 12 Algebraic Expressions

Prime Mathematics Book - 8

e. Addition and Subtraction of Rational Expressions having different denominators:

Can you solve 4 + 7 ?
5 6
4 7
Here, 5 + 6 , First we take LCM of 5 and 6.

The LCM of 5 and 6 is 30.

Then, 30 ÷ 5 = 6 ∴ 4 × 6 = 24
∴ 75 × 65 = 3350
30 ÷ 6 = 5 6 5 30

Now, 4 7 24 35 24 + 35 59
5 6 30 30 30 30
+ = + = = .

The above same process can be applied for addition and subtraction of rational
expressions having different denominators.

The following points are used to solve such problems.

- Factorise the different denominators and find their LCM.
- Divide LCM of denominators by each denominator of rational expressions.
- Multiply the quotient by its numerator and then simplify.

To understand the above process, Study the following examples.

Example 37: Simplify : a) 2x + x b) 4 - x 3 2
3 5 x-2 +

Solution: a) Here, b) Here,

2x + x 4 - x 3
3 5 x-2 +2

LCM of 3 and 5 is 15 LCM of (x - 2) and (x + 2) is (x - 2)(x + 2)
∴ (x - 2)(x + 2) ÷ (x - 2) = x + 2
15 ÷ 3 = 5 and 15 ÷ 5 = 3
(x - 2)(x + 2) ÷ (x + 2) = x - 2

2x 5 x 3
∴ 3 × 5 + 5 × 3

= 10x + 3x Now, x 4 2 - x 3 2
15 15 - +

= 10x + 3x = 4(x + 2) - 3(x - 2)
15 (x - 2)(x + 2)

= 13x = 4x + 8- 3x + 6
15 x2 - 4

= x + 14
x2 - 4

Algebraic Expressions 287

Prime Mathematics Book - 8

Example 38: Simplify : 3 + 2 + 4y
x-y x+y x2 + y2
Solution: Here,

3 2 4y
x - y + x + y + x2 - y2

3(x + y) + 2(x - y) 4y
= (x - y)(x + y) + x2 - y2

= 3x + 3y + 2x - 2y + 4y
x2 - y2 x2 - y2

= 5x + y + 4y
x2 - y2 x2 - y2

= 5x + y + 4y
x2 - y2

= 5x + 5y
x2 - y2

= 5(x + y)
(x + y)(x - y)


= 5
x-y

3 3x
Example 39: Simplify : x2 -x-2 + x2-5x+6

Solution : Here,

3 + 3x = 3x - 9 + 3x2 + 3x
x2-x-2 x2-5x+6 (x + 1) (x - 2) (x - 3)

= x2 - (2 3 - 2 + 3x = (x + 3x2 + 6x - 9 - 3)
- 1)x x2 - (3 + 2)x + 6 1) (x - 2) (x

= 3 + 3x = (x 3(x2 + 2x - 3) 3)
x2 - 2x + x - 2 x2 - 3x - 2x + 6 + 1) (x - 2) (x -

= x(x - 2) 3 1(x - 2) + x(x - 3x - 3) 3(x2 + 3x - x - 3)
+ 3) - 2(x = (x + 1) (x - 2) (x - 3)

= (x + 3 - 2) + (x - 3x - 3) 3{x(x + 3) - 1(x + 3)}
1) (x 2) (x = (x + 1) (x - 2) (x - 3)

= 3(x - 3) + 3x(x + 1) = (x 3(x + 3) (x - 1) 3)
(x + 1) (x - 2) (x - 3) + 1) (x - 2) (x -

288 12 Algebraic Expressions

Prime Mathematics Book - 8

Exercise 1.3(d)

1. Simplify:

a) 2 + x b) 2x + 5x c) m2 + 3m2 d) 4x - x
3 4 7 3 4 6 3 4

e) 3 - 2 f) 7x - 2 g) a + b h) 4x + 37x
m n 11 5 8 12

i) m - 2 j) 6 - x k) 4a - 2b l) 2 - 3
n x 4 3b 3a a ab

2. Simplify:

a) x2 - y2 b) 6 + 4 5q3 5p3 d) x2 - 4y
4 3 5x2 x c) p - q y

e) m - 2-m f) 2 - a 2 2 g) 5 - x 2 y h) 3 + y 4 a
2-m m a-2 + x-y + y-a +

i) x 2 + 1 j) a a - b k) p 5 - 1 l) x 3) + 2
- 2y x + 2y +b a-b 5p - p-1 3(x + 3(x - 3)



3. Simplify:

x1 b) x+6 - x x+3 x+5
a) + c) x-5 - x-3
x2-1 x-1 x-6 6-x

2x+1 e) x - 2 x+3 x-2
d) 6 + 2x 2(x+y) 3(x+y) f) x+2 - x-3

4. Simplify:

7x+3 3 b+2 3 2-x 2x-1
a) x2-9 - x-1 b) - c ) x2+2x-8 + x2+4x
b2+b b2-b-2

2x x2+3 e) m - 2 f) 2 + 5y
d) - m2+3m+2 m2-1
x-1 x2-1 y2+3y+2 y2-y-6

g) 1 - 1 + 2 h) 2 + 2x - x2+3 i) 2 - 1 - 3
x+2 x-2 x2+4 x+1 x-1 x2-1 a-1 a+1 a2-1

j) x + y - 2xy k) 3 + 2 - 6y l) x+1 - x-1 + 4x
x+y x-y x2-y2 x-y x+y x2-y2 x-1 x+1 x2+1

Algebraic Expressions 289

Prime Mathematics Book - 8

Multiplication of Rational Expressions:

When we multiply two or more algebraic rational expressions, we should first

factories the numerators and denominators of the given expressions, if it is possible.

Then, the common factors of the numerator and denominator can be cancelled by

the rule of cancellation. After that we multiply the numerator and numerator of the

expressions and the denominator and denominator of the expressions and to form a

single resultant expression.

Look at the following examples:

3ax2 30b2y3 a2 - b2 4x2 - 5xy
Example 40: Simplify: (a) 5bxy × 12a2x4y (b) 2xy × 2a - 2b

Solution: 3ax2 30b2y3
(a) 5bxy × 12a2x4y

= 3×a×x×x 2×3×5×b×b×y×y×y = 3by
5×b×x×y × 2×2×3×a×a×x×x×x×x×y 2ax3

(b) a2-b2 × 4x2-5xy
2xy 2a-2b

= (a+b) (a-b) × x(4x-5y) = (a+b) × (4x - 5y) (a+b) (4x-5y)
2xy 2(a-b) 2y 2 = 4y

Example 41: Simplify : x2-x-6 2x2-x-1
×
2x-2 x2-9

Solution: Here,

x2-x-6 × 2x2-x-1
2x-2 x2-9

x2-(3-2)x-6 2x2-(2-1)x-1
= 2(x-1) × (x)2-(3)2

= x2-3x+2x-6 2x2-2x+x-1
2(x-1) × (x+3)(x-3)

= x(x-3)+2(x-3) × 2x(x-1)+1(x-1)
2(x-1) (x+3)(x-3)

= (x-3)(x+2) (x-1)(2x+1) = (x+2)(2x+1)
2(x-1) × (x+3)(x-3) 2(x+3)

290 12 Algebraic Expressions

1. Simplify: Exercise 1.3(e) Prime Mathematics Book - 8

3x 2y a2 3 7x2 15x3yz2
(a) × 5a (b) × (c) 5xy2 × 2z
4b
bb
9x2y 7yz 2a2b y4 2a2
(d) × 3xz2 (e) 3b3y × 3a3 × 4by2
14y3

2. Simplify: a2-b2 b3c x2-y2 y
2x 2xy+y2 (b) × (c) ×
ab a3+b3 x x+y
(a) 2x+y × 8y2

a-3 a2-4 x2+xy x-y a2-a 6a
(d) × (a-3)2 (e) × x+y (f) ×
a+2 x2-xy 2a+2 a2-1

3. Simplify:

a2+2ab+b2 a+b x2+x-6 2x2+x-1 (c) a2+10a+24 × a-3
(a) a2-b2 × a-b (b) × a2+2a-8 a+6
x+1 x+3 a2-4 a-3

a2-4b2 a2-3ax (x-y)2+4xy x-y (f) ×
(d) × ax+2bx (e) (x+y)2-4xy × x+y a2-a-6 2y
a2-9x2

(g) x2-3x-10 bx-3b a2-9 a2+2a-8
× (h) a2+4a × a2+a-6
x2-5x+6 cx-5c

Division of Rational Expressions:

We know that, x ÷ a be changed into x × b .
y b y a

So, the division is the reciprocal of multiplication. Then, we apply the same method
as we applied for multiplication.

Algebraic Expressions 291

Prime Mathematics Book - 8

Look at the following examples.

6a2b 3ab2 (b) a2-b2 a+b
Example 42: Simplify: (a) ÷ ab ÷
7xy 14x2y b

Solution: (a) 6a2b 3ab2
÷
7xy 14x2y


6a2b 14x2y 3ab2
= × [ (÷) changes to (×) and 14x2y changes to its
7xy 3ab2 \

6a2b×14x2y reciprocal 14x2y ]
= 7xy×3ab2
3ab2

2×3×a×a×b×2×7×x×x×y 4ax
= =
7×x×y×3×a×b×b b

a2-b2 a+b a2-b2 b (a+b)(a-b) × b a-b
(b) ÷ = × = a×b (a+b) = a
ab b ab a+b

x2+3x+2 x2-x-6

Example 43: Simplify: x2-4x-12 ÷
Solution: x2-9x+18

x2+3x+2 x2-x-6
= x2-4x-12 ÷ x2-9x+18

x2+3x+2 x2-9x+18
=×
x2-4x-12 x2-x-6

x2+(2+1)x+2 x2-(6+3)x+18
= x2-(6-2)x-12 × x2-(3-2)x-6

x2+2x+x+2 x2-6x-3x+18
= x2-6x+2x-12 × x2-3x+2x-6

x(x+2)+1(x+2) x(x-6)-3(x-6)
= x(x-6)+2(x-6) × x(x-3)+2(x-3)

(x+2)(x+1) (x-6)(x-3) x+1
= (x-6)(x+2) × (x-3)(x+2) = x+2

292 12 Algebraic Expressions

Exercise 1.3(f) Prime Mathematics Book - 8

1. Simplify: 27xy3 9y2
(c) 4ab2 ÷ 2b
a2b 2a 3ab 6b
(a) ÷ (b) ÷
m bm 4xy 5y

2x 8y2 a2b2 b2a x4y3
(d) 2x-y ÷ 2xy-y2 (e) x2y2 ÷ x2m × x2am2

2. Simplify:

x2-y2 x-y a2-1 a+1 x2-y2 (x-y)2
(a) a ÷ b (b) ÷ (c) x+y ÷ x-y
b2 b

(d) x2-4x ÷ x2-16 (e) 1 - 1 ÷ x-y
x y x2-y2
3y-4z 9y2-16z2

3. Simplify: 3x2-4x-7 x2-1 (c) x2+2x-15 ÷ 3(x2+4x-5)
(b) ÷ x-2 x2-3x+2
a2-5a+6 a-3 3x2-7x x-4
(a) a2-9 ÷ a+3 x2-6x+9 x2-5x+6
x2-25 x+5 (f) x2-2x-3 ÷ x2-3x+2
2a2+a-1 a+1 (e) ÷
(d) a2+a-6 ÷ a+3
x2-16 x-4

4. Simplify:

x2 1 x-1 x1 ÷ x+1
(a) x+1 - x+1 ÷ (b) -
4 x-1 x+1 x2-1

a-4 a-3 a2-7a+12 a+b a-b a2-b2
(c) × ÷ a2+7a+12 (d) - ÷
a+4 a+3 a-b a+b 4ab

Algebraic Expressions 293

Prime Mathematics Book - 8

Area Revision Test (Algebraic Expression)

1. Resolve into factors:
a. ax2 − bxy b. 12x2 + xy + xz c. x(2x + y) − 3(2x + y) + 4y (2x + y)

2. Resolve into factors: b. x2y − xy + 2x2y − 2xy c. 2x2 + 3ax + 2ax + 3a2
a. ax + by + ay + bx

3. Factorize: b) 4x3y − 81xy3 c) x4 − 1 d) (m + 2)2 − 4
a) a2 − 16b2

4. Find the value of the following by using a2 − b2 = (a + b) (a − b):

a) 45 × 45 - 31 × 31 c) 105 × 95 f) 219×219−181×181
219−181

5. Resolve into factors: b) 4 - 32a + 64a2 c) x2 + 2x + 1 - 9y2
a) x2 + 10x + 25

d) x2 + 8x + 15 e) x2 + 7x -18

6. If a + b = 3 and ab = 4 and the value of a3 + b3.

7.If y − 1 = 6, find the value of y3 - 1
y y3

8. Find the H.C.F.:
a) x2 + 2xy + y2 and s2 - y2 b) 4a2 -16a + 12 and 2a2 - 10a + 12

9. Find the L.C.M of: b) x2 - 4x - 5, 2 x 2 - 11 x + 5 and x2 - 6x + 5
a) a2 - 2ab + b2 and a2 - b2

10. Simplify:

3x 3 ab ab a3 4a
a) x+1 + b) a+b - a+b c) -
x+1 a-2 a-2

2x2 + 24xy + 16y2
(d) 3x+4y
3x+4y 3x+4y

Algebraic Expressions e) 52 f) 2 + 5y
x-y - x+y y2+3y+2 y2-y-6

a2+2ab+b2 a+b 1 1 x-y
g) a2-b2 × a-b h) - ÷

x y x2-y2

294 12 Algebraic Expressions

Prime Mathematics Book - 8

Answers

Exercise - 1.1(a) (Factorization) 7. 2mn(1−2n)
1. m(x − y) 2. x(ax − by) 3. 2xy(2x − a) 4. 5b(a − 2c) 5. m(2n2−m) 6. 6(2p2−q2)
8. x(x2+1) 9. x(12x+y+z) 10. 2x(x2−x+4) 11. 4x(15x2−xy+3) 12. (a+b)(2y) 13. (a+b)(x−y+z)
14. (2x+y)(x−3+4y)

Exercise - 1.1(b)
1. (x−y)(m+y) 2. (m−ny)(x−2) 3. (x+y)(a+b) 4. (m−5)(m+3n) 5. (x+1)(x−y) 6. (q−1)(p−1) 7. (a−3)(2a+5)
8. 3xy(x−1) 9. (a−b)(a−3) 10. (a−c)(ac+b2) 11. (m+n)(mn+1) 12. (x−1)(x2−y+1) 13. (x+a)(2x+3a)
14. b(a+2)(a−1) 15. (4x - y) (1 + 4x + y)

Exercise - 1.1(c)

1. (a) (a+4)(a−4) (b) (3+c)(3−c) (c) (5a+b)(5a−b) (d) (a+4b)(a−4b) (e) (3x+7y)(3x−7y)

(f) (p+ 1 ) (p− 1 ) (g) (x+ 1 ) (x− 1 ) (h) 5(x+2y)(x−2y) (i) 13(m+3n)(m−3n)
q q 3y 3y

(j) ab(b+3a)(b−3a) (k)(5+ 1 ) (5− 1 ) (l) −24xy (m) z(x+y)(x−y) (n) m(m+4) (o) (1+ 191pq) (1− 191pq)
3a 3a

(p) (m−n+3)(m−n−3) (q) (a+3b+3)(a+3b−3)

2. (a) 784 (b) 35600 (c) 9975 (d) 16819 (e) 39951 (f) 400 (g) 600

3. (x+2)(x−2)cm2, 60cm2 4. π (R+r)(R−r), 147.84cm2 5. (x2−0.385)m2

Exercise - 1.1(d)

1. Show to your teacher.

2. (a) (x+6)(x+6) (b) (a−3)(a−3) (c) (p−2)(p−2) (d) (y−7x)(y−7x) (e) (7p+1)(7p+1) (f) 3(x−1)(x−1)

(g) (7q−5)(7q−5) (h) (a+9x)(a+9x) (i) (3a+7b)(3a+7b) (j) (2a−5)(2a−5) (k) (5y−6)(5y−6)

3. (a) (a2+ab+b2)(a2−ab+b2) (b) (a2+a+1)(a2−a+1) (c) (p+3q+5)(p+3q−5) (d) (x+y−z)(x−y+z)

(e) (3x−2y+a−b)(3x−2y−a+b) (f) (a - 1 + b) (a - 1 - b) g) (2x + y -m) (2x + y +m)

(h) (1 + xy + x - y) (1 + xy - x + y)

Exercise - 1.1(e)
1. (a) 2,3 (b) 9,8 (c) 16,3 (d) 14,7 (e) 20,7 (f) 14,11
2. (a) (x+2)(x+3) (b) (x+5)(x+9) (c) (x−8)(x−9) (d) (m+8)(m+4) (e) (a+8)(a+6)

(f) (p+6)(p+11) (g) x(x+11)(x+1) (h) a2(a+8)(a+4) (i) (x−4)(x−2) (j) (m+3)(m+1) (k) (a+7b)(a+15b)
(l) (x−8y)(x−5y) (m) (x + 3) (x - 13) n) (y + 5) (y + 7) o) (m - 3) (m + 3) p) (xy - 4) (xy - 2)
3. (a) (x−12)(x−14) (b) (a+b−7)(a+b−14) (c) (p−q−15)(p−q−3) (d) (m+n+4) (m+n+19) (e) (xy−3z) (xy−9z)

22 Algebraic Expressions 295