Chapter 9: Algebraic Expressions and Identities
Exercise 9.1 (Page 140 of Grade 8 NCERT)
Q1. Identify the terms, their coefficients for each of the following expressions.
(i) 25 3xyz zy−
(ii) 21 x x+ +
(iii) 2 2 2 2 2 24 4x y x y z z− +
(iv) 3 pq qr rp− + −
(v) 2 2
x yxy+ −
(vi) 0.3 0.6 0.5a ab b− +
Difficulty level: Easy
Known:
Expressions
Unknown:
Terms and their coefficients
Reasoning:
The numerical factor of a term is called its numerical coefficient or simply coefficient.
Solution:
The terms and the respective coefficients of the given expressions are as follows.
- Terms Coefficients
(i) 25
- 3
xyz
zy
5
3−
(ii)
2
1
x
x
1
1
1
(iii)
2 2
2 2 2
2
4
4
x y
x y z
z
−
4
4
1
−
(iv) 3
pq
qr
rp
−
−
3
1
1
1
−
−
(v)
2
2
x
y
xy−
1
2
1
2
1−
(vi) 0.3
0.6
0.5
a
ab
b
−
0.3
0.6
0.5
−
Q2. Classify the following polynomials as monomials, binomials, trinomials.
Which
polynomials do not fit in any of these three categories?
2 3 4 2 2 3
2 2 2
, 1000, , 7 5 , 2 3 ,2 3 4 , 5 4 3 ,
4 15 , , , , 2 2
x y x x x x y x y y y y y x y xy
z z ab bc cd da pqr p q pq p q
+ + + + + + − − + − +
− + + + + +
Difficulty level: Easy
Known:
Expression
Unknown:
The degree of the expression.
Reasoning:
1. Expression that contains only one term is called a monomial.
2. Expression that contains two terms is called a binomial.
3. Expression containing three terms is a trinomial and so on.
4. An expression containing, one or more terms with non-zero coefficient (with
variables having non-negative integers as exponents) is called a polynomial.
5. A polynomial may contain any number of terms, one or more than one.
Solution:
The given expressions are classified as
Monomials: 1000, pqr
Binomials: 2 2 2 2, 2 3 , 4 15 , , 2 2x y y y z z p q pq p q+ − − + +
Trinomials: 2 37 5 , 2 3 4 , 5 4 3y x y y y x y xy+ + − + − +
Polynomials that do not fit in any of these categories are
2 3 4 , x x x x ab bc cd da+ + + + + +
Q3. Add the following.
(i) , , ab bc bc ca ca ab− − −
(ii) , , a b ab b c bc c a ac− + − + − +
(iii) 2 2 2 2 2 3 4, 5 7 3p q pq pq p q− + + −
(iv) 2 2 2 2 2 2, , , 2 2 2l m m n n l lm mn nl+ + + + +
Difficulty level: Medium
Known:
Expressions
Unknown:
Addition
Reasoning:
Addition will take place between like terms.
Solution:
The given expressions written in separate rows, with like terms one below the other
and then the addition of these expressions are as follows.
(i)
0
ab bc
bc ca
ab ca
−
+ −
+− +
Thus, the sum of the given expressions is 0.
(ii)
a b ab
b c bc
a c ac
ab bc ac
− +
+ − +
+ − + +
+ +
Thus, the sum of the given expressions is ab bc ac+ + .
(iii) 2 2
2 2
2 2
2 3 4
3 7 5
4 9
p q pq
p q pq
p q pq
− +
+ − + +
− + +
Thus, the sum of the given expressions is 2 2 4 9.p q pq− + +
(iv) 2 2
2 2
2 2
2 2 2
2 2 2
2 2 2 2 2 2
l m
m n
l n
lm mn nl
l m n lm mn nl
+
+ + +
+ +
+ + +
+ + + + +
Thus, the sum of the given expressions is 2(2 2 2l m n lm mn nl+ + + + + ).
Q4. (a) Subtract 4 7 3 12a ab b− + + from 12 9 5 3a ab b− + −
(b) Subtract 3 5 7xy yz zx+ − from 5 2 2 10xy yz zx xyz− − +
(c) Subtract 2 24 3 5 8 7 10p q pq pq p q− + − + − from
2 218 3 11 5 2 5p q pq pq p q− − + − +
Difficulty level: Medium
Known:
Expressions
Unknown:
Subtraction
Reasoning:
Subtraction will take place between like terms.
Solution:
The given expressions in separate rows, with like terms one below the other and
then the subtraction of these expressions is as follows.
(a)
( ) ( ) ( ) ( )
12 9 5 3
4 7 3 12
8 2 2 15
a ab b
a ab b
a ab b
− + −
− + +
− + − −
− + −
(b)
( ) ( ) ( )
5 2 2 10
3 5 7
2 7 5 10
xy yz zx xyz
xy yz zx
xy yz zx xyz
− − +
+ −
− − +
− + +
(c)
( ) ( ) ( ) ( ) ( ) ( )
2 2
2 2
2 2
18 3 11 5 2 5
10 8 7 3 5 4
28 5 18 8 7
p q pq pq p q
p p pq pq p q
p q pq pq p q
− − + − +
− − + − + +
+ + − + − −
+ − + − +
Chapter 9: Algebraic Expressions and Identities
Exercise 9.2 (Page 143 of Grade 8 NCERT)
Q1. Find the product of the following pairs of monomials.
( ) ( ) ( ) ( ) ( )3 4,7p 4 , 7 4 , 7 4 , 3 4 , 0i ii p p iii p pq iv p p v p− − −
Difficulty level: Easy
Known:
Pairs of monomials
Unknown:
Product
Reasoning:
i) By using the distributive law, we can carry out the multiplication term by term.
ii) In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Solution:
The product will be as follows.
( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )
2
2
3 4
4 7 4 7 28
4 7 4 7 4 7 28
4 7 4 7 4 7 28
4 3 4 3 12
4 0 4 0 0
i p p p
ii p p p p p p p
iii p pq p p q p p q p q
iv p p p p p p p
v p p
= =
− = − = − = −
− = − = − = −
− = − = −
= =
Q2. Find the areas of rectangles with the following pairs of monomials as their
lengths and breadths respectively.
( ) ( ) ( ) ( ) ( )2 2 2, ; 10 , 5 ; 20 , 5 ; 4 , 3 ; 3 , 4p q m n x y x x mn np
Difficulty level: Easy
Known:
Lengths and breadths of rectangles
Unknown:
Areas of rectangles
Reasoning:
Area of a rectangle = length × breadthSolution:
We know that,
2 2 2 2 2 2
Area of rectangle Length
Area of 1 rectangle
Area of 2 rectangle 10 5 10 5 50
Area of 3 rectangle 20 5 20 5 100
Area of
st
nd
rd
Breadth
p q pq
m n m n mn
x y x y x y
=
= =
= = =
= = =2 2 3
2
4 rectangle 4 3 4 3 12
Area of 5 rectangle 3 4 3 4 12
th
th
x x x x x
mn np m n n p mn p
= = =
= = =
Q3. Complete the table of products.
First monomial
Secondmonomial
→
2x 5y− 23x 4xy− 27x y 2 29x y−
2x 24x5y− 215x y−
23x4xy−
27x y2 29x y−
Difficulty level: Easy
Known:
Expressions
Unknown:
Product
Solution:
The table can be completed as follows.
First monomial
Secondmonomial
→
2x 5y− 23x 4xy− 27x y 2 29x y−
2x 24x 10xy− 36x 28x y− 314x y 3 218x y−
5y− 10xy− 225y 215x y− 220xy 2 235x y− 2 345x y23x 36x
215x y− 49x312x y− 421x y 4 227x y−
4xy− 28x y− 220xy 312x y− 2 216x y3 228x y− 3 336x y
27x y 314x y 2 235x y− 421x y 3 228x y− 4 249x y 4 363x y−2 29x y− 3 218x y− 2 345x y 4 227x y− 3 336x y 4 363x y− 4 481x y
Q4. Obtain the volume of rectangular boxes with the following length, breadth and
height respectively.
( ) ( ) ( ) ( )2 4 2 25 ,3 ,7 2 , 4 , 8 , 2 , 2 , 2 , 3i a a a ii p q r iii xy x y xy iv a b c
Difficulty level: Easy
Known:
Length, breadth and height respectively of rectangular boxes
Unknown:
Volume of rectangular boxes
Reasoning: volumeof a rectangular box = length × breadth × height
Solution:
We know that,
( )
( )
( )
2 4 2 4 7
2 2 2
Volume Length Breadth Height
Volume 5 3 7 5 3 7 105
Volume 2 4 8 2 4 8 64
Volume 2 2 2 2
i a a a a a a a
ii p q r p q r pqr
iii xy x y xy xy x y
=
= = =
= = =
= =
( )
2 4 4 4
Volume 2 3 2 3 6
xy x y
iv a b c a b c abc
=
= = =
Q5. Obtain the product of
( ) ( ) ( )
( ) ( )
2 3 2 3 , , , , 2, 4 , 8 , 16
, 2 , 3 , 6 , ,
i xy yz zx ii a a a iii y y y
iv a b c abc v m mn mnp
−
−
Difficulty level: Easy
Known:
Expressions
Unknown:
Product
Reasoning:
By using the distributive law, we can carry out the multiplication term by term.
Solution:
( )
( ) ( )( )
( )
( ) ( )
2 2 2
2 3 6
2 3 2 3 6
2 2 2
3 2
2 4 8 16 2 4 8 16 1024
2 3 6 2 3 6 36
i xy yz zx x y z
ii a a a a
iii y y y y y y y
iv a b c abc a b c abc a b c
v m mn mnp m n p
=
− = −
= =
= =
− = −
Chapter 9: Algebraic Expressions and Identities
Exercise 9.3 (Page 146 of Grade 8 NCERT)
Q1. Carry out the multiplication of the expressions in each of the following pairs.
( ) ( ) ( )
( ) ( )
2 2
2
4 , , , 7
9, 4 , 0
i p q r ii ab a b iii a b a b
iv a a v pq qr rp
+ − +
− + +
Difficulty level: Medium
Known:
Expressions
Unknown:
Product
Reasoning:
i) By using the distributive law, we can carry out the multiplication term by term.
ii) In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Solution:
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( )
2 2
2 2 2 2 2 2 3 2 2 3
2 2 3
4 4 4 4 4
7 7 7 7 7
9 4 4 9 4 4 36
i p q r p q p r pq pr
ii ab a b ab a ab b a b ab
iii a b a b a a b b a b a b a b
iv a a a a a a a
v
+ = + = +
− =
−
+ − = −
+ = + = +
− = −+ =
( ) ( ) ( ) ( ) 0 0 0 0 0pq qr rp pq qr rp+ + = + + =
Q2. Complete the table
--- First expression Second Expression Product
(i) a b c d+ + ---
(ii) 5x y+ − 5xy ---
(iii) p 26 7 5p p− + ---
(iv) 2 24p q 2 2p q− ---
(v) a b c+ + abc ---
Difficulty level: Medium
Known:
Expressions
Unknown:
Product
Solution:
The table can be completed as follows.
--- First expression Second Expression Product
(i) a b c d+ + ab ac ad+ +
(ii) 5x y+ − 5xy 2 25 5 25x y xy xy+ −
(iii) p 26 7 5p p− + 3 26 7 5p p p− +
(iv) 2 24p q 2 2p q− 4 2 2 44 4p q p q−
(v) a b c+ + abc 2 2 2a bc ab c abc+ +
Q3. Find the product.
(i) ( ) ( ) ( )2 22 26 2 4a a a
(ii) 2 22 9
3 10xy x y
−
(iii) 3 310 6
3 5pq p q
−
(iv) 2 3 4 x x x x
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Solution:
(i) ( ) ( ) ( )2 22 26 2 22 26 50 2 4 2 4 8a a a a a a a = =
(ii) 2 2 2 2 3 3
5
2 9 2 9 3
3 10 3 10 5xy x y x y x y x y
− − − = =
(iii) 2
3 3 3 3 4 410 6 10 64
3 5 3 5pq p q pq p q p q
− − = = −
(iv) 2 3 4 10 x x x x x =
Q4.
a) Simplify ( )3 4 5 3x x − + and find its values for ( ) ( ) 3,1
2 i x ii x == .
b) ( )2 1 5a a a+ + + and find its values for ( ) ( ) ( )0, 1, 1i a ii a iii a= = = −
Difficulty level: Medium
Known:
Expression with their corresponding values.
Unknown:
Simplification and its result with their corresponding values
Solution:
( ) 2( ) 3 4 5 3 12 15 3a x x x x− + = − +
( )
( ) ( )
2
2
i For 3,
12 15 3
12 3 15 3 3
108 45 3
66
x
x x
=
= − +
= − +
= − +
=
2
2
3
1(ii) For ,
2
12 15 3
1 112 15 3
2 2
1 1512 3
4 2
153 3
2
156
2
12 15
2
3
2
x
x x
=
= − +
= − +
= − +
= − +
= −
−=
−=
( ) ( )
( )
( ) ( ) ( )
( ) ( ) ( ) ( )
2 3 2
3 2
3 23 2
3 23 2
b 1 5 5
i For 0, 5 0 0 0 5 5
ii For 1, 5 1 1 1 5
1 1 1 5 8
iii For 1, 5 1 1 1 5
1 1 1 5 4
a a a a a a
a a a a
a a a a
a a a a
+ + + = + + +
= + + + = + + + =
= + + + = + + +
= + + + =
= − + + + = − + − + − +
= − + − + =
Q5. (a) Add: ( ) ( ) ( ), and p p q q q r r r p− − −
(b) Add: ( ) ( )2 2x z x y and y z y x− − − −
(c) Subtract: ( ) ( )3 4 5 from 4 10 3 2l l m n l n m l− + − +
(d) Subtract: ( ) ( ) ( )3 2 from 4a a b c b a b c c a b c+ + − − + − + +
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Reasoning:
Addition and Subtraction takes place between like terms.
Solution:
(a) First expression = ( ) 2 p p q p pq− = −
Second expression = ( ) 2 q q r q qr− = −
Third expression = ( ) 2r r p r pr− = −
Adding the three expressions, we obtain 2
2
2
2 2 2
p pq
q qr
r pq
p pq q qr r pq
−
+ −
+ −
− + − + −
Therefore, the sum of the given expressions is 2 2 2 .p q r pq qr rp+ + − − −
(b) First expression = ( ) 22 2 2 2x z x y xz x xy− − = − −
Second expression = ( ) 22 2 2 2y z y x yz y yx− − = − −
Adding the two expressions, we obtain 2
2
2 2
2 2 2
2 2 2
2 2 4 2 2
xz x xy
xy yz y
xz x xy yz y
− −
+ − + −
− − + −
Therefore, the sum of the given expressions is 2 22 2 4 2 2 .x y xy yz zx− − − + +
( ) ( )
( )
2
2
c 3 4 5 3 12 15
4 10 3 2 40 12 8
l l m n l lm ln
l n m l ln lm l
− + = − +
− + = − +
Subtracting these expressions, we obtain
( ) ( ) ( )
2
2
2
40 12 8
15 12 3
25 5
ln lm l
l lm l
ln l
− +
− +
− + −
+ +
Therefore, the result is 25 25 .l ln+
( ) ( ) ( )
( )
2 2
2 2
2
3 2
3 3 3 2 2 2
3 2 3 2
4 4 4 4
d a a b c b a b c
a ab ac ba b bc
a b ab ac bc
c a b c ac bc c
+ + − − +
= + + − + −
= + + + −
− + + = − + +
Subtracting these expressions, we obtain
( ) ( ) ( ) ( ) ( )
2
2 2
2 2 2
4 4 4
3 2 3 2
7 6 4 3 2
ac bc c
ac bc a b ab
ac bc c a b ab
− + +
− + + +
− + − − −
− + + − − −
Therefore, the result is 2 2 23 2 4 6 7 .a b c ab bc ac− − + − + −
Chapter 9: Algebraic Expressions and Identities
Exercise 9.4 (Page 148 of Grade 8 NCERT)
Q1. Multiply the binomials.
(i) ( ) ( )2 5 and 4 3x x+ −
(ii) ( ) ( )8 and 3 4y y− −
(iii) ( ) ( )2.5 0.5 and 2.5 0.5l m l m− +
(iv) ( ) ( )3 5a b and x+ +
(v) ( ) ( )2 22 3 and 3 2pq q pq q+ −
(vi) 2 2 2 23 23 and 4
4 3a b a b
+ −
Difficulty level: Medium
Known:
Expressions
Unknown:
Multiplication
Reasoning:
i) By using the distributive law, we can carry out the multiplication term by term.
ii) In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Solution:
(i)
( ) ( ) ( ) ( )
( )
2
2
2 5 4 3 2 4 3 5 4 3
8 6 20 15
8 14 15 By adding like terms
x x x x x
x x x
x x
+ − = − + −
= − + −
= + −
(ii)
( ) ( ) ( ) ( )
( )
2
2
8 3 4 3 4 8 3 4
3 4 24 32
3 28 32 By adding like terms
y y y y y
y y y
y y
− − = − − −
= − − +
= − +
(iii)
( ) ( ) ( ) ( )2 2
2 2
2.5 0.5 2.5 0.5 2.5 2.5 0.5 0.5 2.5 0.5
6.25 1.25 1.25 0.25
6.25 0.25
l m l m l l m m l m
l lm lm m
l m
− + = + − +
= + − −
= −
(iv)
( ) ( ) ( ) ( )3 5 5 3 5
5 3 15
a b x a x b x
ax a bx b
+ + = + + +
= + + +
(v)
( ) ( ) ( ) ( )2 2 2 2 2
2 2 3 3 4
2 2 3 4
2 3 3 2 2 3 2 3 3 2
6 4 9 6
6 5 6
pq q pq q pq pq q q pq q
p q pq pq q
p q pq q
+ − = − + −
= − + −
= + −
(vi)
( )
2 2 2 2 2 2 2 2
2 2 2 2 2 2
22 2 2 2 2 2 2 2
4 2 2 2 2 4
4 2 2 4
3 2 3 83 4 3 4
4 3 4 3
3 8 84 3 4
4 3 3
3 3 8 84 3 4 3
4 4 3 3
3 2 12 8
3 10 8
a b a b a b a b
a a b b a b
a a a b b a b b
a b a b a b
a a b b
+ − = + −
= − + −
= − + −
= − + −
= + −
Q2. Find the product.
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
2 2
2 2
i 5 2 3
ii 7 7
iii
iv 2
x x
x y x y
a b a b
p q p q
− +
+ −
+ +
− +
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Reasoning:
i) By using the distributive law, we can carry out the multiplication term by term.
ii) In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Solution:
(i)
( ) ( ) ( ) ( )2
2
5 2 3 5 3 2 3
15 5 6 2
15 2
x x x x x
x x x
x x
− + = + − +
= + − −
= − −
(ii)
( ) ( ) ( ) ( )2 2
2 2
7 7 7 7 7
7 49 7
7 48 7
x y x y x x y y x y
x xy xy y
x xy y
+ − = − + −
= − + −
= + −
(iii)
( ) ( ) ( ) ( )2 2 2 2 2
3 2 2 3
a b a b a a b b a b
a a b ab b
+ + = + + +
= + + +
(iv)
( ) ( ) ( ) ( )2 2 2 2
3 2 2 3
2 2 2
2 2
p q p q p p q q p q
p p q pq q
− + = + − +
= + − −
Q3. Simplify.
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
2
2 3
2 2
2 2
i 5 5 25
ii 5 3 5
iii
iv 2
v 2 2
vi
vii 1.5 4 1.5 4 3 4.5 12
vii
x x
a b
t s t s
a b c d a b c d ac bd
x y x y x y x y
x y x xy y
x y x y x y
− + +
+ + +
+ −
+ − + − + + +
+ + + + −
+ − +
− + + − +
( ) ( ) ( )i a b c a b c+ + + −
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Solution:
(i)
( ) ( ) ( ) ( )2 2
3 2
3 2
5 5 25 5 5 5 25
5 5 25 25
5 5
x x x x x
x x x
x x x
− + + = + − + +
= + − − +
= + −
(ii)
( ) ( ) ( ) ( )2 3 2 3 3
2 3 2 3
2 3 2 3
5 3 5 3 5 3 5
3 5 15 5
3 5 20
a b a b b
a b a b
a b a b
+ + + = + + + +
= + + + +
= + + +
(iii)
( ) ( ) ( ) ( )2 2 2 2 2
3 2 2 3
t s t s t t s s t s
t st s t s
+ − = − + −
= − + −
(iv)
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2
2
2 2
2 2
4
a b c d a b c d ac bd
a c d b c d a c d b c d ac bd
ac ad bc bd ac ad bc bd ac bd
ac ac ac ad ad bc bc bd bd bd
ac
+ − + − + + +
= − + − + + − + + +
= − + − + + − − + +
= + + + − + − + − −
=
(v)
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
2 2 2 2
2 2 2 2
2 2
2 2
2 2 2
2 2 2 2
2 2 2 2
3 4
x y x y x y x y
x x y y x y x x y y x y
x xy xy y x xy xy y
x x y y xy xy xy xy
x y xy
+ + + + −
= + + + + − + −
= + + + + − + −
= + + − + + − +
= − +
(vi)
( ) ( )
( ) ( )
( ) ( )
2 2
2 2 2 2
3 2 2 2 2 3
3 3 2 2 2 2
3 3
x y x xy y
x x xy y y x xy y
x x y xy x y xy y
x y xy xy x y x y
x y
+ − +
= − + + − +
= − + + − +
= + + − + −
= +
(vii)
( ) ( )
( ) ( )
( ) ( ) ( )
2 2
2 2
2 2
1.5 4 1.5 4 3 4.5 12
1.5 1.5 4 3 4 1.5 4 3 4.5 12
2.25 6 4.5 6 16 12 4.5 12
2.25 6 6 4.5 4.5 16 12 12
2.25 16
x y x y x y
x x y y x y x y
x xy x xy y y x y
x xy xy x x y y y
x y
− + + − +
= + + − + + − +
= + + − − − − +
= + − + − − + −
= −
(viii)
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
2 2 2
2 2 2
2 2 2 2
a b c a b c
a a b c b a b c c a b c
a ab ac ab b bc ca bc c
a b c ab ab bc bc ca ca
a b c ab
+ + + −
= + − + + − + + −
= + − + + − + + −
= + − + + + − + −
= + − +
Chapter 9: Algebraic Expressions and Identities
Exercise 9.5 (Page 151 of Grade 8 NCERT)
Q1. Use a suitable identity to get each of the following products.
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
2 2 2 2
i 3 3
ii 2 5 2 5
iii 2 7 2 7
iv
v 1.1m 0.4 1.1m 0.4
vi
vii 6 7 6 7
viii
ix
1 13 3
2 2
3 3
2 4 2 4
x 7 9 7 9
x x
y y
a a
a
a a
x y x
b a b
x x
a c a c
y
a b a b
− −
+
+ +
+ +
− −
− +
+ − +
− +
− + −
+
+
− −
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Reasoning:
i) By using the distributive law, we can carry out the multiplication term by term.
ii) In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Solution:
The products will be as follows.
(i)
( ) ( ) ( )
( ) ( )( ) ( ) ( )
2
2 2 2 2 2
2
3 3 3
2 3 3 2
6 9
x x x
x x a b a ab b
x x
+ + = +
= + + + = + +
= + +
(ii)
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2
2 2 2 2 2
2
2 5 2 5 2 5
2 2 2 5 5 2
4 20 25
y y y
y y a b a ab b
y y
+ + = +
= + + + = + +
=
+ +
(iii)
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2
2 2 2 2 2
2
2 7 2 7 2 7
2 2 2 7 7 2
4 28 49
a a a
a a a b a ab b
a a
− − = −
= − + − = − +
=
− +
(iv)
( ) ( ) ( )
2
22 2 2 2
2
1 1 13 3 3
2 2 2
1 13 2 3 2
2 2
19 3
4
a a a
a a a b a ab b
a a
− − = −
= − + − = − +
= − +
(v)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2
2
1.1m 0.4 1.1m 0.4 1.1m 0.4
1.21m 0.16
a b a b a b − + = − +
= −
− = −
(vi)
( ) ( ) ( ) ( )
( ) ( ) ( )( )
2 2 2 2 2 2 2 2
2 22 2 2 2
4 4
a b a b b a b a
b a a b a b a b
b a
+ − + = + −
= − + − = −
= −
(vii)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2
2
6 7 6 7 6 7
36 49
x x x a b a b a b
x
− + = − + −
=
= −
−
(viii)
( ) ( ) ( )
( ) ( )( ) ( ) ( )
2
2 2 2 2 2
2 2
2 2
2
a c a c a c
a a c c a b a ab b
a ac c
− + − + = − +
= − + − + + =
−
=
+ +
+
(ix)
( )
2
2 22 2 2
2 2
3 3 3
2 4 2 4 2 4
3 32 2
2 2 4 4
3 9
4 4 16
x y x y x y
x x y ya b a ab b
x xy y
+ + = +
= + + + = + +
= + +
(x)
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2
2 2 2 2 2
2 2
7 9 7 9 7 9
7 2 7 9 9 2
49 126 81
a b a b a b
a a b b a b a ab b
a ab b
− − = −
= − + − = − +
= − +
Q2. Use the identity ( ) ( ) ( )2x a x b x a b x ab+ + = + + + to find the following
products.
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
2 2
i 3 7
ii 4 5 4 1
iii 4 5 4 1
iv 4 5 4 1
v 2 5 2 3
vi 2 9 2 5
vii 4 2
x x
x x
x x
x x
x y x y
a a
xyz xyz
+ +
+ +
− −
+ −
+ +
+ +
− −
Difficulty level: Easy
Known:
( )( ) ( )2 x a x b x a b x ab+ + = + + +
Unknown:
Simplification
Solution:
The products will be as follows.
(i)
( ) ( ) ( ) ( ) ( )2
2
3 7 3 7 3 7
10 21
x x x x
x x
+ + = + + +
= + +
(ii)
( ) ( ) ( ) ( ) ( ) ( ) ( )2
2
4 5 4 1 4 5 1 4 5 1
16 24 5
x x x x
x x
+ + = + + +
= + +
(iii)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2
2
4 5 4 1 4 5 1 4 5 1
16 24 5
x x x x
x x
− − = + − + − + − −
= − +
(iv)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2
2
4 5 4 1 4 5 1 4 5 1
16 16 5
x x x x
x x
+ − = + + + − + + −
= + −
(v)
( ) ( ) ( ) ( ) ( ) ( ) ( )2
2 2
2 5 2 3 2 5 3 2 5 3
4 16 15
x y x y x y y x y y
x xy y
+ + = + + +
= + +
(vi)
( ) ( ) ( ) ( ) ( ) ( ) ( )2
2 2 2 2
4 2
2 9 2 5 2 9 5 2 9 5
4 28 45
a a a a
a a
+ + = + + +
= + +
(vii)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2
2 2 2
4 2 4 2 4 2
6 8
xyz xyz xyz xyz
x y z xyz
− − = + − + − + − −
= − +
Q3. Find the following squares by using the identities.
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
2
2
2
2
2
2
2
2 3
3 2
i 7
ii 3
iii 6 5
iv
v 0.4 0.5
vi 2 5
b
xy z
x y
p
n
x
m
q
y y
−
+
−
−
+
+
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Reasoning: 2 2 2
2 2 2
2 2
( ) 2
( ) 2
( )( )
a b a ab b
a b a ab b
a b a b a b
+ = + +
− = − +
+ − = −
Solution:
(i)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2
2
7 2 7 7 2
14 49
b b b a b a ab b
b b
− = − + − = − +
= − +
(ii)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2
2 2 2
3 2 3 3 2
6 9
xy z xy xy z z a b a ab b
x y xyz z
+ = + + + = + +
=
+ +
(iii)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 22 2 2 2 2
4 2 2
6 5 6 2 6 5 5 2
36 60 25
x y x x y y a b a ab b
x x y y
− = − + − = − +
− +
=
(iv)
( )2 2 2
2 2
2 2
22 3 2 2 3 32
3 2 3 3 2 2
4 92
9 4
2m n m m n n
m m
a a a b
n n
b b + = + +
+
+
+
+
=
= +
(v)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2
2 2
0.4 0.5 0.4 2 0.4 0.5 0.5 2
0.16 0.4 0.25
p q p p q q a b a ab b
p pq q
− = − + − = − +
− +
=
(vi)
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2
2 2 2 2
2 5 2 2 2 5 5 2
4 20 25
xy y xy xy y y a b a ab b
x y xy y
+ = + + + = + +
= +
+
Q4. Simplify.
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
22 2
2 2
2 2
2 2
2 2
2 2
22 2 3 2
i
ii 2 5 2 5
iii 7 8 7 8
iv 4 5 5 4
v 2.5 1.5 1.5 2.5
vi 2
vii 2
a b
x x
m n m n
m n m n
p q p q
ab bc ab c
m n m m n
−
+ − −
− + +
+ + +
− − −
+ −
− +
Difficulty level: Medium
Known:
Expressions
Unknown:
Simplification
Reasoning: 2 2 2
2 2 2
2 2
( ) 2
( ) 2
( )( )
a b a ab b
a b a ab b
a b a b a b
+ = + +
− = − +
+ − = −
Solution:
( ) ( )
( ) ( )( ) ( ) ( )
22 2
2 2 22 2 2 2 2 2
4 2 2 4
2 2
2
i a b
a a b b a b a ab b
a a b b
−
= − + − = − +
= − +
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )( ) ( )
( )
( )
2 2
2 2 2 2
2 2 2
2 2 2
2 2
2 2
2 5 2 5
2 2 2 5 5 2 2 2 5 5
2
2
4 20 25 4 20 25
4 20 25 4 20 25
40
ii x x
x x x x
a b a ab b
a b a ab b
x x x x
x x x x
x
+
+ − −
= + + − − +
− = − +
+ = +
= + + −
=
− +
+ + −
+ −
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
2 2
2 2 2 2
2 22 2 2 2
2
7 8 7 8
7 2 7 8 8 7 2 7 8 8
2 2
49 112
iii m n m n
m m n n m m n n
a b a ab b and a b a ab b
m
−
− + +
= + + + +
− = − + + = + +
= − 2 2 64 49 112mn n m+ + + 2
2 2
64
98 128
mn n
m n
+
= +
( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )
2 2
2 2 2 2 2 2 2
2 2 2 2
2 2
4 5 5 4
4 2 4 5 5 5 2 5 4 4 2
16 40 25 25 40 16
41 80 41
iv m n m n
m m n n m m n n a b a ab b
m mn n m mn n
m mn n
+ + +
= + + + + + + = + +
= + + + + +
= + +
( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )( ) ( )
( )
2 2
2 2 2 2
2 2 2
2 2 2 2
2
2.5 1.5 1.5 2.5
2.5 2 2.5 1.5 1.5 1.5 2 1.5 2.5 2.5
2
6.25 7.5 2.25 2.25 7.5 6.25
6.25 7.5
v p q p q
p p q q p p q q
a b a ab b
p pq q p pq q
p
− − −
= − + − − +
− = − +
= − + − − +
=
−
2.25pq + 2 2.25q − 2 7.5p + 2
2 24
] 6.25
4
pq q
p q
−
= −
( ) ( )
( ) ( )( ) ( ) ( )
2 2
2 2 22 2 2
2 2 2
2
2 2 2
2
vi ab bc ab c
ab ab bc bc ab c a b a ab b
a b ab c
+ −
= + + − + =
= +
+ +
2 2 2 2b c ab c+ −2 2 2 2 a b b c= +
( ) ( )
( ) ( ) ( ) ( ) ( )
22 2 3 2
2 2 22 2 2 2 3 2 2 2
4 3 2
2
2 2 2
2
vii m n m m n
m m n m n m m n a b a ab b
m m n
− +
= − + + − = − +
= − 4 2 3 22n m m n+ +4 4 2 m n m= +
Q5. Show that
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 2
2 2
2 2 2
2
2 2
i 3 7 84 3 7
ii 9 5 180 9 5
ii4 3 16 9
23 4 9 1
i
iv 4 3 4 3 48
0
6
v
m n mn
x x x
p q pq p q
pq q pq q pq
a b a b b c b
m
c c a c a
n
+ − = −
− + = +
+ − − =
− + + − + + −
− + = +
+
=
Difficulty level: Hard
Known:
LHS and RHS expression
Unknown:
Verification of LHS = RHS
Solution:
( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
2
2 2
2
2
2
2 2
2
i . . 3 7 84
3 2 3 7 7 84
9 42 49 84
9 42 49
. . 3 7
3 2 3 7 7
9 42 49
. . . .
L H S x x
x x x
x x x
x x
R H S x
x x
x x
L H S R H S
= + −
= + + −
= + + −
= − +
= −
= − +
= − +
=
( ) ( )
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
2
2 2
2 2
2 2
2
2 2
2 2
ii . . 9 5 180
9 2 9 5 5 180
81 90 25 180
81 90 25
. . 9 5
9 2 9 5 5
81 90 25
. . . .
L H S p q pq
p p q q pq
p pq q pq
p pq q
R H S p q
p p q q
p pq q
L H S R H S
= − +
= − + +
= − + +
= + +
= +
= + +
= + +
=
( )2
2 2
2
4 3iii . . 2
3 4
4 4 3 32 2
3 3 4 4
162
9
L H S m n mn
m m n n mn
m mn
= − +
= − + +
= − 292
16n mn+ +
2 216 9
9 16
. . . .
m n
L H S R H S
= +
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
2 2
2 2 2 2
2 2 2 2 2 2 2 2
2 2
iv . . 4 3 4 3
4 2 4 3 3 4 2 4 3 3
16 24 9 16 24 9
16
L H S pq q pq q
pq pq q q pq pq q q
p q pq q p q pq q
p q
= + − −
= + + − − +
= + + − − +
= 2 224 9pq q+ + 2 216 p q− 2 224 9pq q+ −
248
. . . .
pq
L H S R H S
=
=
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )2 2 2 2 2 2
2
v . . L H S a b a b b c b c c a c a
a b b c c a
a
= − + + − + + − +
= − + − + −
= 2b− 2b+ 2c− 2c+ 2a−
0
. . . . .L H S R H S
=
=
Q6.Using identities, evaluate.
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
2
2
2
2
2
2
i 71
ii 99
iii 102
iv 998
v 5.2
vi 297 303
vii 78 82
viii 8.9
ix 1.05 9.5
Difficulty level: Hard
Known:
Expressions
Unknown:
Values of the expressions
Reasoning: 2 2 2
2 2 2
2 2
( ) 2
( ) 2
( )( )
a b a ab b
a b a ab b
a b a b a b
+ = + +
− = − +
+ − = −
Solution:
( ) ( )
( ) ( ) ( ) ( ) ( )
22
2 2 2 2 2
i 71 70 1
70 2 70 1 1 2
4900 140 1
5041
a b a ab b
= +
= + + + = + +
= + +
=
( )
( ) ( ) ( ) ( ) ( )
2 2
2 2 2 2 2
ii 99 1 00 1
100 2 100 1 1 2
10000 200 1
980
(
1
)
a b a ab b
= −
= − + − = − +
= − +
=
( ) ( )
( ) ( ) ( ) ( ) ( )
22
2 2 2 2 2
iii 102 100 2
100 2 100 2 2 2
10000 400 4
10404
a b a ab b
= +
= + + + =
=
+
=
+ +
+
( ) ( )
( ) ( ) ( ) ( ) ( )
22
2 2 2 2 2
iv 998 1000 2
1000 2 1000 2 2 2
1000000 4000 4
996004
a b a ab b
= −
= − + − = − +
= − +
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2 2
2 2 2 2 2
v 5.2 5.0 0.2
5.0 2 5.0 0.2 0.2 2
25 2 0.04
27.04
a b a ab b
= +
= + + + = + +
= + +
=
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2 2
vi 297 303 300 3 300 3
300 3
90000 9
89991
a b a b a b
= − +
= − + − = −
= −
=
( ) ( ) ( )
( ) ( ) ( ) ( )2 2 2 2
vii 78 82 80 2 80 2
80 2
6400 4
6396
a b a b a b
= − +
= − + − = −
= −
=
( ) ( )
( ) ( ) ( ) ( ) ( )
22
2 2 2 2 2
viii 8.9 9.0 0.1
9.0 2 9.0 0.1 0.1 2
81 1.8 0.01
79.21
a b a ab b
= −
= − + − = − +
= −
=
+
( )
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
ix 1.05 9.5 1.05 0.95 10
1 0.05 1 0.05 10
1 0.05 10
1 0.0025 10
0.9975 10
9.975
a b a b a b
=
= + −
= −
= − + − =
=
=
−
Q 7. Using ( ) ( )2 2 ,a b a b a b− = + − find
( )
( ) ( ) ( )
( )
( )
2 2
2 2
2 2
2 2
i 51 – 49
ii 1.02 0.98
iii 153 147
iv 12.1 7.9
−
−
−
Difficulty level: Medium
Known:
( ) ( )2 2 ,a b a b a b− = + −
Unknown:
Results of the given expression with their corresponding values
Solution:
( ) ( ) ( )
( ) ( )
2 2i 51 49 51 49 51 49
100 2 200
− = + −
= =
( ) ( ) ( ) ( ) ( )
( ) ( )
2 2ii 1.02 0.98 1.02 0.98 1.02 0.98
2 0.04
0.08
− = + −
=
=
( ) ( ) ( )
( ) ( )
2 2iii 153 147 153 147 153 147
300 6
1800
− = + −
=
=
( ) ( ) ( )
( ) ( )
2 2iv 12.1 7.9 12.1 7.9 12.1 7.9
20.0 4.2
84
− = + −
=
=
Q8. Using ( ) ( ) ( )2x a x b x a b x ab+ + = + + + , find
( )
( )
( )
( )
i 103 104
ii 5.1 5.2
iii 103 98
iv 9.7 9.8
Difficulty level: Medium
Known:
( ) ( ) ( )2x a x b x a b x ab+ + = + + +
Unknown:
Results of the given expression with their corresponding values
Solution:
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )2
i 103 104 100 3 100 4
100 3 4 100 3 4
10000 700 12
10712
= + +
= + + +
= + +
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )2
ii 5.1 5.2 5 0.1 5 0.2
5 0.1 0.2 5 0.1 0.2
25 1.5 0.02
26.52
= + +
= + + +
= + +
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )2
iii 103 98 100 3 100 2
100 3 2 100 3 2
10000 100 6
10094
= + −
= + + − + −
= + −
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( )
2
iv 9.7 9.8 10 0.3 10 0.2
10 0.3 0.2 10 0.3 0.2
100 0.5 10 0.06
100 5 0.06
95 0.06
95.06
= − −
= + − + − + − −
= + − +
= − +
= +
=
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