# Algebraic Expressions – Full Chapter – Class VIII

Let us recall what we have studied in earlier classes about algebraic expressions.

• Constant: A symbol which has a fixed value.

ex: 5,-8,√6,2+√3 and etc.

• Variable: A symbol which does not have any fixed value but may be assigned value according to the requirements.

ex: p,q,x,y etc.

• Term: A number , a variable or a combination of numbers and variables.

ex: 8,x,3x,5ab etc.

• Algebraic expressions: A single term or a combination of two or more terms connected by additive(both addition and subtraction)  and multiplication(both multiplication and division) symbols.

ex: 7-y, 3x2+4y etc.

• monomial: An algebraic expression which contains only one term

ex: x, 7y, 8xy, x/y etc.

• binomial: An algebraic expression which contains two terms

ex: x+2x, 7y-x2, 8xy+9, x/y-xy etc.

• trinomial: An algebraic expression which contains three terms

ex: x+2x+3,7y-8xy+x/y etc.

• polynomial: An algebraic expression which contains only the variable(s) with non negative integer exponent(s)

ex: x2-4x, x-4xy+y2, 6-5y+xy+x2y.

## Algebraic Expressions – Exercise 2. 1. 1

1. Separate the constants and variables from the following.

12+z,–x/5, -3/7, √x, √3, (2/3)xy, (5xy)/2, 7, 7-x, 6x+4y, -7z, (8yz)/3x, y+4, y/4 and (2x)/(8yz).

Solution:

constants: -3/7, √3, 7

Variables: 12+z,–x/5, √x, √3, (2/3)xy, (5xy)/2, 7-x, 6x+4y, -7z, (8yz)/3x, y+4, y/4 and (2x)/(8yz)

1. Separate the monomials, binomials and trinomials from the following.

7xyz, 9-4y, 4y2-xz, x-2y+3z, 7x+z2, 8x/y, (8/5)x2y2, 4+5y-6z

Solution:

monomials:

7xyz, 8x/y, (8/5)x2y2

binomials:

7x+z2, 4y2-xz, 9-4y

trinomials:

x-2y+3z, 4+5y-6z

## Algebraic expressions:

These are the expressions which contain ordinary numbers, variables (like x or y) and operators (like add, subtract, multiply and divide).

Ex:

a+b, x+1 etc.

### Like terms:

The terms that have the same variables and powers.

ex: 5x, 2x, -9x, (1/3)x etc.

### Unlike terms:

The terms terms that have the different variables and same/different powers.

Ex:  -x, xy, xy2   etc.

### Addition and subtraction of polynomials – Algebraic expressions

Let us first review the properties of addition and multiplication.

1. Sum of two positive integer is a positive integer; (+7)+(+5)=+7+5=+12

1. Sum of two negative integer is a negative integer; (-7)+(-5)=-12

1. Sum of positive integer and negative integer is positive if the absolute value of the negative integer is smaller than the positive integer; (+7)+(-5)=+2

1. Sum of positive integer and negative integer is positive if the absolute value of the negative integer is larger than the positive integer; (-7)+(+5)=-2

1. Product of two positive integer is also a positive integer; (+7)X(+5)=+35

1. Product of two negative integer is also a positive integer; (-7)X(-5)=+35

1. Product of a positive integer and a negative integer is a negative integer; (+7)X(-5)=-35
2. Product of a negative integer and a positive integer is a negative integer; (-7)X(+5)=-35

Example 1: Add 5x2y, -7x2y, 9x2y

Solution:

Recall rule 1:We have,

(5x2y)+(-7x2y)+(9x2y)=(5+(-7)+9)x2y

=(5-7+9)x2y

=7x2y

+5x2y

-7x2y

+9x2y

_______________________

7x2y

_______________________

Note: We are adding coefficients and retaining the variables as it is.

Examples 2: Add 7x2-4x+5 and 9x-10.

Solution:

Note: here we have unlike terms, we can add only like terms . We have to write like terms one below other to easy addition.

7x2      -4x      +5

9x        -10

______________________

7x2        +5x     -5

______________________

Example 3: Add 8xy+4yz-7zx, 6yz+11zx-6y, -5xz+6x-2yx.

Solution:

Note: here we have unlike terms, we can add only like terms . We have to write like terms one below other to easy addition.

+8xy   +4yz         -7zx

+ 6yz       +11zx                        -6y

-2yx                     -5xz           +6x

_____________________________

6xy     +10yz        -xz       +6x     -6y

_____________________________

## Algebraic Expressions – Exercise 2. 1. 3

1. Classify into like terms.

4x2, (1/3)x, -9x3, xy, 6x3, 4y, -74x3, 8xy, 7xyz, 3x2

Solution:

Like terms:

group1: (1/3)x,

group2: xy, 8xy

group 4: 4x2, 3x2

group 5: -9x3, 6x3, -74x3

group 6: 7xyz.

1. 7x-9y+3-3x-5y+8

Solution:

(7+3)x+(-9-5)y+(3+8)=10x+(-14)y+11

=10x-14y+1

3. 3x2+5xy-4y2+x2-8xy-5y2

Solution:

3x2+5xy-4y2+x2-8xy-5y2  =(3x2+x2)+(-8xy+5xy)+(-4y2-5y2 )

=(4x2)+(-3xy)+(-9y2 )

=4x2-3xy-9y2

5a+3b, a-2b and 3a+5b

Solution:

=(5a+3b)+( a-2b)+(3a+5b)

=(5+1+3)a+(3-2+5)b

=9a+6b

4. x3-x2y+5xy2+y3, –x3-9xy2+y3 and 3x2y+9xy2

Solution:

=(x3-x2y+5xy2+y3)+( –x3-9xy2+y3 )+(3x2y+9xy2 )

=(x3-x3)+(-x2y+3x2y)+(5xy2-9xy2+9xy2)+(y3+y3)

=(0)+(2x2y)+(5xy2)+(2y3)

=2x2y+5xy2+2y3

5. -2x2y+3xy2 from 8x2y

Solution:

-2x2y  +3xy2

+8x2y

(-)        (-)

———————

-2x2y  -5xy

———————

6. a-b-2c from 4a+6b-2c

Solution:

a          -b        -2c

4a       +6b     -2c

(-)        (-)        (+)

———————

-3a      -7b      +0

## Multiplication of polynomials

### Multiplying a monomial by a monomial:

Example 6: find the product of 4x*5y*7z

Solution:

we have 4x*5y*7z=(4*5*7)xyz

=140xyz

### Multiplying a monomial by a binomial:

example 7: determine the product (8y+3)*4x

Solution:

(8y+3)*4x=(8y*4x)+(3*4x)

=32xy+12x

### Multiplying a binomial by a binomial:

Example 8: find the product (4a+6b) and (5a+7b)

Solution:

(4a+6b)*(5a+7b)=4a*(5a+7b)+6b*(5a+7b)

=[4a*5a+4a*7b]+[6b*5a+6b*7b]

=[20a2+28ab]+[30ab+42b2]

=20a2+58ab+42b2

## Algebraic Expressions – Exercise 2. 1.4

1. Complete the following table of products of two monomials.
 * 3x -6y 4x2 -8xy 9x2y -11x3y2 3x -6y 4x2 -8xy 9x2y -11x3y2

Solution:

 * 3x -6y 4x2 -8xy 9x2y -11x3y2 3x 9x2 -18xy 12x3 -24yx2 27yx3 -33x4y2 -6y -18xy +36y2 -24yx2 +48xy2 -54x2y2 +66y3x3 4x2 12x3 -24yx2 16x4 -32yx3 36yx4 -44y2x5 -8xy -24x2y +48xy2 -32x3y +64x2y2 -72x3y2 +88x4y3 9x2y 27yx3 -54x3y 36x4y -72x3y2 81x4y2 -99x5y3 -11x3y2 -33x4y2 +66x3y3 -44x5y2 +88x4y3 -99x5y3 +121x6y4

1. Find the products:

(i) (5x+8)3

Solution:

(5x+8)3=5x*3+8*3

=15x+24

(ii) (-3pq)(-15p3q2-q3)

Solution:

(-3pq)(-15p3q2-q3)

=(-3pq)* (-15p3q2)+(-3pq)*(-q3)

=+45p4q3+3pq4

(iii) (a3-b3)

Solution:

(a3-b3)= a3 – b3

(iv) –x(x-15)

Solution:

–x(x-15)=( -x)*x+(-15)*(-x)

=-x2+15x

1. Simplify the following

(i) (2x2y-xy)(3xy-5)

Solution:

(2x2y-xy)(3xy-5)= (2x2y)*(3xy-5)+(-xy)(3xy-5)

=6x3y2-10x2y-3x2y2+5xy

=6x3y2-13x2y2+5xy

(ii) (3x2y2+1)(4xy-6xy2)

Solution:

(3x2y2+1)(4xy-6xy2)= (3x2y2)(4xy-6xy2)+(1)(4xy-6xy2)

=12x3y3-18x3y4+4xy-6xy2

(iii) (3x2+2x)(2x2+3)

Solution:

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

=6x4+9x2+4x3+6x

(iv) (2m3+3m)(5m-1)

Solution:

(2m3+3m)(5m-1)= (2m3)(5m-1)+3m(5m-1)

=10m4-2m3+15m2-3m

## SPECIAL PRODUCT – Algebraic Expressions

In this let us study a special product, product of two binomials. Consider the product:

(x+a)(x+b)=x(x+b)+a(x+b)

=x2+xb+xa+ab

=x2+x(a+b)+ab

We have used commutative property and the distributive property, xb=bx and (ax+bx)=x(a+b)

We say (x+a)(x+b)=x2+(a+b)x+ab is an Identity.

Example 9: Find the product (x+6)(x+7)

Solution:

We observe that, given problem (x+6)(x+7) is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here a=6 and b=7

Let us substitute the value of a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(x+6)(x+7)=x2+x(6+7)+(6*7)

=x2+13x+42

Example 10:Compute (2x+5)(2x+3)

Solution:

We observe that, given problem (2x+5)(2x+3) is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here a=5 and b=3 and also x=2x.

Let us substitute the value of a, b and x in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(x+6)(x+7)=(2x)2+2x(5+3)+(5*3)

=4x2+16x+15

Example 11: Find the product 103*96 using the above identity.

Solution:

It is mentioned that we have bring the given problem 103*96 to the form of identity (x+a)(x+b)=x2+(a+b)x+ab.

[Note: We know that, in the above identity, we have (x+a)(x+b), that means there should be a common value x and two different values for a and b]

So, let us consider (100+3)(100-2),which satisfies the condition.

Here x=100,a=3 and b=-2

Let us substitute the value of x, a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

103*96=(100+3)(100-2)=1002+100(3-2)+(3*(-2))

=10000+100(1)+(-6)

=10000+100-6

=10094

Example 13.Find the product (p2-5)(p2-3)

Solution:

We observe that, given problem (x+6)(x+7) is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here x=p2 and a=-5, b=-3

Let us substitute the value of x, a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(p2-5)(p2-3)=(p2)2+p2((-5)+(-3))+((-5)*(-3))

=(p2)2+(-8)p2+(+15)

=p4-8p2+15

## IDENTITIES – Algebraic Expressions:

We have some special identities which help are helpful in solving problems. Consider,

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

Now let us multiply (a+b)(a+b),

therefore, (a+b)2=(a+b)(a+b)=a*(a+b)+b*(a+b)

=(a2+ab)+(ab+b2)

=a2+2ab+b2, where ab=ba.

We have got another identity,

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

Let us consider, (a-b)2

We know, (a-b)2=(a-b)(a-b)

Now let us multiply (a-b)(a-b),

therefore, (a-b)2=(a-b)(a-b)=a*(a-b)-b*(a-b)

=(a2-ab)-(ab-b2)

=a2-2ab+b2, where ab=ba.

We have got another identity,

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

These are called standard Identities.

Example 14. Find (2x+3y)2

Solution:

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=2x and b=3y

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

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

=(4x2+12xy+9y2)

Example 15.            Compute (4.9)2

Solution:

Let us use the identity, (a-b)2=(a2-2ab+b2)

(4.9)2=(5.0-0.1)2

here a=5 and b=0.1

let us substitute the value of a and b in the identity (a-b)2=(a2-2ab+b2)

(5-0.1)2=((5)2-2(5)(0.1)+(0.1)2)

=(52-10.1+0.01)

=25-10.1+0.01

=24.01

Example 14. Compute 54*46

Solution:

Here let us the identity (a+b)(a-b)=a2-b2

54*46=(50+4)(59-4)

here a=50 and b=4

Substitute the value of  and b in the identity (a+b)(a-b)=a2-b2

54*46=(50+4)*(50-4)

=(50)2-(4)2

=2500-16

=2484

## Algebraic Expression – Exercise 2.1.5

1. Find the product:

(i) (a+3)(a+5)

Solution:

We observe that, given problem (a+3)(a+5) is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here a=3 and b=5 and also x=a.

Let us substitute the value of a, b and x in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(a+3)(a+5)=(a)2+a(5+3)+(5*3)

=a2+8a+15

(ii) (3t+1)(3t+4)

Solution:

We observe that, given problem (3t+1)(3t+4 is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here a=1 and b=4 and also x=3t.

Let us substitute the value of a, b and x in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(3t+1)(3t+4)=(3t)2+3t(4+1)+(4*1)

=9t2+15t+4

(iii) (a-8)(a+2)

Solution:

We observe that, given problem (a-8)(a+2) is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here a=-8 and b=2 and also x=a.

Let us substitute the value of a, b and x in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(a-8)(a+2)=(a)2+a((-8)+(2))+((-8)*2)

=a2-6a-16

(iv) (a-6)(a-2)

Solution:

We observe that, given problem (a-6)(a-2) is in the form of the identity (x+a)(x+b)=x2+(a+b)x+ab.

Here a=-6 and b=-2 and also x=a.

Let us substitute the value of a, b and x in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

(a-6)(a-2)=(a)2+a((-6)+(-2))+((-6)*(-2))

=a2-8a+12

1. Evaluate:

(i) 53*55

Solution:

we have bring the given problem 103*96 to the form of identity (x+a)(x+b)=x2+(a+b)x+ab.

[Note: We know that, in the above identity, we have (x+a)(x+b), that means there should be a common value x and two different values for a and b]

So, let us consider (50+3)(50+5),which satisfies the condition.

Here x=50,a=3 and b=5

Let us substitute the value of x, a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

53*55=(50+3)(50+5)=502+50(3+5)+(3*5)

=2500+50(8)+(15)

=2500+400+15

=2915

(ii) 102*106

Solution:

we have bring the given problem 102*102 to the form of identity (x+a)(x+b)=x2+(a+b)x+ab.

[Note: We know that, in the above identity, we have (x+a)(x+b), that means there should be a common value x and two different values for a and b]

So, let us consider (100+2)(100+6),which satisfies the condition.

Here x=100,a=2 and b=6

Let us substitute the value of x, a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

102*106=(100+2)(100+6)=1002+100(2+6)+(2*6)

=10000+100(8)+(12)

=10000+800+12

=10812

(iii) 34*36

Solution:

we have bring the given problem 34*36 to the form of identity (x+a)(x+b)=x2+(a+b)x+ab.

[Note: We know that, in the above identity, we have (x+a)(x+b), that means there should be a common value x and two different values for a and b]

So, let us consider (30+4)(30+6,which satisfies the condition.

Here x=30,a=4 and b=6

Let us substitute the value of x, a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

34*36=(30+4)(30+6)=302+30(4+6)+(4*6)

=900+30(10)+(24)

=900+30+24

=954

(iv) 103*96

Solution:

we have bring the given problem 103*96 to the form of identity (x+a)(x+b)=x2+(a+b)x+ab.

[Note: We know that, in the above identity, we have (x+a)(x+b), that means there should be a common value x and two different values for a and b]

So, let us consider (100+3)(100-4),which satisfies the condition.

Here x=100,a=3 and b=-2

Let us substitute the value of x, a and b in the identity (x+a)(x+b)=x2+(a+b)x+ab.

We have,

103*96=(100+3)(100-4)=1002+100(3-4)+(3*(-4))

=10000+100(-1)+(-12)

=10000-100-12

=9888

1. Can you imitate the case of identity (x+a)(x+b)=x2+x(a+b)+ab, and get a similar expression for the product (x+a)(x+b)(x+c)?

Solution:

We have, (x+a)(x+b)=x2+x(a+b)+ab.

Thus, (x+a)(x+b)(x+c)=[x2+x(a+b)+ab](x+c)

=(x)*[x2+x(a+b)+ab]+c*[x2+x(a+b)+ab]

=x3+x2(a+b)+xab+cx2+xc(a+b)+abc

= x3+x2a+x2b+xab+cx2+xca+cxb+abc

= x3+x2(a+b+c)+x(ca+cb+ab)+abc

1. Using the identity (a+b)2=a2+2ab+b2, simplify the following.

(i) (a+6)2

Solution:

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=a and b=6

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(2x+3y)2=(a)2+2(a)(6)+(6)2)

=(a2+12a+362)

(ii)(3x+2y)2

Solution:

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=3x and b=2y

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

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

=(9x2+12xy+4y2)

(iii) (2p+3q)2

Solution:

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=2p and b=3q

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(2p+3q)2=((2p)2+2(2p)(3q)+(3q)2)

=(4p2+12pq+9q2)

(iv) (x2+5)2

Solution:

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=x2 and b=5

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(x2+5)2=(x2)2+2(x2)(5)+(5)2)

=(x4+10x2+25)

1. Evaluate using the identity (a+b)2=(a2+2ab+b2).

(i) (34)2

Solution:

(34)2=(30+4)2

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=302 and b=4

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(30+4)2=(30)2+2(30)(4)+(4)2)

=(900+240+16)

=1156

(ii) (10.2)2

Solution:

(10.2)2=(10+0.2)2

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=10 and b=0.2

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(10+0.2)2=(10)2+2(10)(0.2)+(0.2)2)

=(100+4+0.04)

=104.04

(iii) (53)2

Solution:

(53)2=(50+3)2

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=50 and b=3

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(50+3)2=(50)2+2(3)(50)+(3)2)

=(2500+300+9)

=2809

(iv) (41)2

Solution:

(41)2=(40+1)2

Let us use the identity, (a+b)2=(a2+2ab+b2)

here a=40 and b=1

let us substitute the value of a and b in the identity (a+b)2=(a2+2ab+b2)

(40+1)2=(40)2+2(40)(1)+12)

=(1600+80+1)

=1681

1. Use the identity to (a-b)2=a2-2ab+b2,to compute.

(i) (x-6)2

Solution:

Let us use the identity, (a-b)2=a2-2ab+b2

here a=x and b=6

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(x-6)2=x2-2(x)(6)+62

=x2-12x+36

(ii) (3x-5y)2

Solution:

Let us use the identity, (a-b)2=a2-2ab+b2

here a=3x and b=5y

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(x-6)2=(3x)2-2(3x)(5y)+(5y)2

=9x2-30xy+25y2

(iii) (5a-4b)2

Solution:

Let us use the identity, (a-b)2=a2-2ab+b2

here a=5a and b=4b

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(5a-4b)2=(5a)2-2(5a)(4b)+(4b)2

=25a2-20ab+16b2

(iv) (p2-q2)2

Solution:

Let us use the identity, (a-b)2=a2-2ab+b2

here a=p2 and b=q2

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(x-6)2=(p2)2-2(p2)(q2)+(q2)2

=p4-2p2q2+q4

1. Evaluate using the identity (a-b)2=(a2-2ab+b2).

(i) (49)2

Solution:

492=(50-1)2

Let us use the identity, (a-b)2=a2-2ab+b2

here a=50 and b=1

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(50-1)2=502-2(50)(1)+12

=502-100+1

=2500-100+1

=2401

(ii) (9.8)2

Solution:

(9.8)2=(10-0.2)2

Let us use the identity, (a-b)2=a2-2ab+b2

here a=10 and b=0.2

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(10-o.2)2=102-2(10)(0.2)+(0.2)2

=102-8+0.04

=100-4+0.04

=96.04

(iii) (59)2

Solution:

(59)2=(60-1)2

Let us use the identity, (a-b)2=a2-2ab+b2

here a=60 and b=1

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(60-1)2=602-2(60)(1)+12

=602-120+1

=3600-120+1

=3481

(iv) (198)2

Solution:

1982=(200-2)2

Let us use the identity, (a-b)2=a2-2ab+b2

here a=200 and b=2

let us substitute the value of a and b in the identity (a-b)2=a2-2ab+b2

(200-1)2=2002-2(200)(2)+22

=2002-800+4

=39204

1. use the identity (a+b)(a-b)=a2-b2 to find the products.

(i) (x+6)(x-6)

Solution:

Let us use the identity, (a+b)(a-b)=a2-b2

here a=a and b=6

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(x+6)(x-6)=x2-62

=x2-36

(ii) (3x+5)(3x-5)

Solution:

Let us use the identity, (a+b)(a-b)=a2-b2

here a=3x and b=5

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(3x+5)(3x-5)=(3x)2-52

=9x2-25

(iii) (2a+4b)(2a-4b)

Solution:

Let us use the identity, (a+b)(a-b)=a2-b2

here a=2a and b=4b

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

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

=4a2-16b2

(iv) (2x/3  + 1)( 2x/3 – 1)

Solution:

Let us use the identity, (a+b)(a-b)=a2-b2

here a= 2x/3  and b=1

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

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

1. Evaluate these using identity:

(i) 55*45

Solution:

55*45=(50+5)*(50-5)

Let us use the identity, (a+b)(a-b)=a2-b2

here a=50 and b=5

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(50+5)(50-5)=(50)2-(5)2

=2500-25

=2475

(ii) 33*27

solution:

33*27=(30+3)(30-3)

Let us use the identity, (a+b)(a-b)=a2-b2

here a=30 and b=3

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(30+3)(30-3)=(30)2-(3)2

=900-9

=891

(iii) 8.5*9.5

Solution:

8.5*9.5=(9+0.5)(9-0.5)

Let us use the identity, (a+b)(a-b)=a2-b2

here a=9 and b=0.5

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(9+0.5)(9-0.5)=(9)2-(0.5)2

=81-(0.25)

=80.75

(iv) 102*98

Solution:

102*98=(100+2)(100-2)

Let us use the identity, (a+b)(a-b)=a2-b2

here a=100 and b=2

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(100+2)(100-2)=(100)2-(2)2

=10,000-4

=9996

1. Find the product:

(i) (x+3)(x-3)(x2+9)

Solution:

let us select (x+3)(x-3) from the given problem (x+3)(x-3)(x2+9),

Then, Let us use the identity, (a+b)(a-b)=a2-b2 for (x+3)(x-3)

here a=x and b=3

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

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

=(x2-9)—————(1)

Then, substitute equation (1) in the given problem,

(x+3)(x-3)(x2+9)=(x2-9)(x2+9)

[note: (x+3)(x-3)=(x2-9)]

Let us use the identity, (a+b)(a-b)=a2-b2 for (x+3)(x-3)(x2+9)=(x2-9)(x2+9)

here a=x2 and b=9

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(x+3)(x-3)(x2+9)=(x2-9)(x2+9)

=(x2)2-(9)2

So, (x+3)(x-3)(x2+9)=x4-81

(ii) (2a+3)(2a-3)(4a2+9)

Solution:

let us select (2a+3)(2a-3) from the given problem (2a+3)(2a-3)(4a2+9),

Then, Let us use the identity, (a+b)(a-b)=a2-b2 for (2a+3)(2a-3)

here a=2a and b=3

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

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

=((2a)2-9)

=(4a2-9) —————(1)

Then, substitute equation (1) in the given problem,

(2a+3)(2a-3)(4a2+9)=(4a2-9)(4a2+9)

[note: (4a+3)(4a-3)=(4a2-9)]

Let us use the identity, (a+b)(a-b)=a2-b2 for (2a+3)(2a-3)(4a2+9)=(4a2-9)(4a2+9)

here a=4a2 and b=9

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(2a+3)(2a-3)(4a2+9)=(4a2-9)(4a2+9)

=(4a2)2-(9)2

So, (2a+3)(2a-3)(4a2+9)=16a4-81

(iii) (p+2)(p-2)(p2+4)

Solution:

let us select (p+2)(p-2) from the given problem (p+2)(p-2)(p2+4),

Then, Let us use the identity, (a+b)(a-b)=a2-b2 for (p+2)(p-2)

here a=p and b=2

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(p+2)(p-2)=(p)2-(2)2

=(p2-4)—————(1)

Then, substitute equation (1) in the given problem,

(p+2)(p-2)(p2+4)=(p2-4)(p2+4)

[note: (p+2)(p-2)=(p2-4)]

Let us use the identity, (a+b)(a-b)=a2-b2 for (p+2)(p-2)(p2+4)=(p2-4)(p2+4)

here a=p2 and b=4

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(p+2)(p-2)(p2+4)=(p2-4)(p2+4)

=(p2)2-(4)2

So, (p+2)(p-2)(p2+4)=p4-16

(iv) ( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) ( 1/4 m2+ 1/9 )

Solution:

let us select ( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) from the given problem ( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) ( 1/4 m2+ 1/9 )

Then, Let us use the identity, (a+b)(a-b)=a2-b2 for( 1/2 m– 1/3 ) ( 1/2 m + 1/3 )

Here a = 1/2 m   and b = 1/3

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) = ( 1/4 m21/9 ) —————(1)

Then, substitute equation (1) in the given problem,

( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) ( 1/4 m2+ 1/9 )

Let us use the identity, (a+b)(a-b)=a2-b2 for

( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) ( 1/4 m2+ 1/9 ) = ( 1/4 m21/9 ) ( 1/4 m2+ 1/9 )

Here a= 1/4 m2 and b = 1/9

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) ( 1/4 m2+ 1/9 )= ( 1/4 m21/9 ) ( 1/4 m2+ 1/9 ) = ( 1/4 m2)2+ (1/9 )2

So, ( 1/2 m– 1/3 ) ( 1/2 m + 1/3 ) ( 1/4 m2+ 1/9 )= ( 1/4 m2)2+ (1/9 )2

(v) (2x-y)(2x+y)(4x2+y2)

Solution:

Let us select (2x-y)(2x+y) from the given problem (2x-y)(2x+y)(4x2+y2),

Then, Let us use the identity, (a+b)(a-b)=a2-b2 for (2x+y)(2x-y)

here a=2x and b=y

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(2x+y)(x-y)=(2x)2-(y)2

=(4x2-y2)—————(1)

Then, substitute equation (1) in the given problem,

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

[note: (2x+y)(2x-y)=(4x2-y2)]

Let us use the identity, (a+b)(a-b)=a2-b2 for (2x+y)(2x-y)(4x2+y2)=(4x2-y2)(4x2+y2)

here a=4x2 and b=y2

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

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

=(4x2)2-(y2)2

So, (2x+y)(2x-y)(4x2+y2)=16x4-y4

(vi) (2x-3y)(2x+3y)(4x2+9y2)

Solution:

let us select (2x-3y)(2x+3y) from the given problem (2x-3y)(2x+3y)(4x2+9y2),

Then, Let us use the identity, (a+b)(a-b)=a2-b2 for (2x+3y)(2x-3y)

here a=2x and b=3y

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

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

=(4x2-9y2)—————(1)

Then, substitute equation (1) in the given problem,

(2x+3y)(2x-3y)(4x2+9y2)=(4x2-9y2)(4x2+9y2)

[note: (2x+3y)(2x-3y)=(4x2-9y2)]

Let us use the identity, (a+b)(a-b)=a2-b2 for (2x+3y)(2x-3y)(4x2+9y2)=(4x2-9y2)(4x2+9y2)

here a=4x2 and b=9y2

let us substitute the value of a and b in the identity (a+b)(a-b)=a2-b2

(2x+3y)(2x-3y)(4x2+9y2)=(4x2-9y2)(4x2+9y2)

=(4x2)2-(9y2)2

So, (2x+3y)(2x-3y)(4x2+9y2)= (16)x4-(81)y4

## Glossary – Algebraic Expression:

• Constant: A symbol which has a fixed value.
• Variable: A symbol which does not have any fixed value but may be assigned value according to the requirements.
• Term: A number , a variable or a combination of numbers and variables.
• Algebraic expressions: A single term or a combination of two or more terms connected by additive(both addition and subtraction)  and multiplication(both multiplication and division) symbols.
• monomial: An algebraic expression which contains only one term
• binomial: An algebraic expression which contains two terms
• trinomial: An algebraic expression which contains three terms
• polynomial: An algebraic expression which contains only the variable(s) with non negative integer exponent(s)

### Addition and subtraction of polynomials

Let us first review the properties of addition and multiplication.

1. Sum of two positive integer is a positive integer; (+7)+(+5)=+7+5=+12

1. Sum of two negative integer is a negative integer; (-7)+(-5)=-12

1. Sum of positive integer and negative integer is positive if the absolute value of the negative integer is smaller than the positive integer; (+7)+(-5)=+2

1. Sum of positive integer and negative integer is positive if the absolute value of the negative integer is larger than the positive integer; (-7)+(+5)=-2

1. Product of two positive integer is also a positive integer; (+7)X(+5)=+35

1. Product of two negative integer is also a positive integer; (-7)X(-5)=+35

1. Product of a positive integer and a negative integer is a negative integer; (+7)X(-5)=-35
2. Product of a negative integer and a positive integer is a negative integer; (-7)X(+5)=-35

### Formulae:

• (x+a)(x+b)=x2+(a+b)x+ab.
• (a+b)2= a2+2ab+b2
• (a-b)2= a2-2ab+b2
• (a+b)(a-b)=a2-b2