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

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)

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.

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

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

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

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

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

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

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

This is horizontal addition.

Let us have vertical addition:

+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

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.

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

Add:

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+3xy²from 8x²y

Solution:

-2x²y +3xy²

+8x²y

(-) (-)

———————

-2x²y -5xy²

———————

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

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

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

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

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

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

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

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)

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

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

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

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

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

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

Solution:

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

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

Here a = ¹/₂m and b = ¹/₃

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

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

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

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

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

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

Here a= ¹/₄m² and b = ¹/₉

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

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

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

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

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

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

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

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

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

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

Algebraic Expressions – Full Chapter – Class VIII

Algebraic Expressions – Exercise 2. 1. 1

Algebraic expressions:

Like terms:

Unlike terms:

Addition and subtraction of polynomials – Algebraic expressions

Algebraic Expressions – Exercise 2. 1. 3

Multiplication of polynomials

Multiplying a monomial by a monomial:

Multiplying a monomial by a binomial:

Multiplying a binomial by a binomial:

Algebraic Expressions – Exercise 2. 1.4

SPECIAL PRODUCT – Algebraic Expressions

IDENTITIES – Algebraic Expressions:

Algebraic Expression – Exercise 2.1.5

Glossary – Algebraic Expression:

Addition and subtraction of polynomials

Formulae:

Published by Ashoora Arif

1 thought on “Algebraic Expressions – Full Chapter – Class VIII”

Algebraic Expressions – Exercise 2. 1. 1

Algebraic expressions:

Like terms:

Unlike terms:

Addition and subtraction of polynomials – Algebraic expressions

Algebraic Expressions – Exercise 2. 1. 3

Multiplication of polynomials

Multiplying a monomial by a monomial:

Multiplying a monomial by a binomial:

Multiplying a binomial by a binomial:

Algebraic Expressions – Exercise 2. 1.4

SPECIAL PRODUCT – Algebraic Expressions

IDENTITIES – Algebraic Expressions:

Algebraic Expression – Exercise 2.1.5

Glossary – Algebraic Expression:

Addition and subtraction of polynomials

Formulae:

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Published by Ashoora Arif

1 thought on “Algebraic Expressions – Full Chapter – Class VIII”