Algebraic Expressions

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.

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

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

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

_____________________________

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

  1. 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

  1. Add:
  2. 5a+3b, a-2b and 3a+5b

Solution:

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

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

=9a+6b

 

  1. 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

  1. -2x2y+3xy2 from 8x2y

Solution:

-2x2y  +3xy2

+8x2y

(-)        (-)

———————

-2x2y  -5xy

———————

 

  1. 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

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

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:

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

 

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:

  • 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

 

 

 

 

 

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