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+3xy2 from 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) ( 1/2 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 m2– 1/9 ) —————(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 m2 – 1/9 ) ( 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 m2 – 1/9 ) ( 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.
- 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
1 thought on “Algebraic Expressions – Full Chapter – Class VIII”
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