# Algebraic Expressions and Identities - Class 8 : Notes Algebraic Expressions:
Any expression containing constants, variables, and the operations like addition, subtraction, etc. is called as an algebraic expression.
Example: 5x, 2x – 3, x2 + 1, etc.

Relation between number line and expression:
For any given expression of the form (a + b) ,where a is variable and b is constant then the value of this expression will always lie at b units after the point a on the number line.
Example 1: The following figure shows a number line drawn for the expression x + 5. Here, X represents the variable x which is unknown.
Thus, the final point will definitely be at 5 units from X which is denoted by P.

1. Term:
A term is either a single number or variable and it can be combination of numbers and variable.
They are usually separated by different operators like +, -, etc.

Example 1: Some example of terms are y, 5, 2x, etc.

Example 2: Consider an expression 6x - 7 = 2.
Then, the terms in this expression are 6x, -7 and 2.

Example 3: Identify the terms for 0.7a – 1.2b + 0.5ab.
Solution: The terms for given expression are 0.7a, -1.2b and 0.5ab.

2. Factors:
Factors can be product of numbers or number and variable.

Example 1: Term 7x is made of two factors 7 and x.

Example 2: Number 6 is made of two factors 2 and 3, 1 and 6.

3. Coefficient
The number multiplied to variable is called as coefficient.

Example 1: The coefficient of the term 2x will be 2.

Example 2: The coefficient of the term 5ab will be 5.

Example 3: Identify the coefficients for 0.7a – 1.2b + 0.5ab.
Solution: The coefficients for the given expression are 0.7, -1.2 and 0.5.

4. Monomials:
The expressions which have only one term are called as monomials.

Example: 10, 3x, 5xy, 2x2, etc. are some monomials.

5. Binomials:
The expressions which have two terms are called as binomials.

Example: x + 10, 3x + 1, a + b , 7x2 + y2 etc. are some binomials.

6. Trinomials:
The expressions which have three terms are called as trinomials.

Example: 2x + y+ 10, 3y + 3x, a + b + c , 7x2 + y2 + 7 etc. are some trinomials.

7. Polynomials:
The expression which contains one or more terms with non-zero coefficient is called a polynomial. A polynomial can have any number of terms.

Example 1: 10, a + b, 7x + y + 5, w + x + y + z, etc.

Example 2: Classify following polynomials into monomials, binomials, trinomials or others:
(a) a +b         (b) 7             (c) ab + bc + cd + da          (d) 5x – 5y + 13xy
Solution: (a) Binomial        (b) Monomial          (c) Polynomial  (d) Trinomial

8. Like terms:
The terms which have same variables are known as like terms.

Example
:
5x and 7x; 2xy and 3yx; 4x2, 7x2, 9x2 and x2; etc. are some like terms.

9. Unlike terms:
The terms which do not have the same variables are known as unlike terms.
Example: 5x and 7y; 2xy and 3ax; 4x2, 7y2and 9z2; etc. are some unlike terms.

Addition and Subtraction of Algebraic Expressions:

When performing addition or subtraction, we can perform the operations only for the like terms.
Let us understand it by an example:

Example 1: Add 7x + y + 7 to 3x + 2y + 1.
SolutionWrite down both the given expression into separate rows such that like terms fall below each other

7x +  y +  7
+ 3x + 2y + 1
10x + 3y + 8    Ans.

Example 2: Subtract 2x2 + 5xy + 1 from 7x2 + 2xy + 2y + 3.
Solution:
7x2 + 2xy + 2y + 3
- 2x2 + 5xy         + 1
5x2 - 3xy + 2y +  2      Ans

Example 3: Add a – b + ab, b – c + bc and c – a + ac.
Solution:
a  - b + ab
+                b           -c + bc
+        -a                   c +        + ac
ab       + bc   + ac         Ans

Example 4: Subtract 4a2b – 3ab + 5ab2 – 8a + 7b – 10 from 18 – 3a – 11b + 5ab – 2ab2+5a 2b.
Solution:
18 – 3a – 11b + 5ab – 2ab2 + 5a 2b
-10 – 8a – 7b  – 3ab + 5 ab2 + 4a2b
28 + 5a – 4b + 8ab – 7ab2 +  a2b      Ans

Multiplication of Algebraic Expressions:
(i) Take note of following points for like terms:
(a) The coefficients will get multiplied.
(b) The power of the resultant variable will be the addition of the individual powers.

Example 1: Product of 2x and 3x will be 6x2.

Example 2: Product of 2x, 3x and 4x will be 24x3.

(ii) Take note of following points for unlike terms:
(a) The coefficients will get multiplied.
(b) If all the variables are different then there will be no change in the power of variables.
(c) If some of the variables are same then the respective power of variables will be added.

Example 1: Product of 2x and 3y will be 6xy.

Example 2: Product of 2x, 3y and 4z will be 24xyz.

Example 3: Product of 2x2, 3x and 4y will be 24x3y.

1. Multiplying a Monomial by a Monomial:

(a) Multiplication of two monomials:
Let us look at some examples:

Example 1: Multiplication of terms 4 and y will be 4y.

Example 2: Multiplication of terms 4x and 3y will be 12xy.

Example 3: Multiplication of terms 4x and x will be 4x2.

(b) Multiplication of three or more monomials:
Let us look at some examples:

Example 1: Multiplication of terms 4, x, and y will be 4xy.

Example 2: Multiplication of terms 4x, 3y, 2 and z will be 24xyz.

Example 3: Multiplication of terms 4x3, x4, y4 and 2y will be 8x7y5

2. Multiplying a Monomial by a Polynomial:

(a) Multiplication of Monomial by a Binomial
Let us look at some examples:

Example 1: Multiplication of 4 and (x + y) will be (4x + 4y).

Example 2: Multiplication of 5x and (3y + 2) will be (15xy + 10x).

Example 3: Multiplication of 7x3 and (2x4+ y4) will be (14x7+7x3y4).

(b) Multiplication of Monomial by a Binomial:
Let us look at some examples:

Example 1: Multiplication of 4 and (x + y + z) will be (4x + 4y + 4z).

Example 2: Multiplication of 2x and (2x + y + z) will be (4x2 + 2xy + 2xz).

Example 3: Multiplication of 7x3 and (2x4+ y4+ 2) will be (14x7+7x3y4+14x3).

Examples based on Multiplying a Monomial by a Polynomial:

Example 1: Simplify 2a(4a – 2) + 7 and find its values for    a) x = 2            b) x =1/2
Solution: On simplifying, 2a(4a – 2) + 7 ,we get, 8a2 – 4a + 7
(a) For x = 2, 8a2 – 4a + 7 = 8(2)2 – 4(2) + 7
= 31
(b) For x =1/2, 8a2 – 4a + 7 = 8(1/2)2 – 4(1/2) + 7
= 7

Example 2: Multiply (5/7 x ab) and (-21/10 x a2b2).
Solution(5/7 x ab) x(-21/10 x a2b2) = (5/7) x (-21/10) x ab x a2b2
= (-3/2) a3b3

3. Multiplying a Polynomial by a Polynomial:

(a) Multiplication of Binomial by a Binomial:
Let us look at some examples:

Example 1: Multiplication of (4x + y) and (x + y) will be (4x2 +5xy + y2).

Example 2: Multiplication of (5x2 + 3y) and (3y + 2) will be (15x2y + 10x2 + 9y2 + 6y).

(b) Multiplication of Binomial by a Trinomial:
Let us look at some examples:

Example 1: Multiplication of (4x + 2) and (x + y + z) will be (4x2 + 4xy + 4xz + 2x + 2y + 2z).

Example 2: Multiplication of (2x2 +2xy) and (2x + y + z) will be (4x3 + 6x2y + 2x2z + 2xy2 +2xyz).

Examples based on Multiplying a Polynomial by a Polynomial

Example 1: Multiply the binomials (2ab + 3b2) and (3ab – 2b2).
Solution(2ab + 3b2) x (3ab – 2b2) = 2ab x (3ab – 2b2) +3b2 x (3ab – 2b2)
= 6a2b2 – 4ab3 + 9ab3 – 6b4
=  6a2b2 + 5ab3 – 6b4

Example 2: Simplify (a + b + c)(a + b – c)
Solution(a + b + c)(a + b – c) = a(a + b – c) + b(a + b – c) + c(a + b – c)
= a2 + ab – ac + ab + b2 –bc +ac +bc –c2
=  a2 + b2 – c2+ 2ab

Identity:
It is a relation which satisfies A =B, where A and B will contain some variables and for any values of these variables the relation A = B will always be true.

Example: Consider (x + 1) (x + 3) = x2 + 4x + 3.
Let us take x = 2,
LHS = (2 + 1) (2 + 3) = 3 x 5 = 15.
RHS = 22 + 4x2 + 3 = 4 + 8 + 3 = 15.
Hence, LHS = RHS.
Similarly, for any values of x the relation will always be true i.e. LHS = RHS.

Standard Identities:
(i) (a +b) 2 = (a2 + 2ab + b2)
(ii) (a – b) 2 = (a2 - 2ab + b2)
(iii) (a + b) (a – b) = (a2 – b2)

Example 1: Find square of 102.
SolutionWe can use (a +b) 2 = (a2 + 2ab + b2) identity to simplify the problem.
We can split 102 as (100+2). Let a = 100 and b =2.
Substituting these values in identity, we have,
LHS = (100 + 2) 2 = (102) 2
RHS = (1002 + 2x100x2 + 22) = (10000 + 400 + 4) = 10404.
Thus, square of 102 is 10404.

Example 2: Using (x + a) (x + b) = x2 + (a + b)x + ab, find 105 x 107.
SolutionUsing given identity, we can write
105 x 107 = (100 + 5) (100 + 7)
= 1002 + (5 + 7) x100 + 5 x 7
=   11235

Example 3: Prove that (3a + 7)2 – 84a = (3a – 7)2.
SolutionLHS    =  (3a + 7)2 – 84a
= (3a)2 + 2(3a)(7) + (7) 2 – 84a
= 9a2 + 42a + 49 – 84a
= 9a2 – 42a + 49
RHS    = (3a – 7)2
= (3a) 2 – 2(3a)(7) + (7) 2
= 9a2 – 42a + 49
Since, LHS = RHS, it is proved that (3a + 7)2 – 84a = (3a – 7)2.

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