Learn About Algebraic Identities

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Algebra is one of the most important chapters of basic mathematics. Students get to know about Algebraic Identities in the lower grades, at the high school level, and then move up to the upper grades and learn higher levels of algebraic Identities. Algebraic identification is a broad topic and is useful in all areas of a student’s life. An algebraic identifier is an algebraic equation that applies to all variable values ​​in it. An algebraic equation is a mathematical expression consisting of numbers, variables (unknown values), and mathematical functions (addition, subtraction, multiplication, division, etc.)  they are mainly used to find elements of polynomials.

Everything About Algebraic Identities

If the equation is true for all the values ​​of the variables in it, it is called an identifier. An algebraic identifier is an equation where the value of the left-hand side of the equation is equal to the value of the right-hand side of the equation for all variable values. We have several standard identifiers that we can use in different branches of mathematics. All standard Identities are obtained by the Binomial statement.

An algebraic equation that refers to all the values ​​of a variable in it is called an algebraic identifier. It is also used to factor polynomials. Thus, algebraic identifiers are used in the calculation of algebraic expressions and in the solution of various polynomials. You must have learned about some Algebraic Identities in the younger classes. In this class, you will revise those concepts and enhance your learning.

What are Algebraic Identities for Class 8?

• Algebraic identities are algebraic equations which are always true for every value of variables in them.

• Algebraic identities have their application in the factorization of polynomials.

• They contain variables and constants on both sides of the equation.

• In an algebraic identity, the left-side of the equation is equal to the right-side of the equation.

• For example, (a+b)2 = a2+2ab+b2 , which is true for all the values of and b.

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Methods to Verify Algebraic Identities 

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Using Substitution Method

• Substitution generally means putting numbers or values in the place of variables or letters.

• In the substitution method, an arithmetic operation is performed by substituting the values for the variables.

• For example, when we have x-2=4

When we substitute x= 6, 

On the Right-hand side,

4

On the left-hand-side,

x-2 = 6 – 2 = 4

Here, Right hand side = Left hand side which means (x-2) is an identity.

Suppose, (a+3) (a-3) = (a2-9) 

Substituting a= 1

On the Right- hand side,

(a2-9) = (1-9) = -8

On the Left- hand side,

(a+3) (a-3) = (1+3) (1-3) = (4) (-2) = -8

Here, Right hand side = Left hand side which means that (a+3) (a-3) is an identity.

Using Activity Method

• In this method, the algebraic identity is verified geometrically by taking different values of a x and y.

• In the activity method, the identities are verified by cutting and pasting paper.

• To verify an identity using this method, you need to have a basic knowledge of Geometry.

The standard identities class 8 are derived from the Binomial Theorem. The table below consists of some Standard identities in maths class 8.

Identities Class 8 –

Now, you Might Think What a Binomial Theorem is!

• In algebra, the Binomial Theorem is defined as a way of expanding a binomial expression raised to a large power which might be troublesome.

• A polynomial equation with just two terms generally having a plus or a minus sign in between is known as a Binomial expression.

A Small Explanation for the Above Algebraic Identities for Class 8

For example, let us take one of the basic identities,

(a+b)2 = a2+2ab+b2, which holds for all the values of a and b.

• An identity holds true for all the values of a and b.

• We can possibly substitute one instance of one side of the equality with its other side.

• In simple words, (a+b)2 can be replaced by a2+2ab+b2 and vice versa.

• These can be used as shortcuts which make manipulating algebra easier.

Factoring Identities

The identities listed below in the table are factoring formulas for identities of algebraic expressions class 8.

Three – Variable Identities –

By manipulation of the various discussed identities

entities of algebraic expressions class 8 we get these three- variable identities.

Important Algebraic Expressions and Identities Class 8 Formula –

The Four Basic Identities in Maths Class 8 have Been Listed Below

Questions to be Solved on Identities Class 8

Question 1) Find the product of (x-1) (x-1)

Solution) We need to find the product (x-1) (x-1),

(x-1) (x-1) can also be written as (x-1)2.

We know the formula for (x-1)2, expand it

(a-b)2 = a2- 2ab+b2 where a= x, b=1

(x-1)2 = x2- 2x+1

Therefore, the product of (x-1) (x-1) is x2- 2x+1 

Question 2) Find the product of (x+1) (x+1) as well as the value of it using x = 2.

Solution) We need to find the product (x+1) (x+1),

(x+1) (x+1) can also be written as (x+1)2.

We know the formula for (x+1)2, expand it

(a+b)2 = a2+ 2ab+b2 where a= x, b=1

(x+1)2 = x2+ 2x+1

Putting the value of x = 2 in equation 1,

(2)2+ 2(2) +1 = 9

Therefore, the product of (x+1) (x+1) is x2+ 2x+1 and the value of the expression is 9.

Question 3) Separate the constants and the variables from the given question.

-4, 4+x, 3x+4y, -5, 4.5y, 3y2+z

Solution) Variables are the ones which include any letter such as x, y, z etc along with the numbers.

In the given question, 

Constants = -4, -5

Variables = 3x+4y, 4+x, 4.5y, 3y2+z

Question 4) Find the value of \[\frac{{{x^2} – 1}}{5}\],at x = -1.

Solution) At x = -1,  \[x =  – 1,\frac{{{x^2} – 1}}{5}\]

                                     = \[\frac{{{(-1)^2} – 1}}{5}\]

                                      = 0

Question 5) Find the value of x2+y2 – 10 at x=0 and y=0?

Solution) At x= 0 and y = 0,

x2+y2 – 10 = (0)2+(0)2 – 10 

= -10

Question 6) Solve the following (x+2)2 using the concept of identities.

Solution) According to the identities and algebraic expression class 8,

We know the formula,

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

Where, a= x, b= 2

Let’s expand the given (x+2)2,

Therefore, (x+2)2 = x2+4x+4 is the solution.

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