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Standard Algebraic Identities | Class 9 Maths

  • Last Updated : 11 Feb, 2021

An algebraic identity is an equality that holds for any value of its variables. They are generally used in the factorization of polynomials or simplification of algebraic calculations. Many everyday situations can be formulated in the form of mathematical equations. Identities help us by giving away to factorize them.

While operating a mobile phone. There are millions of chips under the screen of your phone, they do their job perfectly so that you rarely feel any glitch. Thousands of engineers brainstorming on complex equations have made this possible. They deal with much bigger equations and problems. Algebraic identities then become small tools to help solve those problems. So maybe they don’t come in front of us directly, but somewhere behind the scenes, someone has definitely used them and making life easier for us.

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Types of Polynomials

Based on the number of terms present in an Algebraic Expression, For example, there can be 1 term, 2 terms, 3 terms, 4 terms, and so on, they are classified in different categories. The terms are separated by either a positive or a negative sign:



Types of PolynomialsDefinitionExample
MonomialsA Polynomial containing only one term

10,

 2x2,

 4abc

BinomialsA Polynomial containing two terms

 x+y,

 3p2-5,

 x3y+8z

TrinomialsA Polynomial containing three terms

p+q+r,

x5+5x+3,



QuadrinomialA Polynomial containing four terms

p+q+r+s,

m2n-2mn+3m+7

QuintinomialA Polynomial containing five terms5x3y+ 6x2– 9xy+8y-7

 Note: 

  • Quadrinomials and Quintinomials are not very famous terms and are often referred as Polynomials.
  • The most common Standard Identities are from Binomials and Trinomials.

Standard Identities of Algebraic Expression

All standard Algebraic Identities are derived from the Binomial Theorem. There are a number of algebraic identities but a few are standard that are listed below:

Standard Identities
(a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 -2ab
a2 – b2 = (a + b)(a – b)
(ax + b)(ax – b) = ax2 – b2
(x + a) (x + b) = x2 + (a + b)x + ab
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Let’s look at different applications of these identities. 

Applications of Identities

Identity 1: (a + b)2 = a2  + b2 + 2ab

Question: Find the value of (x + 6)(x + 6) using algebraic identities when x = 3. 

Solution:

(x+6)(x+6) can be re-written as (x + 6)2

It can be rewritten in this form, (a + b)2 = a2 + b2 + 2ab. 

(x + 6)2 = x2 + 62 + 2(6x) 



             = x2 + 36 + 12x 

Given, x = 3. 

(x + 6)2 = 32 + 36 + 12(3) 

             = 9 + 36 + 36 

             = 81

Identity 2: (a – b)2 = a2 + b2  -2ab

Question: Expand (5x – 3y)2.

Solution:

This is similar to expanding (a – b)2 = a2 + b2 – 2ab.

where a = 5x and b = 3y,

So (5x – 3y)2 = (5x)2 + (3y)2 – 2(5x)(3y)

                    = 25x2 + 9y2 – 30xy 

Identity 3: a2 – b2 = (a + b)(a – b)

Question: Factorize (x6 – 1) using the identities mentioned above. 

Solution:

(x6 – 1) can be written as (x3)2 – 12

This resembles the identity a2 – b2 = (a + b)(a – b). 

where a = x3, and b = 1. 

So, x6 – 1 = (x3)2 – 1 = (x3 + 1) (x3 – 1). 

Some other questions and applications

Question 1: If a +b = 12 and ab = 35, what is a4 + b4

Solution:

a4 + b4 can be written as (a2)2 + (b2)2



and we know, (x + y)2 = x2 + y2 + 2xy

                   ⇒ x2 + y2 = (x + y)2 -2xy 

So, in this case, x = a2, y = b2 ;

a4 + b4 = (a2 + b2)2 -2(a2)(b2

⇒ ((a+b)2 – 2ab)2 – 2(a2)(b2

⇒ ((12)2 – 2(35))2 – 2(35)2 

⇒ 5475 – 2450

⇒ 3026

Question 2: The identity 4(z+7)(2z-1)=Az2+Bz+C holds for all real values of z. What is A+B+C?

Solution:

Multiplying out the left side of the identity, we have

4(x+7)(2x−1)=8x2 +52x−28.

This expression must be equal to the right-hand side of the identity, implying

8x2+52x-28=Ax2+Bx+C,

So now comparing both sides of the equation. 

A = 8, B = 52 ad C -28. 

A + B + C  = 8 + 52 -28 = 32

Question 3: If a+b+c=6, a2+b2+c2 = 14 and ab+bc+ca=11 what is a3+b3+c3-3abc?

Solution:

We know this identity,

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab -bc -ca)

Substituting the given values, 

a3 + b3 + c3 -3abc = (6)(14 -11)

⇒ (6)(3) = 18




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