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Standard identities can be determined by multiplying one binomial with any other binomial. Identification is described as an equation that holds or is legitimate no matter the value chosen for its variables.
- Standard identities are used to rearrange algebraic expressions.
- The definition states that two identities are replaceable, so we will replace one with another identity at any time.
- Equality refers to the expression where the left side is equivalent to the right-hand side.
- It is a set of algebraic expressions that are used to simplify a mathematical problem.
- When equality holds true for all values of variables, then it is called algebraic identity.
- Activity and substitution are two ways of verifying standard identities.
Key Terms: Standard identities, Binomial, Binomial Theorem, Factoring, Trigonometry, Algebraic Identities, Variables, Linear Equation, Algebra, Trigonometric Ratios, Algebraic Identities of Trinomial
What are Standard Identities?
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Algebraic Identities that are derived from the Binomial Theorem are known as standard algebraic identities or standard identities. It considered two sets of expressions that are equivalent to one another.
- LHS and RHS are two parts of standard identities.
- The process involves the multiplication of one binomial with another binomial value.
- It helps in the factorization of polynomials.
- All constant values should be defined accurately in the standard identity.
- It is used in the field of engineering, physics, computer science, and economics.
Example of What are Standard IdentitiesExample: Simplify (x – 4)(x – 4). (x – 4)(x – 4) = (x – 4)2 Using the identity (a – b)2 = a2 – 2ab + b2, (x – 4)2 = x2 – 2(x) (4) + (4)2 = x2 – 8x + 16 |
Types of Standard Identities
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Standard Identities are divided into two categories which are as follows:
Binomial Standard Identities
Binomial Standard Identities is a form of identity that consists of the sum of two terms which are equivalent to a monomial. These identities can be explained with the help of the binomial theorem.
- Binomial Standard Identities is the simplest kind of polynomial that forms the powers of a binomial.
Standard identities under Algebra under binomial theorem are as follows:
- Identity 1: (p + q) ² = p² + 2pq + q²
- Identity 2: (p - q) ² = p² + q² - 2pq
- Identity 3: (p + q) ³ = p³ + q³ + 3pq (p + q)
- Identity 4: (p - q) ³ = p³ - q³ - 3pq (p - q)
Example of Binomial Standard IdentitiesExample: Using identities, solve 27 × 33. Ans: 27 × 33 can be written as ( 30 - 3 ) × ( 30 + 3 )
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Trinomial Standard Identities
Trinomial Standard Identities is a form of identity that consists of three monomials which are separated by some algebraic operations. It consist of expansions of the powers of trinomials.
Example of Trinomial Standard IdentitiesExample: (a+b+c)2=a2+b2+c2+2ab+2ac+2bc |
Standard Identities in Algebra
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Algebra is defined as the combination of symbols and analysis of those symbols with the help of mathematical formulas. They are basically algebraic equations made up of algebraic expressions.
There is a wide variety of algebraic identities but few are standard which can be listed under.
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 + b2 – 2ab
- (a + b)(a – b) = a2 – b2
- (a + b)3 = a3 + b3 + 3ab(a + b)
- (a – b)3= a3 – b3 – 3ab(a – b)
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Example of Standard Identities in AlgebraExample: Solve (2p – 3q) (2p + 3q) using a suitable identity. (2p – 3q)(2p + 3q) Using the identity (a – b)(a + b) = a2 – b2 So, (2p – 3q)(2p + 3q) = (2p)2 – (3q)2 = 4p2 – 9q2 |
Also Read:
| Chapter Related Concepts | ||
|---|---|---|
| Direct proportion | Mensuration formula | Area of rhombus |
| Compound interest | Area of square | Ordered Pair |
Standard Identities of Trigonometry
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Trigonometric identities are used to specify the relationship between the six trigonometric ratios: sin, cos, tan, cot, sec and cosec. There exist a few general identities in trigonometry that contain all six trigonometric ratios.Some important Standard Identities of Trigonometry are as follows:
- sin2θ + cos2θ = 1
- sec2θ – tan2θ = 1
- cosec2θ – cot2θ = 1
Example of Standard Identities of TrigonometryExample: If the value of sec θ + tan θ = 36, then evaluate sec θ – tan θ? Ans: We know, sec2θ – tan2θ =1 Or, (sec θ + tan θ)( sec θ – tan θ)=1 Or, 36(sec θ – tan θ)=1 Or, sec θ – tan θ =1/36 So, the value of sec θ – tan θ =1/36. |
Standard Identities for Algebraic Factorization
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Algebraic Factorization is used to express identities, which consist of two binomials in multiplication. It is used for factorization of the polynomials. Individuals can use it to solve quadratic equations.
Standard algebraic identities under factoring formula are as follows:
- Identity 1: p² - q² = ( p + q)(p - q)
- Identity 2: p³ + q³ = (p + q)(p² + q² - pq)
- Identity 3: p³ - q³ = ( p - q)(p² + q² + pq)
Example of Standard Identities for Algebraic FactorizationExample: Solve (2p – q) (2p + q) using a suitable identity. (2p – q)(2p + q) Using the identity (a – b)(a + b) = a2 – b2 So, (2p – q)(2p + q) = (2p)2 – (q)2 = 4p2 – q2 |
Standard Algebraic Identities of Trinomials
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Standard identities under three variables are obtained by factoring and manipulation of some terms. These are as follows;
- Identity 1: (p + q)(p + r)(q + r) = (p + q + r)(pq + pr +qr) - pqr
- Identity 2: p² + q² + r² = (p + q + r)² - 2 (pq + pr + qr)
- Identity 3: p + q + r - 3pqr = (p + q + r)(p² + q² + r² - pq - pr - qr)
Example of Standard Algebraic Identities of TrinomialsExample: Help Rim find the factors of x2 - 3x + 2. Solution: x2 - 3x + 2
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Proof of Standard Identities for Binomials
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Proof of Standard Identities for Binomials are as follows:
Identity 1
Formula: (a+b)2=a2+2ab+b2
Proof: (a+b)2
- (a+b)(a+b)
- (a+b)a+(a+b)b
- a2+ab+ab+b2
- a2+2ab+b2
- LHS = RHS
- Hence, Proved
Identity 2
Formula: (a – b)2=a2 - 2ab+b2
Proof: (a – b)2
- (a – b)(a – b)
- (a- b)a-(a-b)b
- a2 – ab-ab+b2
- a2 – 2ab+b2
- LHS = RHS
- Hence, Proved
Identity 3
Formula: (a+b)3=a3+b3+3ab(a+b)
Proof: (a+b)3
- LHS = (a+b)(a+b)(a+b)
- LHS = [(a+b)a+(a+b)b](a+b)
- LHS = [a2+ab+ab+b2](a+b)
- LHS = [a2+2ab+b2](a+b)
- LHS = a[a2+2ab+b2]+b[a2+2ab+b2]
- LHS = [a3+2a2b+ab2]+[ba2+2ab2+b3]
- LHS = a3+3a2b+3ab2+b3
- LHS = a3+b3+3ab(a+b)
- LHS = RHS
- Hence, Proved
Proof of Standard Algebraic Identities of Trinomials
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Proof of Standard Algebraic Identities of Trinomials are as follows:
Identity 1
Formula: (a+b)(a+c)(b+c)=(a+b+c)(ab+ac+bc)–abc
Proof: (a+b)(a+c)(b+c)
- [a(a+b)+c(a+b)](b+c)
- [a2+ab+ac+bc](b+c)
- b[a2+ab+ac+bc]+c[a2+ab+ac+bc]
- a2b+ab2+abc+b2c+a2c+abc+ac2+bc2
- a2b+ab2+b2c+a2c+ac2+bc2+2abc …….. (1)
- (a+b+c)(ab+ac+bc)–abc
- [a(ab+ac+bc)+b(ab+ac+bc)+c(ab+ac+bc)]–abc
- [a2b+a2c+abc+ab2+abc+b2c+abc+ac2+bc2]–abc
- a2b+ab2+b2c+a2c+ac2+bc2+2abc …….. (2)
- From (1) & (2)
- RHS = LHS
- Hence, Proved.
Identity 2
Formula: a2+b2+c2=(a+b+c)2–2(ab+ac+bc)
Proof: (a+b+c)2–2(ab+ac+bc)
- RHS = [(a+b)+c]2–2(ab+ac+bc)
- RHS = [(a+b)2+c2+2(a+b)c]–2(ab+ac+bc)
- RHS = [a2+b2+2ab+c2+2ac+2bc]–2(ab+ac+bc)
- RHS = a2+b2+2ab+c2+2ac+2bc–2ab–2ac–2bc
- RHS = a2+b2+c2
- RHS = LHS
- Hence, Proved.
- a2+b2+c2=(a+b+c)2–2(ab+ac+bc)
Things to Remember
- Standard identities are obtained by multiplying one binomial by another which makes calculation much easier
- These identities are derived from the binomial theorem.
- There are numerous methods to solve standard identities.
- It is also known as standard algebraic identities.
- It consists a number of identities under the binomial theorem, under factoring and under trinomials.
Also Read:
Sample Questions
Ques. What are standard identities? (1 Mark)
Ans. in Mathematics, identity is defined as an equation that holds or is valid irrespective of the value chosen for its variables. The identities that are obtained by multiplying one binomial with another binomial are known as standard identities.
Ques. How many standard identities are there? (2 Marks)
Ans. The standard algebraic identities are:
- (x + a)(x + b) = x2 + (a + b) x + ab.
- (a + b + c) 2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
- (a + b)3 = a3 + b3 + 3ab (a + b)
- (a – b)3 = a3 – b3 – 3ab (a – b)
Ques. Use a suitable identity to simplify (x – 4)(x – 4). (2 Marks)
Ans. (x – 4) (x – 4) = (x – 4)2
Using the identity (a – b)2 = a2 – 2ab + b2,
(x – 4)2 = x2 – 2(x) (4) + (4)2
= x2 – 8x + 16
Ques. Solve (2p – 3q) (2p + 3q) using a suitable identity. (2 Marks)
Ans. (2p – 3q) (2p + 3q)
Using the identity (a – b)(a + b) = a2 – b2
So,
(2p – 3q)(2p + 3q) = (2p)2 – (3q)2
= 4p2 – 9q2
Ques. Using identity, (p + q)2 = p2 + q2 + 2pq
Find (2m + 3n) ² (1 Marks)
Ans. As we know, (p + q)2 = p2 + q2 + 2pq
Accordingly, (2m + 3n)2 = (2m)2 + (3n)2 + 2(2m)(3n)
Hence, (2m + 3n)2 = 4m2 + 9n2 + 12mn
Ques. Using identity, (p - q)2 = p2 + q2 - 2pq
Find (2m - 3n)2 (1 Marks)
Ans. As we know, (p - q)2 = p2 + q2 - 2pq
Accordingly, (2m - 3n)2 = (2m)2 +(3n)2 - 2(2m)(3n)
Hence, (2m + 3n)2 = 4m2+ 9n2 - 12mn
Ques. Using Identity, (p + q)(p - q) = p2 - q2
Find (2m + 3n) (2m - 3n) (1 Marks)
Ans. As we know, (p + q) (p - q) = p2 - q2
Accordingly, (2m + 3n) (2m - 3n) = (2m)2 - (3n)2
(2m + 3n)(2m - 3n) = 4m2 - 9n2
Ques. Find the product of (x + 1) (x + 1) using standard algebraic identities. (1 Marks)
Ans. (x + 1) (x + 1) can be written as (x + 1)2. So we have,
(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1
Ques. Factorize (x4 – 1) using standard algebraic identities. (2 Marks)
Ans. (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,
(x4 – 1) = ((x2)2– 12) = (x2 + 1) (x2 – 1)
The factor (x2 – 1) can be further factorized using the same Identity III where a = x and b = 1. So,
(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)
Ques. Factorize 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities. (2 Marks)
Ans. 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx are of the form Identity V. So we have,
6x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)
Ques. Expand (3x – 4y)3using standard algebraic identities. (1 Marks)
Ans. (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,
(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2
Ques. Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities. (1 Marks)
Ans. (x3 + 8y3 + 27z3 – 18xyz) is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,
(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)
Ques. Use a suitable identity to simplify (x – 9)(x – 9). (2 Marks)
Ans. (x – 9) (x – 9) = (x – 9)2
Using the identity (a – b)2 = a2 – 2ab + b2,
(x – 9)2 = x2 – 2(x) (9) + (9)2
= x2 –18x + 81
Ques. Use the standard identities to solve the following: (3 Marks)
(A) (8)2
(B) 7 × 13.
Ans. (A) (8)2
8 can be expressed as (7 + 1)
So, (8)2 = (7 + 1)2
Using the identity (a + b)2= a2 + 2ab + b2
Here, a = 7 and b = 1
(7 + 1)2 = 72 + 2(7)(1) + 12
49 + 14 + 1
64
(B) 7 × 13
This can be expressed as (10 – 3) (10 + 3)
Using the identity (a - b) (a + b) = a2 - b2
Here, a = 10 and b = 3
(10 - 3) (10 + 3) = (10)2 – (3)2
7 × 13 = 100 – 9
97 × 103 = 91
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