The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. You have already learned about a few of them in the junior grades. In this article, we will recall them and introduce you to some more standard algebraic identities, along with examples.
All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:
Some Standard Algebraic Identities list are given below:
Identity I: (a + b)^{2} = a^{2} + 2ab + b2
Identity II: (a – b)^{2} = a^{2} – 2ab + b2
Identity III: a^{2} – b^{2}= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x^{2} + (a + b) x + ab
Identity V: (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
Identity VI: (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)
Identity VII: (a – b)^{3} = a3 – b^{3} – 3ab (a – b)
Identity VIII: a^{3} + b^{3} + c^{3 }– 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)
Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.
Solution: (x + 1)(x + 1) can be written as (x + 1)^{2}. Thus, it is of the form Identity I where a = x and b = 1. So we have,
(x + 1)^{2} = (x)^{2} + 2(x)(1) + (1)^{2} = x^{2} + 2x + 1
Example 2: Factorise (x4 – 1) using standard algebraic identities.
Solution: (x^{4} – 1) is of the form Identity III where a = x^{2} and b = 1. So we have,
(x^{4} – 1) = ((x^{2})^{2}– 1^{2}) = (x^{2} + 1)(x^{2} – 1)
The factor (x^{2} – 1) can be further factorised using the same Identity III where a = x and b = 1. So,
(x^{4} – 1) = (x^{2} + 1)((x)^{2} –(1)^{2}) = (x^{2} + 1)(x + 1)(x – 1)
Eample 3: Factorise 16x^{2} + 4y^{2} + 9z^{2} – 16xy + 12yz – 24zx using standard algebraic identities.
Solution: 16x^{2} + 4y^{2} + 9z^{2}– 16xy + 12yz – 24zx is of the form Identity V. So we have,
16x^{2} + 4y^{2} + 9z^{2} – 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)
Example 4: Expand (3x – 4y)^{3 }using standard algebraic identities.
Solution: (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) = 27x^{3} – 64y^{3} – 108x^{2}y + 144xy^{2}
Example 5: Factorize (x^{3} + 8y^{3} + 27z^{3} – 18xyz) using standard algebraic identities.
Solution: (x^{3} + 8y^{3} + 27z^{3} – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,
(x^{3} + 8y^{3} + 27z^{3} – 18xyz) = (x)^{3} + (2y)^{3} + (3z)^{3} – 3(x)(2y)(3z)= (x + 2y + 3z)(x^{2} + 4y^{2} + 9z^{2} – 2xy – 6yz – 3zx)
57 videos399 docs65 tests

Test: Algebraic Identities Cubic Type Test  20 ques 
Factorization of Polynomials Doc  6 pages 
Polynomial in One variable Video  11:13 min 
1. What is the Remainder Theorem in algebra? 
2. How can the Remainder Theorem be applied to find remainders? 
3. Can the Remainder Theorem be used to find factors of a polynomial? 
4. What is the relationship between the Remainder Theorem and the Factor Theorem? 
5. Can the Remainder Theorem be used to solve polynomial equations? 
Test: Algebraic Identities Cubic Type Test  20 ques 
Factorization of Polynomials Doc  6 pages 
Polynomial in One variable Video  11:13 min 

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