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Algebraic Identities

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.

Algebraic Identities | Mathematics (Maths) Class 9

Standard Algebraic Identities List

All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:

Algebraic Identities | Mathematics (Maths) Class 9

Some Standard Algebraic Identities list are given below:
Identity I: (a + b)2 = a2 + 2ab + b2
Identity II: (a – b)2 = a2 – 2ab + b2
Identity III: a2 – b2= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
Identity VIII: a3 + b3 + c– 3abc = (a + b + c)(a2 + b2 + c2 – 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 = x2 + 2x + 1

Example 2: Factorise (x4 – 1) using standard algebraic identities.
Solution: (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 factorised 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)

Eample 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.
Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,
16x2 + 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)

Example 4: Expand (3x – 4y)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) = 27x3 – 64y3 – 108x2y + 144xy2

Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.
Solution: (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)

The document Algebraic Identities | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Algebraic Identities - Mathematics (Maths) Class 9

1. What is the Remainder Theorem in algebra?
Ans. The Remainder Theorem states that if you divide a polynomial f(x) by x-a, then the remainder will be equal to f(a). In other words, if you substitute the value of a into the polynomial, the resulting value will be the remainder of the division.
2. How can the Remainder Theorem be applied to find remainders?
Ans. To apply the Remainder Theorem, you need to divide a polynomial by a linear factor of the form x-a. Substitute the value of a into the polynomial and calculate the result, which will give you the remainder of the division.
3. Can the Remainder Theorem be used to find factors of a polynomial?
Ans. Yes, the Remainder Theorem can be used to find factors of a polynomial. If you have a polynomial f(x) and you know that f(a) = 0, then you can conclude that (x-a) is a factor of the polynomial. This allows you to factorize the polynomial and find its other roots.
4. What is the relationship between the Remainder Theorem and the Factor Theorem?
Ans. The Remainder Theorem and the Factor Theorem are closely related. The Remainder Theorem states that if you divide a polynomial by x-a, the remainder will be equal to f(a). On the other hand, the Factor Theorem states that if f(a) = 0, then (x-a) is a factor of the polynomial. In other words, the Remainder Theorem helps us find the remainder when dividing by a linear factor, while the Factor Theorem helps us determine if a given value is a root or factor of the polynomial.
5. Can the Remainder Theorem be used to solve polynomial equations?
Ans. Yes, the Remainder Theorem can be used to solve polynomial equations. By dividing the polynomial by a linear factor, you can find the remainder and determine if a particular value is a solution to the equation. This can help in finding the roots or solutions of the polynomial equation.
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