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Exploring Algebraic Identities - Factorisation of Algebraic Expressions Using Identities

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Factorisation is the process of writing an algebraic expression as the product of two or more irreducible factors. It is the reverse process of expanding an expression using algebraic identities.

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A polynomial is said to be factorised completely if it is expressed as a product of factors that cannot be further factorised.

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Identities like (a+b)2(a+b)^2, (a−b)2(a-b)^2, and a2−b2a^2 - b^2 are frequently used to factorise quadratic expressions.

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For expressions involving three variables, the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca is utilized.

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Cubic factorisation involves identities like (a+b)3(a+b)^3, (a−b)3(a-b)^3, a3+b3+c3−3abca^3 + b^3 + c^3 - 3abc, and the sum/difference of cubes.

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The identity x2+(a+b)x+ab=(x+a)(x+b)x^2 + (a+b)x + ab = (x+a)(x+b) is used for factorising trinomials by splitting the middle term.

📐Formulae

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

a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a-b)^2

a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b)

x2+(a+b)x+ab=(x+a)(x+b)x^2 + (a+b)x + ab = (x+a)(x+b)

a2+b2+c2+2ab+2bc+2ca=(a+b+c)2a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a+b+c)^2

a3+b3+3a2b+3ab2=(a+b)3a^3 + b^3 + 3a^2b + 3ab^2 = (a+b)^3

a3−b3−3a2b+3ab2=(a−b)3a^3 - b^3 - 3a^2b + 3ab^2 = (a-b)^3

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)

💡Examples

Problem 1:

Factorise the expression: 9x2+24xy+16y29x^2 + 24xy + 16y^2

Solution:

The given expression can be written as: (3x)2+2(3x)(4y)+(4y)2(3x)^2 + 2(3x)(4y) + (4y)^2 Using the identity a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2, where a=3xa = 3x and b=4yb = 4y, we get: (3x+4y)2=(3x+4y)(3x+4y)(3x + 4y)^2 = (3x + 4y)(3x + 4y)

Explanation:

We identify that the first and last terms are perfect squares, and the middle term matches 2ab2ab.

Problem 2:

Factorise: 25a2−4b225a^2 - 4b^2

Solution:

The expression can be rewritten as: (5a)2−(2b)2(5a)^2 - (2b)^2 Using the identity a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b), where A=5aA = 5a and B=2bB = 2b: (5a+2b)(5a−2b)(5a + 2b)(5a - 2b)

Explanation:

This is a difference of two squares. We express each term as a square and apply the identity.

Problem 3:

Factorise: 8x3+27y3+36x2y+54xy28x^3 + 27y^3 + 36x^2y + 54xy^2

Solution:

We observe the terms and rewrite them: (2x)3+(3y)3+3(2x)2(3y)+3(2x)(3y)2(2x)^3 + (3y)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 Comparing this with the identity a3+b3+3a2b+3ab2=(a+b)3a^3 + b^3 + 3a^2b + 3ab^2 = (a+b)^3, where a=2xa = 2x and b=3yb = 3y, we get: (2x+3y)3(2x + 3y)^3

Explanation:

The expression matches the expanded form of a cubic binomial.

Problem 4:

Factorise: x2+5x+6x^2 + 5x + 6

Solution:

We need two numbers aa and bb such that a+b=5a+b = 5 and ab=6ab = 6. These numbers are 22 and 33. x2+(2+3)x+(2×3)x^2 + (2+3)x + (2 \times 3) =(x+2)(x+3)= (x+2)(x+3)

Explanation:

We use the identity x2+(a+b)x+ab=(x+a)(x+b)x^2 + (a+b)x + ab = (x+a)(x+b) by finding factors of the constant term that sum to the coefficient of the middle term.