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Exploring Algebraic Identities - Factorisation Without Using Algebra Tiles

Grade 9CBSE

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

🔑Concepts

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Factorisation is the process of expressing an algebraic expression as a product of two or more simpler expressions (factors).

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It is the reverse process of expanding brackets using algebraic identities.

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For expressions of the form x2+(a+b)x+abx^2 + (a+b)x + ab, we use the technique of splitting the middle term to obtain (x+a)(x+b)(x+a)(x+b).

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An expression of the form a2−b2a^2 - b^2 is called the difference of two squares and is factorised as (a+b)(a−b)(a+b)(a-b).

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Perfect square trinomials like a2+2ab+b2a^2 + 2ab + b^2 and a2−2ab+b2a^2 - 2ab + b^2 are factorised as (a+b)2(a+b)^2 and (a−b)2(a-b)^2 respectively.

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To factorise successfully, always look for common factors first before applying an identity.

📐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=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2), a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2)

💡Examples

Problem 1:

Factorise the following expression using algebraic identities: 9x2+24xy+16y29x^2 + 24xy + 16y^2.

Solution:

The expression can be written as: (3x)2+2(3x)(4y)+(4y)2(3x)^2 + 2(3x)(4y) + (4y)^2 Comparing this with the identity a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2, where a=3xa = 3x and b=4yb = 4y: (3x+4y)2(3x+4y)^2 Thus, the factors are (3x+4y)(3x+4y)(3x+4y)(3x+4y).

Explanation:

We identify that the first term 9x29x^2 is a perfect square of 3x3x and the last term 16y216y^2 is a perfect square of 4y4y. We then verify if the middle term is 2ab2ab.

Problem 2:

Factorise: 49a2−25b249a^2 - 25b^2.

Solution:

Rewrite the terms as squares: (7a)2−(5b)2(7a)^2 - (5b)^2 Using the identity a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b), we substitute a=7aa = 7a and b=5bb = 5b: (7a+5b)(7a−5b)(7a + 5b)(7a - 5b) So, 49a2−25b2=(7a+5b)(7a−5b)49a^2 - 25b^2 = (7a + 5b)(7a - 5b).

Explanation:

This expression is in the form of a difference of two squares. Taking the square root of each term gives the components for the identity.

Problem 3:

Factorise x2+9x+18x^2 + 9x + 18 by splitting the middle term.

Solution:

We need to find two numbers pp and qq such that p+q=9p+q = 9 and pq=18pq = 18. The numbers are 33 and 66. x2+6x+3x+18x^2 + 6x + 3x + 18 x(x+6)+3(x+6)x(x + 6) + 3(x + 6) (x+3)(x+6)(x + 3)(x + 6) For verification, the constant term calculation: 18−180\begin{array}{r} 18 \\ - 18 \\ \hline 0 \end{array}

Explanation:

By comparing the expression to x2+(a+b)x+abx^2 + (a+b)x + ab, we split the middle term 9x9x into 6x6x and 3x3x and then use grouping to find the factors.