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Exploring Algebraic Identities - Simplify rational algebraic expressions using factorisation and cancellation rules

Grade 9CBSE

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

🔑Concepts

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A rational algebraic expression is an expression of the form p(x)q(x)\frac{p(x)}{q(x)}, where p(x)p(x) and q(x)q(x) are polynomials and q(x)≠0q(x) \neq 0.

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Simplification of rational expressions involves factorising both the numerator and the denominator into their irreducible factors.

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The cancellation rule states that if kk is a non-zero common factor of the numerator and the denominator, then k⋅ak⋅b=ab\frac{k \cdot a}{k \cdot b} = \frac{a}{b}.

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Algebraic identities such as difference of squares (a2−b2a^2 - b^2), perfect square trinomials ((a±b)2(a \pm b)^2), and sum/difference of cubes (a3±b3a^3 \pm b^3) are essential tools for factorisation.

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Splitting the middle term is a common method used to factorise quadratic trinomials of the form ax2+bx+cax^2 + bx + c.

📐Formulae

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

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

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:

Simplify the rational expression: x2−25x2+10x+25\frac{x^2 - 25}{x^2 + 10x + 25}

Solution:

x2−25x2+10x+25=(x−5)(x+5)(x+5)2=(x−5)(x+5)(x+5)(x+5)=x−5x+5\frac{x^2 - 25}{x^2 + 10x + 25} = \frac{(x-5)(x+5)}{(x+5)^2} = \frac{(x-5)(x+5)}{(x+5)(x+5)} = \frac{x-5}{x+5}

Explanation:

First, factorise the numerator using the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b), which gives (x−5)(x+5)(x-5)(x+5). Next, factorise the denominator using the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2, which gives (x+5)2(x+5)^2. Finally, cancel the common factor (x+5)(x+5) from both the numerator and the denominator.

Problem 2:

Simplify the rational expression: x2+5x+6x2−4\frac{x^2 + 5x + 6}{x^2 - 4}

Solution:

x2+5x+6x2−4=(x+2)(x+3)(x−2)(x+2)=x+3x−2\frac{x^2 + 5x + 6}{x^2 - 4} = \frac{(x+2)(x+3)}{(x-2)(x+2)} = \frac{x+3}{x-2}

Explanation:

The numerator x2+5x+6x^2 + 5x + 6 is factorised by splitting the middle term: x2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)x^2 + 2x + 3x + 6 = x(x+2) + 3(x+2) = (x+2)(x+3). The denominator x2−4x^2 - 4 is factorised as (x−2)(x+2)(x-2)(x+2) using the difference of squares. The common factor (x+2)(x+2) is then cancelled.

Problem 3:

Simplify: x3−8x−2\frac{x^3 - 8}{x - 2}

Solution:

x3−8x−2=(x−2)(x2+2x+4)x−2=x2+2x+4\frac{x^3 - 8}{x - 2} = \frac{(x-2)(x^2 + 2x + 4)}{x - 2} = x^2 + 2x + 4

Explanation:

Using the identity a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2), we can write x3−23x^3 - 2^3 as (x−2)(x2+2x+4)(x-2)(x^2 + 2x + 4). By cancelling the common factor (x−2)(x-2) in the numerator and denominator, we get x2+2x+4x^2 + 2x + 4.