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Exploring Algebraic Identities - Simplifying Rational Expressions

Grade 9CBSE

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

🔑Concepts

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A rational expression is defined as a fraction 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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To simplify a rational expression, we factorize both the numerator and the denominator completely using algebraic identities or splitting the middle term.

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Once factorized, common factors appearing in both the numerator and denominator can be cancelled: k⋅Ak⋅B=AB\frac{k \cdot A}{k \cdot B} = \frac{A}{B}, provided k≠0k \neq 0.

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The domain of a rational expression is the set of all real numbers except those that make the denominator zero.

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Important identities for simplification include the 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.

📐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)

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

💡Examples

Problem 1:

Simplify the rational expression: x2−16x2+8x+16\frac{x^2 - 16}{x^2 + 8x + 16}

Solution:

Step 1: Factorize the numerator using the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b). x2−16=x2−42=(x−4)(x+4)x^2 - 16 = x^2 - 4^2 = (x - 4)(x + 4) Step 2: Factorize the denominator using the identity a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2. x2+8x+16=x2+2(x)(4)+42=(x+4)2x^2 + 8x + 16 = x^2 + 2(x)(4) + 4^2 = (x + 4)^2 Step 3: Write the expression with factors and cancel common terms. (x−4)(x+4)(x+4)(x+4)=x−4x+4\frac{(x - 4)(x + 4)}{(x + 4)(x + 4)} = \frac{x - 4}{x + 4}

Explanation:

The numerator is a difference of two squares and the denominator is a perfect square trinomial. Cancelling the common factor (x+4)(x+4) simplifies the expression.

Problem 2:

Simplify x2−5x+6x2−9\frac{x^2 - 5x + 6}{x^2 - 9}

Solution:

Step 1: Factorize the numerator by splitting the middle term. x2−5x+6=x2−3x−2x+6=x(x−3)−2(x−3)=(x−3)(x−2)x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x(x - 3) - 2(x - 3) = (x - 3)(x - 2) Step 2: Factorize the denominator using a2−b2a^2 - b^2. x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3) Step 3: Divide out the common factor (x−3)(x - 3). (x−3)(x−2)(x−3)(x+3)=x−2x+3\frac{(x - 3)(x - 2)}{(x - 3)(x + 3)} = \frac{x - 2}{x + 3}

Explanation:

We use quadratic factorization for the numerator and the difference of squares identity for the denominator to identify the common factor (x−3)(x-3).

Problem 3:

Simplify y3−8y2−4\frac{y^3 - 8}{y^2 - 4}

Solution:

Step 1: Use the difference of cubes identity a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2) for the numerator. y3−23=(y−2)(y2+2y+4)y^3 - 2^3 = (y - 2)(y^2 + 2y + 4) Step 2: Use the difference of squares identity for the denominator. y2−22=(y−2)(y+2)y^2 - 2^2 = (y - 2)(y + 2) Step 3: Cancel the common factor (y−2)(y - 2). (y−2)(y2+2y+4)(y−2)(y+2)=y2+2y+4y+2\frac{(y - 2)(y^2 + 2y + 4)}{(y - 2)(y + 2)} = \frac{y^2 + 2y + 4}{y + 2}

Explanation:

The numerator is factorized as a difference of cubes and the denominator as a difference of squares. The common binomial factor (y−2)(y-2) is then removed.