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Q3(iii):
Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$

Solution :

Initial Setup & Given Expression

We are tasked with factorising the following algebraic expression:

$x^2 - \frac{y^2}{100}$

Step 1: Structural Analysis of the Polynomial

To factorise the expression, we must first analyze its algebraic structure. The polynomial consists of exactly two terms separated by a subtraction operator. This structural pattern suggests the potential application of the Difference of Two Squares identity, provided both terms can be expressed as perfect squares.

  • First Term: The term $x^2$ is already expressed as the square of $x$. Thus, we can write it as $(x)^2$.
  • Second Term: The term $\frac{y^2}{100}$ is a fraction where both the numerator and the denominator are perfect squares. Knowing that $100 = 10^2$, we can apply the exponent quotient rule [specifically, $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$] to rewrite the term:

    $\frac{y^2}{100} = \frac{y^2}{10^2} = \left(\frac{y}{10}\right)^2$

Rewriting the original polynomial with these perfect squares yields:

$(x)^2 - \left(\frac{y}{10}\right)^2$

Step 2: Selection of the Algebraic Identity

The expression is now strictly in the form of $a^2 - b^2$. [Per the Fundamental Algebraic Identities], the difference of two squares can be factored into the product of two binomials:

$a^2 - b^2 = (a - b)(a + b)$

Step 3: Variable Substitution & Factorisation

By mapping our rewritten expression to the standard identity, we establish the following equivalencies:

  • $a = x$
  • $b = \frac{y}{10}$

Substituting these specific values into the identity $(a - b)(a + b)$, we obtain the factorised form:

$(x)^2 - \left(\frac{y}{10}\right)^2 = \left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right)$

Geometric Verification of the Identity

The algebraic identity $a^2 - b^2 = (a - b)(a + b)$ can be rigorously proven using geometric area models. Below is a precise spatial representation demonstrating how removing a square of area $\left(\frac{y}{10}\right)^2$ from a larger square of area $x^2$ leaves an area that can be rearranged into a rectangle of dimensions $\left(x - \frac{y}{10}\right)$ by $\left(x + \frac{y}{10}\right)$.

Removed (y/10)² x x y/10 y/10 Area = x² - (y/10)² Rearrange x - y/10 x y/10 x + y/10 Area = (x - y/10)(x + y/10)

Final Solution: The completely factorised form of the polynomial is $\left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right)$.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4


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