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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 $\left(x + \frac{y}{10}\right)$.
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
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(ii): Factorise the following using appropriate identities: (ii) $4y^2 – 4y + 1$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(i): Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q8(v): Factorise each of the following: (v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
- Q9(i): Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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