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Q3(ii):
Factorise the following using appropriate identities:
(ii) $4y^2 – 4y + 1$
Solution :
Step 1: Initial Expression Analysis
We are given the quadratic trinomial:
$P(y) = 4y^2 - 4y + 1$
Our objective is to factorise this expression by identifying it as a perfect square. A quadratic trinomial takes the general form $ax^2 + bx + c$. We must analyze the first and last terms to determine if they are perfect squares.
Step 2: Identifying the Appropriate Algebraic Identity
We observe that the signs of the terms alternate ($+, -, +$). This suggests the use of the standard binomial square identity for subtraction [Per the fundamental axioms of polynomial expansion]:
$(a - b)^2 = a^2 - 2ab + b^2$
Step 3: Term-by-Term Transformation & Verification
We must map the terms of our given polynomial to the components of the identity $a^2 - 2ab + b^2$.
- First Term ($a^2$): The term $4y^2$ can be rewritten as the square of a single monomial.
$4y^2 = (2y)^2 \implies a = 2y$ - Third Term ($b^2$): The constant $1$ is a perfect square.
$1 = (1)^2 \implies b = 1$ - Middle Term ($-2ab$): We must verify that the middle term of our polynomial matches $-2ab$ using our derived values for $a$ and $b$.
$-2ab = -2(2y)(1) = -4y$
Since the middle term perfectly matches the given expression ($-4y$), the polynomial is confirmed to be a perfect square trinomial.
Step 4: Applying the Identity
By substituting $a = 2y$ and $b = 1$ back into the structural framework of the identity, we can rewrite the expanded polynomial as a squared binomial:
$4y^2 - 4y + 1 = (2y)^2 - 2(2y)(1) + (1)^2$
$4y^2 - 4y + 1 = (2y - 1)^2$
Step 5: Geometric Verification (Area Model)
To rigorously prove this factorization, we can use an area model. Consider a large square with a total side length of $2y$. Its total area is $(2y)^2 = 4y^2$. If we partition this square into segments of length $(2y - 1)$ and $1$, we can geometrically derive the identity.
From the geometric model, the total area of the square is the sum of its four internal regions:
$Area_{Total} = (2y - 1)^2 + (2y - 1) + (2y - 1) + 1$
$4y^2 = (2y - 1)^2 + 4y - 2 + 1$
$4y^2 = (2y - 1)^2 + 4y - 1$
Isolating the primary region $(2y - 1)^2$ yields:
$(2y - 1)^2 = 4y^2 - 4y + 1$
[This confirms our algebraic factorization is geometrically absolute].
Step 6: Final Factorisation
To express the polynomial fully in its factorised form, we write the squared binomial as the product of two identical linear binomials.
$(2y - 1)^2 = (2y - 1)(2y - 1)$
Final Solution: The factorised form of the polynomial $4y^2 - 4y + 1$ is $(2y - 1)(2y - 1)$.
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(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- 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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