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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.

2y - 1 1 2y - 1 1 (2y - 1)² 1(2y - 1) 1(2y - 1) 1² Total Side = 2y Total Side = 2y

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


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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