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Q4(vi):
Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$

Solution :

Given Expression & Algebraic Identity

We are tasked with expanding the following trinomial squared:

$ \left( \frac{1}{4}a - \frac{1}{2}b + 1 \right)^2 $

To expand this expression systematically, we utilize the standard algebraic identity for the square of a trinomial [Derived from the distributive property of multiplication over addition]:

$ (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx $

x y z x y z x2 xy xz xy y2 yz xz yz z2 Geometric Area Model of (x + y + z)²

Step 1: Variable Mapping

By comparing our given expression $\left( \frac{1}{4}a - \frac{1}{2}b + 1 \right)^2$ with the standard identity $(x + y + z)^2$, we can establish a direct mapping of the terms. It is critical to include the negative sign in our mapping to maintain algebraic integrity.

  • $ x = \frac{1}{4}a $
  • $ y = -\frac{1}{2}b $
  • $ z = 1 $

Step 2: Substitution into the Identity

Substituting the mapped variables into the expanded form of the identity, we get:

$ \left( \frac{1}{4}a \right)^2 + \left( -\frac{1}{2}b \right)^2 + (1)^2 + 2\left( \frac{1}{4}a \right)\left( -\frac{1}{2}b \right) + 2\left( -\frac{1}{2}b \right)(1) + 2(1)\left( \frac{1}{4}a \right) $

Step 3: Term-by-Term Simplification

We will now apply the exponent rules [specifically $(uv)^n = u^n v^n$] and perform scalar multiplication for each distinct term.

Component Operation Simplified Result
$x^2$ $\left( \frac{1}{4}a \right)^2$ $\frac{1}{16}a^2$
$y^2$ $\left( -\frac{1}{2}b \right)^2$ $\frac{1}{4}b^2$
[Note: The square of a negative is positive]
$z^2$ $(1)^2$ $1$
$2xy$ $2 \cdot \left( \frac{1}{4}a \right) \cdot \left( -\frac{1}{2}b \right)$ $-\frac{1}{4}ab$
$2yz$ $2 \cdot \left( -\frac{1}{2}b \right) \cdot (1)$ $-b$
$2zx$ $2 \cdot (1) \cdot \left( \frac{1}{4}a \right)$ $\frac{1}{2}a$

Step 4: Final Assembly

Combining all the simplified terms from Step 3 yields the fully expanded polynomial. We write the terms in descending order of degree where applicable, though standard expansion order is perfectly rigorous:

$ \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a $

Final Solution: The expanded form of the given polynomial is $ \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a $


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