`(3x+((-1)/(4))+2y)^(2)=` ___

Options

A. `9x^(2)-1/(16)+4y^(2)-(6x)/(4)-y+12xy`
B. `9x^(2)+1/(16)+4y^(2)+(6x)/(4)-y-12xy`
C. `9x^(2)+1/(16)+4y^(2)-(3x)/(2)-y+12xy`
D. `9x^(2)-1/(16)+4y^(2)+(3x)/(2)+y-12xy`

Answer

To solve the expression \((3x + (-\frac{1}{4}) + 2y)^2\), we can use the algebraic identity for the square of a sum. The identity states that:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \]

In our case, we can identify:
- \(a = 3x\)
- \(b = -\frac{1}{4}\)
- \(c = 2y\)

Now, we will apply the identity step by step.

### Step 1: Calculate \(a^2\)
\[ a^2 = (3x)^2 = 9x^2 \]

**Hint:** Square the coefficient and the variable separately.

### Step 2: Calculate \(b^2\)
\[ b^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16} \]

**Hint:** Remember that squaring a negative number results in a positive number.

### Step 3: Calculate \(c^2\)
\[ c^2 = (2y)^2 = 4y^2 \]

**Hint:** Square the coefficient and the variable separately.

### Step 4: Calculate \(2ab\)
\[ 2ab = 2 \cdot (3x) \cdot \left(-\frac{1}{4}\right) = -\frac{3}{2}x \]

**Hint:** Multiply the coefficients and remember to include the negative sign.

### Step 5: Calculate \(2bc\)
\[ 2bc = 2 \cdot \left(-\frac{1}{4}\right) \cdot (2y) = -\frac{1}{4} \cdot 4y = -y \]

**Hint:** The \(2\) and \(2y\) will cancel out the fraction.

### Step 6: Calculate \(2ca\)
\[ 2ca = 2 \cdot (2y) \cdot (3x) = 12xy \]

**Hint:** Simply multiply the coefficients and combine with the variables.

### Step 7: Combine all parts
Now we combine all the results from the previous steps:
\[ (3x + (-\frac{1}{4}) + 2y)^2 = 9x^2 + \frac{1}{16} + 4y^2 - \frac{3}{2}x - y + 12xy \]

### Final Result
Thus, the expression \((3x + (-\frac{1}{4}) + 2y)^2\) simplifies to:
\[ 9x^2 + 4y^2 + 12xy - \frac{3}{2}x - y + \frac{1}{16} \]

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