The equation
`x^(2)y^(2)-2xy^(2)-3y^(2)-4x^(2)y+8xy+12y=0` represents

Options

A. a pair of lines
B. a pair of lines and a circle
C. a pair of lines and a parabola
D. four lines forming a square

Answer

To solve the equation \( x^2y^2 - 2xy^2 - 3y^2 - 4x^2y + 8xy + 12y = 0 \) and determine what it represents, we can follow these steps:

### Step 1: Factor out common terms
We start by grouping the terms in the equation. Notice that we can factor out \( y^2 \) from the first three terms and \( y \) from the last three terms:

\[ y^2(x^2 - 2x - 3) - 4y(x^2 - 2x - 3) = 0 \]

### Step 2: Factor out the common binomial
Now, we can factor out the common binomial \( (x^2 - 2x - 3) \):

\[ (x^2 - 2x - 3)(y^2 - 4y) = 0 \]

### Step 3: Factor the quadratic expressions
Next, we will factor both quadratic expressions:

1. For \( x^2 - 2x - 3 \):
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]

2. For \( y^2 - 4y \):
\[ y^2 - 4y = y(y - 4) \]

### Step 4: Write the complete factorization
Now we can write the complete factorization of the original equation:

\[ (x - 3)(x + 1)y(y - 4) = 0 \]

### Step 5: Identify the solutions
From the factored form, we can identify the solutions:

1. \( x - 3 = 0 \) gives \( x = 3 \)
2. \( x + 1 = 0 \) gives \( x = -1 \)
3. \( y = 0 \)
4. \( y - 4 = 0 \) gives \( y = 4 \)

### Step 6: Interpret the solutions
The solutions \( x = 3 \), \( x = -1 \), \( y = 0 \), and \( y = 4 \) represent four lines in the coordinate plane:

- The lines \( x = 3 \) and \( x = -1 \) are vertical lines.
- The lines \( y = 0 \) and \( y = 4 \) are horizontal lines.

### Conclusion
These four lines intersect to form a square in the coordinate plane with vertices at the points \( (3, 0) \), \( (3, 4) \), \( (-1, 0) \), and \( (-1, 4) \).

### Final Answer
The equation represents four lines forming a square.

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