(7) \( x^{2}-4 y^{2}-x+6 y-2 \)

To factor the expression \( x^{2}-4 y^{2}-x+6 y-2 \), we first rearrange and group the terms: 1. Group the \( x \) terms and constants: \[ (x^{2} - x) + (-4y^{2} + 6y - 2) \] 2. Complete the square for the \( x \) terms: \[ x^{2} - x \] becomes \( \left( x - \frac{1}{2} \right)^{2} - \frac{1}{4} \). 3. For the \( y \) terms, we factor out \(-1\): \[ -1(4y^{2} - 6y + 2) \]. Completing the square gives us: \[ 4\left( y^{2} - \frac{3}{2}y + \frac{1}{4} \right) - 1 = 4(y - \frac{3}{4})^{2} - 4 \] Putting it all together, we can express the original equation as: \[ \left( x - \frac{1}{2} - 2(y - \frac{3}{4}) \right)\left( x - \frac{1}{2} + 2(y - \frac{3}{4}) \right) \] And that leads us to the final factorization of \( (x - 2y + \frac{3}{2})(x + 2y - \frac{5}{2}) \) or similar variations depending on adjustments in constants. So, you can see how factoring is much like solving a puzzle, and with a little practice, it becomes an exciting challenge!