If `x^(2)- y^(2) + 4x-6y + k` is resolvable into two linear factors, then k =

Options

A. -1
B. 4
C. 6
D. -5

Answer

To determine the value of \( k \) such that the expression \( x^2 - y^2 + 4x - 6y + k \) can be factored into two linear factors, we can follow these steps:

### Step 1: Rearranging the Expression
We start with the expression:
\[ x^2 - y^2 + 4x - 6y + k \]
We can group the terms involving \( x \) and \( y \):
\[ (x^2 + 4x) + (-y^2 - 6y) + k \]

### Step 2: Completing the Square
Next, we complete the square for the \( x \) terms and the \( y \) terms.

For \( x^2 + 4x \):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

For \( -y^2 - 6y \):
\[ -y^2 - 6y = - (y^2 + 6y) = -((y + 3)^2 - 9) = - (y + 3)^2 + 9 \]

### Step 3: Substitute Back into the Expression
Now substituting back into the expression:
\[ ((x + 2)^2 - 4) - ((y + 3)^2 - 9) + k \]
This simplifies to:
\[ (x + 2)^2 - (y + 3)^2 + k + 5 \]

### Step 4: Setting Up for Factorization
For the expression to be factored into two linear factors, the constant term must equal zero:
\[ k + 5 = 0 \]

### Step 5: Solve for \( k \)
Now, we solve for \( k \):
\[ k = -5 \]

### Conclusion
Thus, the value of \( k \) that allows the expression to be factored into two linear factors is:
\[ \boxed{-5} \]

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