Factorise : `16-x^(2)-2xy-y^(2)` .The factors are

Options

A. `(4+x-y)(4+x+y)`
B. `(4-x-y)(4-x+y)`
C. `(2+x+y)(2-x-y)`
D. `(4+x+y)(4-x-y)`

Answer

To factorise the expression \( 16 - x^2 - 2xy - y^2 \), we can follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ 16 - x^2 - 2xy - y^2 \]

### Step 2: Group the quadratic terms
Rearrange the expression to group the quadratic terms:
\[ 16 - (x^2 + 2xy + y^2) \]

### Step 3: Recognize a perfect square
Notice that \( x^2 + 2xy + y^2 \) is a perfect square:
\[ x^2 + 2xy + y^2 = (x + y)^2 \]
So we can rewrite the expression as:
\[ 16 - (x + y)^2 \]

### Step 4: Recognize the difference of squares
Now, we have a difference of squares:
\[ 16 - (x + y)^2 = 4^2 - (x + y)^2 \]

### Step 5: Apply the difference of squares formula
Using the difference of squares formula \( a^2 - b^2 = (a + b)(a - b) \), we can factor this as:
\[ (4 + (x + y))(4 - (x + y)) \]

### Step 6: Simplify the factors
This simplifies to:
\[ (4 + x + y)(4 - x - y) \]

### Final Answer
Thus, the factors of the expression \( 16 - x^2 - 2xy - y^2 \) are:
\[ (4 + x + y)(4 - x - y) \]

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