Factorise
` 2(ab + cd) - a^(2) - b^(2) +c^(2) +d^(2)`

Answer

To factorize the expression \( 2(ab + cd) - a^2 - b^2 + c^2 + d^2 \), we will follow these steps:

### Step 1: Rewrite the expression
Start by expanding the expression:
\[ 2(ab + cd) - a^2 - b^2 + c^2 + d^2 = 2ab + 2cd - a^2 - b^2 + c^2 + d^2 \]

### Step 2: Rearrange the terms
Rearranging the terms, we can group them as follows:
\[ (2cd + c^2 + d^2) - (a^2 + b^2 - 2ab) \]

### Step 3: Recognize the perfect square identities
We can recognize that:
- \( c^2 + d^2 + 2cd = (c + d)^2 \)
- \( a^2 + b^2 - 2ab = (a - b)^2 \)

So we can rewrite the expression as:
\[ (c + d)^2 - (a - b)^2 \]

### Step 4: Apply the difference of squares formula
Now we can use the difference of squares formula, which states that \( x^2 - y^2 = (x + y)(x - y) \):
Let \( x = c + d \) and \( y = a - b \):
\[ (c + d)^2 - (a - b)^2 = (c + d + (a - b))(c + d - (a - b)) \]

### Step 5: Simplify the factors
Now, simplify the factors:
\[ (c + d + a - b)(c + d - a + b) \]

### Final Answer
Thus, the factorization of the expression \( 2(ab + cd) - a^2 - b^2 + c^2 + d^2 \) is:
\[ (c + d + a - b)(c + d - a + b) \]

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