The factorisaton of
`25 - p ^(2) - q ^(2) - 2 pq` is

Options

A. `(5 + p +q) ( 5 - p -q)`
B. `(5 + p-q) ( 5 - p + q)`
C. `(5 + p - q) ( 5 - p -q)`
D. `(5 + p + q) ( 5 - p+ q)`

Answer

To factor the expression \( 25 - p^2 - q^2 - 2pq \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 25 - p^2 - q^2 - 2pq \]
We can rearrange it as:
\[ 25 - (p^2 + q^2 + 2pq) \]

### Step 2: Recognize a perfect square
Notice that \( p^2 + q^2 + 2pq \) can be rewritten using the identity for the square of a binomial:
\[ p^2 + q^2 + 2pq = (p + q)^2 \]
Thus, we can rewrite our expression as:
\[ 25 - (p + q)^2 \]

### Step 3: Identify the difference of squares
Now we have a difference of squares:
\[ 25 - (p + q)^2 \]
We can express \( 25 \) as \( 5^2 \), so we rewrite it as:
\[ 5^2 - (p + q)^2 \]

### Step 4: Apply the difference of squares formula
We can apply the difference of squares formula:
\[ a^2 - b^2 = (a + b)(a - b) \]
Here, \( a = 5 \) and \( b = (p + q) \). Thus, we can factor it as:
\[ (5 + (p + q))(5 - (p + q)) \]
This simplifies to:
\[ (5 + p + q)(5 - p - q) \]

### Final Answer
The factorization of \( 25 - p^2 - q^2 - 2pq \) is:
\[ (5 + p + q)(5 - p - q) \]

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