`x^(2)+6xy -z^(2)+9y^(2)=`_______

Options

A. `(x+3y-z)(x+3y-z)`
B. `(x-3y+z)(x-3y+z)`
C. `(x+3y+z)(x+3y-z)`
D. `(x-3y-z)(x-3y-z)`

Answer

To factorize the expression \( x^2 + 6xy - z^2 + 9y^2 \), we can follow these steps:

### Step 1: Rearrange the expression
We can rearrange the expression to group the terms conveniently:
\[ x^2 + 6xy + 9y^2 - z^2 \]

### Step 2: Identify perfect squares
Notice that \( x^2 + 6xy + 9y^2 \) can be recognized as a perfect square. We can rewrite it as:
\[ (x + 3y)^2 \]
This is because:
\[ (x + 3y)^2 = x^2 + 2 \cdot x \cdot 3y + (3y)^2 = x^2 + 6xy + 9y^2 \]

### Step 3: Substitute back into the expression
Now substitute this back into our rearranged expression:
\[ (x + 3y)^2 - z^2 \]

### Step 4: Apply the difference of squares formula
We can now use the difference of squares formula, which states that \( a^2 - b^2 = (a + b)(a - b) \). Here, let \( a = (x + 3y) \) and \( b = z \):
\[ (x + 3y)^2 - z^2 = (x + 3y + z)(x + 3y - z) \]

### Final Answer
Thus, the factorized form of the expression \( x^2 + 6xy - z^2 + 9y^2 \) is:
\[ (x + 3y + z)(x + 3y - z) \]

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