Factorise
` (2x- y)^(2) - 14 x + 7 y - 18 `

Answer

To factorise the expression \( (2x - y)^2 - 14x + 7y - 18 \), we will follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ (2x - y)^2 - 14x + 7y - 18 \]

### Step 2: Substitute for convenience
Let \( t = 2x - y \). Then, we can rewrite the expression as:
\[ t^2 - 14x + 7y - 18 \]

### Step 3: Express \( 14x \) and \( 7y \) in terms of \( t \)
From the substitution \( t = 2x - y \), we can express \( y \) in terms of \( t \) and \( x \):
\[ y = 2x - t \]
Substituting this back into the expression gives:
\[ t^2 - 14x + 7(2x - t) - 18 \]
This simplifies to:
\[ t^2 - 14x + 14x - 7t - 18 = t^2 - 7t - 18 \]

### Step 4: Factor the quadratic expression
Now we need to factor \( t^2 - 7t - 18 \). We can do this by finding two numbers that multiply to \(-18\) and add to \(-7\). The numbers \(-9\) and \(2\) work:
\[ t^2 - 9t + 2t - 18 = (t - 9)(t + 2) \]

### Step 5: Substitute back for \( t \)
Now, substitute back \( t = 2x - y \):
\[ (2x - y - 9)(2x - y + 2) \]

### Final Answer
Thus, the factorised form of the expression \( (2x - y)^2 - 14x + 7y - 18 \) is:
\[ (2x - y - 9)(2x - y + 2) \]