Factorise
` 8( 3x - 2y ) ^(2) - 6x + 4y -1`

Answer

To factorize the expression \( 8(3x - 2y)^2 - 6x + 4y - 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 8(3x - 2y)^2 - 6x + 4y - 1 \]

### Step 2: Group the terms
Notice that we can group the linear terms together:
\[ 8(3x - 2y)^2 - (6x - 4y + 1) \]

### Step 3: Factor out the negative sign
We can factor out a negative sign from the linear terms:
\[ 8(3x - 2y)^2 - (6x - 4y + 1) = 8(3x - 2y)^2 - 1(6x - 4y + 1) \]

### Step 4: Substitute \( t \)
Let \( t = 3x - 2y \). Then, we can rewrite the expression as:
\[ 8t^2 - (6x - 4y + 1) \]

### Step 5: Rewrite the linear terms in terms of \( t \)
We need to express \( 6x - 4y + 1 \) in terms of \( t \):
\[ 6x - 4y = 2(3x - 2y) = 2t \]
Thus, we can rewrite the expression as:
\[ 8t^2 - (2t - 1) \]

### Step 6: Combine the terms
Now we have:
\[ 8t^2 - 2t + 1 \]

### Step 7: Factor the quadratic
To factor \( 8t^2 - 2t + 1 \), we can look for two numbers that multiply to \( 8 \times 1 = 8 \) and add to \( -2 \). The factors are \( -4 \) and \( 2 \):
\[ 8t^2 - 4t + 2t - 1 \]
Now, we can group the terms:
\[ (8t^2 - 4t) + (2t - 1) \]
Factoring out common factors gives:
\[ 4t(2t - 1) + 1(2t - 1) \]
Now we can factor out \( (2t - 1) \):
\[ (4t + 1)(2t - 1) \]

### Step 8: Substitute back \( t \)
Now, substituting back \( t = 3x - 2y \):
\[ (4(3x - 2y) + 1)(2(3x - 2y) - 1) \]
This simplifies to:
\[ (12x - 8y + 1)(6x - 4y - 1) \]

### Final Answer
Thus, the factorized form of the expression is:
\[ (12x - 8y + 1)(6x - 4y - 1) \]