Factorise :
`12(3x - 2y)^(2) - (3x - 2y)-1`

Answer

To factorise the expression \( 12(3x - 2y)^2 - (3x - 2y) - 1 \), we can follow these steps:

### Step 1: Substitute
Let \( z = 3x - 2y \). This substitution simplifies our expression.

So, we rewrite the original expression as:
\[ 12z^2 - z - 1 \]

### Step 2: Factor the Quadratic
Now we need to factor the quadratic \( 12z^2 - z - 1 \). We will use the method of splitting the middle term.

We need to find two numbers that multiply to \( 12 \times (-1) = -12 \) and add to \(-1\). The numbers that satisfy this are \(-4\) and \(3\).

### Step 3: Rewrite the Middle Term
We can rewrite \(-z\) as \(-4z + 3z\):
\[ 12z^2 - 4z + 3z - 1 \]

### Step 4: Group the Terms
Now, we group the terms:
\[ (12z^2 - 4z) + (3z - 1) \]

### Step 5: Factor by Grouping
Now we factor out the common factors from each group:
\[ 4z(3z - 1) + 1(3z - 1) \]

### Step 6: Factor Out the Common Binomial
Now we can factor out the common binomial \((3z - 1)\):
\[ (3z - 1)(4z + 1) \]

### Step 7: Substitute Back
Now we substitute back \( z = 3x - 2y \):
\[ (3(3x - 2y) - 1)(4(3x - 2y) + 1) \]

### Step 8: Simplify
Now we simplify each factor:
1. For \( 3(3x - 2y) - 1 \):
\[ 9x - 6y - 1 \]
2. For \( 4(3x - 2y) + 1 \):
\[ 12x - 8y + 1 \]

### Final Result
Thus, the factorised form of the expression is:
\[ (9x - 6y - 1)(12x - 8y + 1) \]