Factorise: `4(2x-3y)^(2) - 8x + 12y -3`

Answer

To factorise the expression \( 4(2x - 3y)^2 - 8x + 12y - 3 \), we can follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ 4(2x - 3y)^2 - 8x + 12y - 3 \]

### Step 2: Expand the square term
Recognize that \( (2x - 3y)^2 \) can be expanded, but in this case, we will keep it as is for easier manipulation.

### Step 3: Group the terms
Notice that \( -8x + 12y - 3 \) can be grouped with the square term:
\[ 4(2x - 3y)^2 - (8x - 12y + 3) \]

### Step 4: Substitute for simplicity
Let \( a = 2x - 3y \). Then the expression becomes:
\[ 4a^2 - 8x + 12y - 3 \]

### Step 5: Rewrite \( -8x + 12y \)
We can express \( -8x + 12y \) in terms of \( a \):
\[ -8x + 12y = -4(2x - 3y) = -4a \]
Thus, the expression now looks like:
\[ 4a^2 - 4a - 3 \]

### Step 6: Factor the quadratic expression
Now we need to factor \( 4a^2 - 4a - 3 \). We can do this by finding two numbers that multiply to \( 4 \times -3 = -12 \) and add to \( -4 \). The numbers are \( -6 \) and \( 2 \):
\[ 4a^2 - 6a + 2a - 3 \]

### Step 7: Group the terms
Group the terms:
\[ (4a^2 - 6a) + (2a - 3) \]
Factor out the common terms:
\[ 2a(2a - 3) + 1(2a - 3) \]

### Step 8: Factor out the common binomial
Now factor out the common factor \( (2a - 3) \):
\[ (2a + 1)(2a - 3) \]

### Step 9: Substitute back for \( a \)
Replace \( a \) back with \( 2x - 3y \):
\[ (2(2x - 3y) + 1)(2(2x - 3y) - 3) \]
This simplifies to:
\[ (4x - 6y + 1)(4x - 6y - 3) \]

### Final Result
Thus, the factorised form of the given expression is:
\[ (4x - 6y + 1)(4x - 6y - 3) \]

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