Factorise : `4x^2-9y^2-2x-3y`

Explanation

To factorise the expression \(4x^2 - 9y^2 - 2x - 3y\), we will follow these steps:

### Step 1: Rearranging the Expression
First, we can rearrange the expression to group the terms in a more manageable way:
\[ 4x^2 - 2x - 9y^2 - 3y \]

### Step 2: Grouping Terms
Next, we can group the first two terms and the last two terms:
\[ (4x^2 - 2x) + (-9y^2 - 3y) \]

### Step 3: Factoring Out Common Factors
Now, we will factor out the common factors from each group:
1. From \(4x^2 - 2x\), we can factor out \(2x\):
\[ 2x(2x - 1) \]
2. From \(-9y^2 - 3y\), we can factor out \(-3y\):
\[ -3y(3y + 1) \]

So, we rewrite the expression as:
\[ 2x(2x - 1) - 3y(3y + 1) \]

### Step 4: Recognizing the Difference of Squares
Now we notice that we can rewrite \(4x^2 - 9y^2\) as a difference of squares:
\[ (2x)^2 - (3y)^2 \]
Thus, we can use the difference of squares formula \(a^2 - b^2 = (a + b)(a - b)\):
\[ (2x + 3y)(2x - 3y) \]

### Step 5: Combining the Factors
Now, we combine the factors we factored out:
\[ (2x + 3y)(2x - 3y) - (2x + 3y)(1) \]

### Step 6: Factoring Out the Common Binomial
We can now factor out the common binomial \((2x + 3y)\):
\[ (2x + 3y)((2x - 3y) - 1) \]

### Final Step: Simplifying
This gives us the final factorized form of the expression:
\[ (2x + 3y)(2x - 3y - 1) \]

### Final Answer
The factorised form of \(4x^2 - 9y^2 - 2x - 3y\) is:
\[ (2x + 3y)(2x - 3y - 1) \]