Factorise:
`6ab - b ^(2) + 12 ac - 2bc`

Answer

To factorise the expression \( 6ab - b^2 + 12ac - 2bc \), we can follow these steps:

### Step 1: Rearrange the terms
We can rearrange the expression to group similar terms together:
\[ 6ab + 12ac - b^2 - 2bc \]

### Step 2: Factor by grouping
Now, we will group the first two terms and the last two terms:
\[ (6ab + 12ac) + (-b^2 - 2bc) \]

### Step 3: Factor out common factors from each group
From the first group \( 6ab + 12ac \), we can factor out \( 6a \):
\[ 6a(b + 2c) \]
From the second group \( -b^2 - 2bc \), we can factor out \(-b\):
\[ -b(b + 2c) \]

### Step 4: Combine the factored groups
Now we can combine the two factored parts:
\[ 6a(b + 2c) - b(b + 2c) \]

### Step 5: Factor out the common binomial factor
Notice that \( (b + 2c) \) is a common factor:
\[ (b + 2c)(6a - b) \]

### Final Answer
Thus, the factorised form of the expression \( 6ab - b^2 + 12ac - 2bc \) is:
\[ (b + 2c)(6a - b) \]

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