Factorise : `a^(2)-ac +xc - xa +6a - 6c`

Answer

To factorise the expression \( a^2 - ac + xc - xa + 6a - 6c \), we can follow these steps:

### Step 1: Rearrange the terms
Rearranging the expression can help us see common factors more clearly. We can write it as:
\[ a^2 - ac - xa + xc + 6a - 6c \]

### Step 2: Group the terms
Next, we can group the terms in pairs:
\[ (a^2 - ac) + (-xa + xc) + (6a - 6c) \]

### Step 3: Factor out the common factors from each group
1. From the first group \( a^2 - ac \), we can factor out \( a \):
\[ a(a - c) \]

2. From the second group \( -xa + xc \), we can factor out \( -x \):
\[ -x(a - c) \]

3. From the third group \( 6a - 6c \), we can factor out \( 6 \):
\[ 6(a - c) \]

Now, we can rewrite the expression as:
\[ a(a - c) - x(a - c) + 6(a - c) \]

### Step 4: Factor out the common binomial factor
Now we can see that \( (a - c) \) is a common factor in all the terms:
\[ (a - c)(a - x + 6) \]

### Final Answer
Thus, the factorised form of the expression \( a^2 - ac + xc - xa + 6a - 6c \) is:
\[ (a - c)(a - x + 6) \]

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