Factorise :
`x^(2) + 7x + 6 + px + 6p`

Answer

To factorise the expression \( x^2 + 7x + 6 + px + 6p \), we can follow these steps:

### Step 1: Rearrange the expression
We start with the expression:
\[ x^2 + 7x + 6 + px + 6p \]
We can group the terms related to \( x \):
\[ x^2 + (7 + p)x + (6 + 6p) \]

### Step 2: Identify the quadratic form
Now, we can see that the expression is in the form of a quadratic equation:
\[ x^2 + (7 + p)x + (6 + 6p) \]

### Step 3: Factor by grouping
We need to factor this expression. To do this, we will look for two numbers that multiply to \( 6 + 6p \) (the constant term) and add up to \( 7 + p \) (the coefficient of \( x \)).

Let's rewrite the expression:
\[ x^2 + (7 + p)x + (6 + 6p) \]

### Step 4: Split the middle term
We can express \( 7 + p \) as \( 6 + 1 + p \) to help with factoring:
\[ x^2 + (6 + 1 + p)x + (6 + 6p) \]
This can be rearranged as:
\[ x^2 + 6x + (1 + p)x + (6 + 6p) \]

### Step 5: Group the terms
Now, we can group the first two terms and the last two terms:
\[ (x^2 + 6x) + ((1 + p)x + (6 + 6p)) \]

### Step 6: Factor out common terms
From the first group \( (x^2 + 6x) \), we can factor out \( x \):
\[ x(x + 6) + (1 + p)(x + 6) \]

### Step 7: Factor out the common binomial
Now we notice that \( (x + 6) \) is common in both terms:
\[ (x + 6)(x + 1 + p) \]

### Final Answer
Thus, the factorised form of the expression \( x^2 + 7x + 6 + px + 6p \) is:
\[ (x + 6)(x + 1 + p) \]