If `alpha,beta` are zeroes of polynomial f(x)`=x^(2)-p(x+1)-c`, then find the value of `(alpha+1)(beta+1)`.

Explanation

To find the value of \((\alpha + 1)(\beta + 1)\) given that \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(f(x) = x^2 - p(x + 1) - c\), we can follow these steps:

### Step 1: Identify the polynomial
The polynomial is given as:
\[ f(x) = x^2 - p(x + 1) - c \]
We can rewrite this as:
\[ f(x) = x^2 - px - p - c \]

### Step 2: Compare with standard form
The standard form of a quadratic polynomial is:
\[ ax^2 + bx + c \]
From our polynomial, we can identify:
- \(a = 1\)
- \(b = -p\)
- \(c = -(p + c)\)

### Step 3: Use Vieta's formulas
According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-p}{1} = p\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{-(p + c)}{1} = -(p + c)\)

### Step 4: Calculate \((\alpha + 1)(\beta + 1)\)
We can expand \((\alpha + 1)(\beta + 1)\) as follows:
\[ (\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 \]

### Step 5: Substitute values from Vieta's formulas
Substituting the values we found:
\[ (\alpha + 1)(\beta + 1) = \alpha \beta + (\alpha + \beta) + 1 \]
\[ = -(p + c) + p + 1 \]

### Step 6: Simplify the expression
Now, simplifying the expression:
\[ -(p + c) + p + 1 = -p - c + p + 1 = 1 - c \]

### Final Answer
Thus, the value of \((\alpha + 1)(\beta + 1)\) is:
\[ \boxed{1 - c} \]