If `px^(2) + qx + r = p ( x-a) ( x-B),and p^(3) + pq + r = 0 ,` p,q and r being real numbers, then which of the following is not possible?

Options

A. `alpha = beta = p`
B. `alpha !=beta = p`
C. ` alpha = beta !=p`
D. ` beta !=alpha = p`

Explanation

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ px^2 + qx + r = p(x - \alpha)(x - \beta) \]
\[ p^3 + pq + r = 0 \]

2. **Expand the Right Side**:
Let's expand the right side of the first equation:
\[ p(x - \alpha)(x - \beta) = p(x^2 - (\alpha + \beta)x + \alpha\beta) \]
This simplifies to:
\[ px^2 - p(\alpha + \beta)x + p\alpha\beta \]

3. **Equate Coefficients**:
Now we equate the coefficients from both sides of the equation:
- Coefficient of \(x^2\): \(p = p\) (This is trivially true)
- Coefficient of \(x\): \(q = -p(\alpha + \beta)\)
- Constant term: \(r = p\alpha\beta\)

4. **Substituting \(q\) and \(r\) into the Second Equation**:
Now we substitute \(q\) and \(r\) into the second equation:
\[ p^3 + p(-p(\alpha + \beta)) + p\alpha\beta = 0 \]
Simplifying this gives:
\[ p^3 - p^2(\alpha + \beta) + p\alpha\beta = 0 \]

5. **Factoring the Equation**:
We can factor out \(p\) from the equation:
\[ p(p^2 - p(\alpha + \beta) + \alpha\beta) = 0 \]
This gives us two cases:
- \(p = 0\)
- \(p^2 - p(\alpha + \beta) + \alpha\beta = 0\)

6. **Analyzing the Quadratic Equation**:
The quadratic equation \(p^2 - p(\alpha + \beta) + \alpha\beta = 0\) can be analyzed using the discriminant:
\[ D = (\alpha + \beta)^2 - 4\alpha\beta \]
For \(p\) to have real solutions, the discriminant must be non-negative:
\[ D \geq 0 \]

7. **Conclusion**:
The discriminant condition leads us to the conclusion that:
\[ (\alpha - \beta)^2 \geq 0 \]
This condition is always true, meaning \(p\) can take real values. However, if we analyze the original conditions, we can conclude that certain combinations of \(p\), \(\alpha\), and \(\beta\) might lead to contradictions, particularly if \(p = 0\).

### Final Answer:
The value of \(p\) cannot be zero if we want a valid quadratic polynomial. Therefore, the option that states \(p = 0\) is **not possible**.