If the sum of the two roots of ` x^3 +px^2 +qx +r=0` is zero then pq=

Options

A. `-r`
B. `r`
C. `2r`
D. `-2r`

Answer

To solve the problem, we need to find the value of \( pq \) given that the sum of two roots of the polynomial \( x^3 + px^2 + qx + r = 0 \) is zero.

### Step-by-Step Solution:

1. **Identify the roots**: Let the roots of the polynomial be \( \alpha, \beta, \) and \( \gamma \). According to the problem, we know that the sum of two roots is zero, which implies:
\[ \alpha + \beta = 0 \]

2. **Express one root in terms of the other**: From the equation \( \alpha + \beta = 0 \), we can express \( \beta \) as:
\[ \beta = -\alpha \]

3. **Use Vieta's formulas**: According to Vieta's formulas for a cubic polynomial \( x^3 + px^2 + qx + r = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -p \).
- Substituting \( \beta = -\alpha \) gives:
\[ \alpha - \alpha + \gamma = -p \implies \gamma = -p \]

4. **Substitute the roots into the polynomial**: Now, we will substitute \( \gamma = -p \) into the polynomial:
\[ f(\gamma) = \gamma^3 + p\gamma^2 + q\gamma + r = 0 \]
Substituting \( \gamma = -p \):
\[ (-p)^3 + p(-p)^2 + q(-p) + r = 0 \]

5. **Simplify the equation**: This expands to:
\[ -p^3 + p(p^2) - qp + r = 0 \]
Simplifying this gives:
\[ -p^3 + p^3 - qp + r = 0 \]
The \( -p^3 + p^3 \) terms cancel out, leading to:
\[ -qp + r = 0 \implies r = qp \]

6. **Find the value of \( pq \)**: We need to express \( pq \) in terms of \( r \):
\[ pq = \frac{r}{p} \]
However, we also know from the earlier step that \( r = p^3 \). Thus:
\[ pq = p^2 \]

7. **Conclusion**: Since \( r = qp \) and \( r = p^3 \), we conclude:
\[ pq = p^2 \]

### Final Result:
The value of \( pq \) is \( p^2 \).