If one of the zeroes of the cubic polynomial `ax^(3) +bx^(2) +cx +d` is zero, the product of the other two zeroes is

Options

A. `(-c)/(a)`
B. `(c)/(a)`
C. 0
D. `(-b)/(a)`

Explanation

To solve the problem, we need to analyze the cubic polynomial given and the implications of one of its zeroes being zero.

### Step-by-Step Solution:

1. **Identify the Polynomial**:
We have a cubic polynomial given by \( P(x) = ax^3 + bx^2 + cx + d \).

2. **Understand the Zeroes**:
A cubic polynomial can have three zeroes, which we can denote as \( \alpha, \beta, \) and \( \gamma \).

3. **Given Condition**:
We are told that one of the zeroes is zero. Let's assume \( \alpha = 0 \).

4. **Using the Relationship of Zeroes**:
According to Vieta's formulas, for a cubic polynomial \( ax^3 + bx^2 + cx + d \), the relationships between the coefficients and the zeroes are:
- Sum of the zeroes: \( \alpha + \beta + \gamma = -\frac{b}{a} \)
- Sum of the product of the zeroes taken two at a time: \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
- Product of the zeroes: \( \alpha\beta\gamma = -\frac{d}{a} \)

5. **Substituting the Known Zero**:
Since we have \( \alpha = 0 \), we can substitute this into the second relationship:
\[ 0 \cdot \beta + \beta \cdot \gamma + 0 \cdot \gamma = \frac{c}{a} \]
This simplifies to:
\[ \beta \cdot \gamma = \frac{c}{a} \]

6. **Conclusion**:
Therefore, the product of the other two zeroes \( \beta \) and \( \gamma \) is given by:
\[ \beta \cdot \gamma = \frac{c}{a} \]

### Final Answer:
The product of the other two zeroes is \( \frac{c}{a} \).