Given that one of the zereos 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)`

Answer

To find the product of the other two zeros of the cubic polynomial \( ax^3 + bx^2 + cx + d \) given that one of the zeros is zero, we can follow these steps:

### Step 1: Identify the zeros
Let the zeros of the polynomial be \( \alpha, \beta, \gamma \). Given that one of the zeros, say \( \gamma \), is zero, we have:
\[ \gamma = 0 \]
Thus, we can denote the other two zeros as \( \alpha \) and \( \beta \).

### Step 2: Use the relationship of the zeros
For a cubic polynomial \( ax^3 + bx^2 + cx + d \), the relationship between the zeros and the coefficients can be expressed as:
\[ \alpha + \beta + \gamma = -\frac{b}{a} \]
Since \( \gamma = 0 \), this simplifies to:
\[ \alpha + \beta = -\frac{b}{a} \]

### Step 3: Use the product of the zeros
Another relationship for the product of the zeros is given by:
\[ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \]
Substituting \( \gamma = 0 \) into this equation gives:
\[ \alpha \beta + \beta \cdot 0 + 0 \cdot \alpha = \frac{c}{a} \]
This simplifies to:
\[ \alpha \beta = \frac{c}{a} \]

### Conclusion
Thus, the product of the other two zeros \( \alpha \) and \( \beta \) is:
\[ \alpha \beta = \frac{c}{a} \]

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

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