If one of the zereos of the cubic polynomial `x^(3)+ax^(2)+bx+c` is -1, then the product of the other two zeroes is

Options

A. `b-a+1`
B. `b-a-1`
C. `a-b+1`
D. `a-b-1`

Answer

To find the product of the other two zeros of the cubic polynomial \( x^3 + ax^2 + bx + c \) given that one of the zeros is -1, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the zeros**: Let the three zeros of the polynomial be \( \alpha, \beta, \gamma \). Given that one of the zeros, say \( \gamma \), is -1, we can write:
\[ \gamma = -1 \]

2. **Use the sum of the zeros**: The sum of the zeros of a polynomial \( x^3 + ax^2 + bx + c \) is given by:
\[ \alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -a \]
Substituting \( \gamma = -1 \):
\[ \alpha + \beta - 1 = -a \]
Rearranging gives:
\[ \alpha + \beta = -a + 1 \]

3. **Use the product of the zeros**: The product of the zeros of the polynomial is given by:
\[ \alpha \beta \gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -c \]
Substituting \( \gamma = -1 \):
\[ \alpha \beta (-1) = -c \implies \alpha \beta = c \]

4. **Use the sum of the products of the zeros**: The sum of the products of the zeros taken two at a time is given by:
\[ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = b \]
Substituting \( \gamma = -1 \):
\[ \alpha \beta + \beta (-1) + (-1) \alpha = b \]
This simplifies to:
\[ \alpha \beta - \beta - \alpha = b \]

5. **Substituting \( \alpha + \beta \)**: We know from step 2 that \( \alpha + \beta = -a + 1 \). We can substitute this into the equation:
\[ \alpha \beta - (-a + 1) = b \]
Rearranging gives:
\[ \alpha \beta + a - 1 = b \implies \alpha \beta = b - a + 1 \]

6. **Conclusion**: Therefore, the product of the other two zeros \( \alpha \) and \( \beta \) is:
\[ \alpha \beta = b - a + 1 \]

### Final Answer:
The product of the other two zeros is \( b - a + 1 \).