If `alpha, beta, gamma` are the zeros of the polynomial ` x^(3)-6x^(2) -x+ 30` then `(alpha beta+beta gamma + gamma alpha)`= ?

Options

A. ` -1`
B. 1
C. `-5`
D. 30

Explanation

To find the value of \( \alpha \beta + \beta \gamma + \gamma \alpha \) for the polynomial \( x^3 - 6x^2 - x + 30 \), we can use Vieta's formulas. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots (zeros).

### Step-by-Step Solution:

1. **Identify the polynomial and its coefficients**:
The given polynomial is \( x^3 - 6x^2 - x + 30 \).
Here, the coefficients are:
- \( a = 1 \) (coefficient of \( x^3 \))
- \( b = -6 \) (coefficient of \( x^2 \))
- \( c = -1 \) (coefficient of \( x \))
- \( d = 30 \) (constant term)

2. **Use Vieta's formulas**:
According to Vieta's formulas, for a cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} \)

3. **Calculate \( \alpha\beta + \beta\gamma + \gamma\alpha \)**:
From Vieta's formulas, we have:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-1}{1} = -1 \]

4. **Final Answer**:
Therefore, the value of \( \alpha\beta + \beta\gamma + \gamma\alpha \) is \( -1 \).