If `alpha, beta, gamma ` are the zeros of the polynomial `p(x) = 6x^(3)+3x^(2)-5x+1,` find the value of `(1/alpha+1/beta+1/gamma).`

Options

A. 5
B. 6
C. 3
D. 1

Answer

To find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) where \( \alpha, \beta, \gamma \) are the zeros of the polynomial \( p(x) = 6x^3 + 3x^2 - 5x + 1 \), we can use the relationship between the roots and the coefficients of the polynomial.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The polynomial is given as \( p(x) = 6x^3 + 3x^2 - 5x + 1 \). Here, we can identify:
- \( a = 6 \) (coefficient of \( x^3 \))
- \( b = 3 \) (coefficient of \( x^2 \))
- \( c = -5 \) (coefficient of \( x \))
- \( d = 1 \) (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 \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{3}{6} = -\frac{1}{2} \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-5}{6} \)
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{1}{6} \)

3. **Calculate \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \)**:
We can rewrite \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) as:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]
Substituting the values from Vieta's formulas:
\[ = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{\frac{-5}{6}}{-\frac{1}{6}} \]

4. **Simplify the Expression**:
Simplifying the above expression gives:
\[ = \frac{-5}{6} \times \frac{-6}{1} = 5 \]

### Final Answer:
Thus, the value of \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) is \( 5 \).