If `alpha` and `beta` are zeros of the polynomial `t^(2)-t-4`, form a quadratic polynomial whose zeros are `(1)/(alpha)` and `(1)/(beta)`.

Answer

To solve the problem, we need to find a quadratic polynomial whose zeros are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \), given that \( \alpha \) and \( \beta \) are the zeros of the polynomial \( t^2 - t - 4 \).

### Step-by-Step Solution:

1. **Identify the Coefficients of the Given Polynomial:**
The polynomial is \( t^2 - t - 4 \). Here, we can identify:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -4 \)

2. **Calculate the Sum and Product of the Roots \( \alpha \) and \( \beta \):**
Using the relationships for the sum and product of roots:
- Sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1 \)
- Product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-4}{1} = -4 \)

3. **Find the Sum and Product of the New Roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \):**
- Sum of the new roots:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{1}{-4} = -\frac{1}{4} \]
- Product of the new roots:
\[ \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{-4} = -\frac{1}{4} \]

4. **Form the Quadratic Polynomial:**
A quadratic polynomial with roots \( p \) and \( q \) can be expressed as:
\[ x^2 - (p + q)x + pq \]
Substituting \( p = \frac{1}{\alpha} \) and \( q = \frac{1}{\beta} \):
\[ t^2 - \left(-\frac{1}{4}\right)t + \left(-\frac{1}{4}\right) \]
This simplifies to:
\[ t^2 + \frac{1}{4}t - \frac{1}{4} \]

5. **Eliminate the Fraction:**
To write the polynomial in a standard form, we can multiply through by 4 to eliminate the fractions:
\[ 4t^2 + t - 1 \]

### Final Result:
The required quadratic polynomial whose zeros are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
\[ 4t^2 + t - 1 \]