Find a quadratic polynomial whose zeros are `(3+sqrt(5))/(5)` and `(3-sqrt(5))/(5)`.

Answer

To find a quadratic polynomial whose zeros are \(\frac{3 + \sqrt{5}}{5}\) and \(\frac{3 - \sqrt{5}}{5}\), we can use the fact that if \(\alpha\) and \(\beta\) are the roots of a quadratic polynomial, then the polynomial can be written in the form:

\[ P(x) = x^2 - (\alpha + \beta)x + \alpha\beta \]

Here, \(\alpha = \frac{3 + \sqrt{5}}{5}\) and \(\beta = \frac{3 - \sqrt{5}}{5}\).

### Step-by-Step Solution:

1. **Calculate the Sum of the Roots:**
\[ \alpha + \beta = \frac{3 + \sqrt{5}}{5} + \frac{3 - \sqrt{5}}{5} \]
Combine the fractions:
\[ \alpha + \beta = \frac{(3 + \sqrt{5}) + (3 - \sqrt{5})}{5} \]
Simplify the numerator:
\[ \alpha + \beta = \frac{3 + \sqrt{5} + 3 - \sqrt{5}}{5} = \frac{6}{5} \]

2. **Calculate the Product of the Roots:**
\[ \alpha \beta = \left(\frac{3 + \sqrt{5}}{5}\right) \left(\frac{3 - \sqrt{5}}{5}\right) \]
Use the identity \((a + b)(a - b) = a^2 - b^2\):
\[ \alpha \beta = \frac{(3 + \sqrt{5})(3 - \sqrt{5})}{25} = \frac{3^2 - (\sqrt{5})^2}{25} = \frac{9 - 5}{25} = \frac{4}{25} \]

3. **Form the Polynomial:**
Using the sum and product of the roots, the polynomial is:
\[ P(x) = x^2 - (\alpha + \beta)x + \alpha \beta \]
Substitute the values:
\[ P(x) = x^2 - \left(\frac{6}{5}\right)x + \frac{4}{25} \]

4. **Simplify the Polynomial (Optional):**
To simplify, multiply through by 25 to clear the fractions:
\[ 25P(x) = 25x^2 - 25 \cdot \frac{6}{5}x + 25 \cdot \frac{4}{25} \]
Simplify each term:
\[ 25P(x) = 25x^2 - 30x + 4 \]
Therefore, the polynomial can also be written as:
\[ P(x) = 25x^2 - 30x + 4 \]

### Final Answer:
The quadratic polynomial whose zeros are \(\frac{3 + \sqrt{5}}{5}\) and \(\frac{3 - \sqrt{5}}{5}\) is:
\[ P(x) = x^2 - \frac{6}{5}x + \frac{4}{25} \]
or equivalently:
\[ P(x) = 25x^2 - 30x + 4 \]