If `alpha` and `beta` are the zeros of the polynomial `f(x) = 4x^2 - 5x +1`, find a quadratic polynomial whose zeros are `alpha^(2)/beta` and `beta^(2)/alpha`

Answer

To find a quadratic polynomial whose zeros are \(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\), we start from the given polynomial \(f(x) = 4x^2 - 5x + 1\). Here are the steps to solve the problem:

### Step 1: Identify the coefficients of the polynomial
The polynomial is given as:
\[ f(x) = 4x^2 - 5x + 1 \]
From this, we can identify:
- \(a = 4\)
- \(b = -5\)
- \(c = 1\)

### Step 2: Calculate the sum and product of the roots \(\alpha\) and \(\beta\)
Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-5}{4} = \frac{5}{4}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{1}{4}\)

### Step 3: Find the sum of the new roots \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\)
We can rewrite the sum of the new roots as:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} \]
Using the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We also know:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{4}\right)^2 - 2 \cdot \frac{1}{4} = \frac{25}{16} - \frac{2}{4} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16} \]
Thus,
\[ \alpha^3 + \beta^3 = \left(\frac{5}{4}\right)\left(\frac{17}{16} - \frac{1}{4}\right) = \left(\frac{5}{4}\right)\left(\frac{17}{16} - \frac{4}{16}\right) = \left(\frac{5}{4}\right)\left(\frac{13}{16}\right) = \frac{65}{64} \]
Now substituting back:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\frac{65}{64}}{\frac{1}{4}} = \frac{65}{64} \cdot 4 = \frac{65}{16} \]

### Step 4: Calculate the product of the new roots \(\frac{\alpha^2}{\beta} \cdot \frac{\beta^2}{\alpha}\)
This can be simplified as:
\[ \frac{\alpha^2}{\beta} \cdot \frac{\beta^2}{\alpha} = \frac{\alpha^2 \beta^2}{\alpha \beta} = \alpha \beta \]
We already found \(\alpha \beta = \frac{1}{4}\).

### Step 5: Form the new quadratic polynomial
Using the sum and product of the new roots, we can write the new quadratic polynomial:
\[ g(x) = x^2 - \left(\frac{65}{16}\right)x + \frac{1}{4} \]

### Step 6: Clear the fractions
To eliminate the fractions, multiply the entire equation by 16:
\[ 16g(x) = 16x^2 - 65x + 4 \]
Thus, the required quadratic polynomial is:
\[ g(x) = 16x^2 - 65x + 4 \]

### Final Answer:
The quadratic polynomial whose zeros are \(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\) is:
\[ \boxed{16x^2 - 65x + 4} \]