If `alpha` and `beta` are the zeros of the quadratic polynomial `f(x) = 3x^2 - 7x - 6`, find a polynomial whose zeros are
(i) `alpha^(2)` and `beta^(2)`, (ii) `2alpha + 3beta` and `3alpha + 2beta`.

Answer

To solve the problem, we need to find a polynomial whose zeros are given based on the zeros of the quadratic polynomial \( f(x) = 3x^2 - 7x - 6 \). Let's go through the steps systematically.

### Step 1: Find the zeros of the polynomial \( f(x) \)

The zeros of the polynomial are given by the quadratic formula:

\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the polynomial \( f(x) = 3x^2 - 7x - 6 \), we identify \( a = 3 \), \( b = -7 \), and \( c = -6 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot (-6) = 49 + 72 = 121 \]

Now applying the quadratic formula:

\[ \alpha, \beta = \frac{7 \pm \sqrt{121}}{2 \cdot 3} = \frac{7 \pm 11}{6} \]

Calculating the two zeros:

1. For the positive sign:
\[ \alpha = \frac{18}{6} = 3 \]

2. For the negative sign:
\[ \beta = \frac{-4}{6} = -\frac{2}{3} \]

### Step 2: Find the polynomial with zeros \( \alpha^2 \) and \( \beta^2 \)

Calculating \( \alpha^2 \) and \( \beta^2 \):

\[ \alpha^2 = 3^2 = 9 \]
\[ \beta^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \]

Now, we need to find the polynomial whose zeros are \( 9 \) and \( \frac{4}{9} \).

**Sum of the zeros**:

\[ \text{Sum} = 9 + \frac{4}{9} = \frac{81}{9} + \frac{4}{9} = \frac{85}{9} \]

**Product of the zeros**:

\[ \text{Product} = 9 \cdot \frac{4}{9} = 4 \]

Using the formula for a quadratic polynomial:

\[ P(x) = x^2 - (\text{Sum})x + (\text{Product}) \]

Substituting the values:

\[ P(x) = x^2 - \frac{85}{9}x + 4 \]

To eliminate the fraction, multiply through by 9:

\[ P(x) = 9x^2 - 85x + 36 \]

### Step 3: Find the polynomial with zeros \( 2\alpha + 3\beta \) and \( 3\alpha + 2\beta \)

Calculating \( 2\alpha + 3\beta \):

\[ 2\alpha + 3\beta = 2(3) + 3\left(-\frac{2}{3}\right) = 6 - 2 = 4 \]

Calculating \( 3\alpha + 2\beta \):

\[ 3\alpha + 2\beta = 3(3) + 2\left(-\frac{2}{3}\right) = 9 - \frac{4}{3} = \frac{27}{3} - \frac{4}{3} = \frac{23}{3} \]

Now we have the zeros \( 4 \) and \( \frac{23}{3} \).

**Sum of the zeros**:

\[ \text{Sum} = 4 + \frac{23}{3} = \frac{12}{3} + \frac{23}{3} = \frac{35}{3} \]

**Product of the zeros**:

\[ \text{Product} = 4 \cdot \frac{23}{3} = \frac{92}{3} \]

Using the formula for a quadratic polynomial:

\[ Q(x) = x^2 - (\text{Sum})x + (\text{Product}) \]

Substituting the values:

\[ Q(x) = x^2 - \frac{35}{3}x + \frac{92}{3} \]

To eliminate the fraction, multiply through by 3:

\[ Q(x) = 3x^2 - 35x + 92 \]

### Final Answers

1. The polynomial whose zeros are \( \alpha^2 \) and \( \beta^2 \) is:
\[ 9x^2 - 85x + 36 \]

2. The polynomial whose zeros are \( 2\alpha + 3\beta \) and \( 3\alpha + 2\beta \) is:
\[ 3x^2 - 35x + 92 \]