If `alpha ` and `beta` are zeros of polynomial `6x^(2)-7x-3`, then form a quadratic polynomial where zeros are `2alpha` and `2beta`

Answer

To solve the problem, we need to find a quadratic polynomial whose zeros are \(2\alpha\) and \(2\beta\), given that \(\alpha\) and \(\beta\) are the zeros of the polynomial \(6x^2 - 7x - 3\).

### Step 1: Identify the coefficients of the given polynomial
The given polynomial is \(6x^2 - 7x - 3\). Here, we can identify:
- \(a = 6\)
- \(b = -7\)
- \(c = -3\)

### Step 2: Calculate the sum and product of the roots \(\alpha\) and \(\beta\)
Using the formulas for the sum and product of the roots of a quadratic polynomial:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}\)

### Step 3: Find the sum and product of the new roots \(2\alpha\) and \(2\beta\)
Now, we need to find the sum and product of the new roots \(2\alpha\) and \(2\beta\):
- The sum of the new roots \(2\alpha + 2\beta = 2(\alpha + \beta) = 2 \times \frac{7}{6} = \frac{14}{6} = \frac{7}{3}\)
- The product of the new roots \(2\alpha \cdot 2\beta = 4(\alpha \beta) = 4 \times -\frac{1}{2} = -2\)

### Step 4: Form the new quadratic polynomial
Using the sum and product of the new roots, we can form the quadratic polynomial:
\[ x^2 - \text{(sum of roots)} \cdot x + \text{(product of roots)} = 0 \]
Substituting the values we found:
\[ x^2 - \frac{7}{3}x - 2 = 0 \]

### Step 5: Eliminate the fraction by multiplying through by 3
To eliminate the fraction, multiply the entire equation by 3:
\[ 3x^2 - 7x - 6 = 0 \]

Thus, the quadratic polynomial whose zeros are \(2\alpha\) and \(2\beta\) is:
\[ \boxed{3x^2 - 7x - 6} \]