Answer :
Answer:
Given that the quadratic polynomial is \(2x^2 - 3x + 1\) and its zeros are α and β. We can find the sum and product of the zeros using Viète's formulas:
The sum of the zeros (α + β) = -(-3)/2 = 3/2
The product of the zeros (αβ) = 1(2) = 2
Now we need to find a quadratic polynomial whose zeros are 5α and 5β. Let the new quadratic polynomial be \(ax^2 + bx + c\).
The sum of the new zeros (5α + 5β) = 5(α + β) = 5 * 3/2 = 15/2
The product of the new zeros (5α)(5β) = 25αβ = 25 * 2 = 50
Now we can form the new quadratic polynomial using the new sum and product of zeros:
\(a x^2 + b x + c = x^2 - \frac{15}{2}x + 50\)
Hence, the quadratic polynomial with zeros 5α and 5β is \(x^2 - \frac{15}{2}x + 50\).