Write a polynomial \( f(x) \) that satisfies the given conditions. Polynomial of lowest degree with zeros of \( \frac{4}{3} \) (multiplicity 2\( ) \) and \( -\frac{1}{2} \) (multiplicity 1 ) and with \( f(0) \)

To create a polynomial \( f(x) \) of the lowest degree while satisfying the given conditions, we start with the zeros and their multiplicities. The zero \( \frac{4}{3} \) has multiplicity 2, contributing the factor \( \left( x - \frac{4}{3} \right)^2 \). The zero \( -\frac{1}{2} \) has multiplicity 1, contributing the factor \( \left( x + \frac{1}{2} \right) \). Combine these factors to express the polynomial: \[ f(x) = k \left( x - \frac{4}{3} \right)^2 \left( x + \frac{1}{2} \right) \] where \( k \) is a constant that will be determined to ensure \( f(0) \) is also defined correctly. Now, let's convert the factor involving the zero \( \frac{4}{3} \): \[ f(x) = k \left( x - \frac{4}{3} \right)^2 (x + \frac{1}{2}) = k \left( x^2 - \frac{8}{3}x + \frac{16}{9} \right) \left( x + \frac{1}{2} \right) \] Now, we will evaluate \( f(0) \) to determine the value of \( k \): \[ f(0) = k \left( -\frac{16}{9} \cdot \frac{1}{2} \right) = k \left( -\frac{8}{9} \right) \] To get a polynomial with integer coefficients, we multiply the whole polynomial by 9: \[ f(x) = 9k \left( x - \frac{4}{3} \right)^2 \left( x + \frac{1}{2} \right) \] Choosing \( k=1 \) gives us an easy variant: \[ f(x) = 9 \left( x - \frac{4}{3} \right)^2 \left( x + \frac{1}{2} \right) = (x - \frac{4}{3})^2 (x + \frac{1}{2}) \] Ultimately after further calculations, we find: \[ f(x) = 18x^2 - 48x + 32 \] Therefore, the polynomial \( f(x) = 18x^2 - 48x + 32 \) satisfies all given conditions.