Write a polynomial \( f(x) \) that satisfies the given conditions. Polynomial of lowest degree with zeros of \( \frac{5}{3} \) (multiplicity 2) and \( -\frac{2}{3} \) (multiplicity 1 ) and with \( f(0)=-150 \). \( f(x)=\square \)

To create the polynomial \( f(x) \) with the specified zeros, we start by incorporating the roots and their multiplicities: 1. The zero \( \frac{5}{3} \) with multiplicity 2 gives us the factor \( \left(x - \frac{5}{3}\right)^2 \). 2. The zero \( -\frac{2}{3} \) with multiplicity 1 gives us the factor \( \left(x + \frac{2}{3}\right) \). Thus, we can express the polynomial as: \[ f(x) = k \left(x - \frac{5}{3}\right)^2 \left(x + \frac{2}{3}\right) \] where \( k \) is a constant to be determined to satisfy the condition \( f(0) = -150 \). Next, we compute \( f(0) \): \[ f(0) = k \left(0 - \frac{5}{3}\right)^2 \left(0 + \frac{2}{3}\right) = k \left(\frac{5}{3}\right)^2 \left(\frac{2}{3}\right) = k \cdot \frac{25}{9} \cdot \frac{2}{3} = k \cdot \frac{50}{27} \] We set this equal to -150: \[ k \cdot \frac{50}{27} = -150 \] To find \( k \), we can multiply both sides by \( 27/50 \): \[ k = -150 \cdot \frac{27}{50} = -81 \] Now, substituting back for \( k \): \[ f(x) = -81 \left(x - \frac{5}{3}\right)^2 \left(x + \frac{2}{3}\right) \] To express this more cleanly, we can eliminate the fractions by multiplying throughout by 27: \[ f(x) = -81 \cdot \frac{1}{27} (3x - 5)^2 (3x + 2) \] This simplifies to: \[ f(x) = -3(3x - 5)^2 (3x + 2) \] Expanding this will confirm the polynomial's form, but for now, we can express the polynomial simply, ensuring all components fit the defined zeros and conditions. Therefore, the polynomial is: \[ f(x) = -81 \left(x - \frac{5}{3}\right)^2 \left(x + \frac{2}{3}\right) \]