Write a polynomial \( f(x) \) that satisfies the given conditions. Polynomial of lowest degree with zeros of \( \frac{1}{6} \) (multiplicity 2) and \( -\frac{5}{3} \) (multiplicity 1\( ) \) and with \( f(0)=-5 \). \( f(x)=\square \)

To construct the polynomial \( f(x) \), we begin by considering its zeros and their multiplicities. The zeros of \( f(x) \) are \( \frac{1}{6} \) with multiplicity 2 and \( -\frac{5}{3} \) with multiplicity 1. Therefore, the polynomial can be expressed as: \[ f(x) = k \left( x - \frac{1}{6} \right)^2 \left( x + \frac{5}{3} \right) \] Here, \( k \) is a constant that we will determine using the condition \( f(0) = -5 \). Now, let's simplify the expression for \( f(x) \): 1. First, calculate \( \left( x - \frac{1}{6} \right)^2 \): \[ \left( x - \frac{1}{6} \right)^2 = x^2 - \frac{1}{3}x + \frac{1}{36} \] 2. Now, compute \( \left( x + \frac{5}{3} \right) \): \[ \left( x + \frac{5}{3} \right) = x + \frac{5}{3} \] 3. Multiply the two expressions: \[ f(x) = k \left( x^2 - \frac{1}{3}x + \frac{1}{36} \right) \left( x + \frac{5}{3} \right) \] Using the distributive property, we can perform the multiplication: \[ = k \left( x^3 + \frac{5}{3} x^2 - \frac{1}{3}x^2 - \frac{5}{9}x + \frac{1}{36}x + \frac{5}{108} \right) \] Combining like terms, we have: \[ = k \left( x^3 + \frac{4}{3} x^2 - \frac{11}{36}x + \frac{5}{108} \right) \] Next, we will apply the condition \( f(0) = -5 \): \[ f(0) = k \left(0^3 + \frac{4}{3} \cdot 0^2 - \frac{11}{36} \cdot 0 + \frac{5}{108}\right) = k \cdot \frac{5}{108} \] Setting this equal to \( -5 \): \[ k \cdot \frac{5}{108} = -5 \] Solving for \( k \): \[ k = -5 \cdot \frac{108}{5} = -108 \] Now substituting back for \( k \): \[ f(x) = -108 \left( x^3 + \frac{4}{3} x^2 - \frac{11}{36}x + \frac{5}{108} \right) \] Distributing \( -108 \): \[ f(x) = -108x^3 - 144x^2 + 33x - 5 \] Thus, the polynomial that satisfies the conditions is: \[ \boxed{-108x^3 - 144x^2 + 33x - 5} \]