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 \( f(0)=-16 \).

To construct the polynomial \( f(x) \), we start with its zeros and their multiplicities. The zeros given are \( \frac{4}{3} \) with multiplicity 2, and \( -\frac{1}{2} \) with multiplicity 1. Therefore, we can express the polynomial in factored form as: \[ f(x) = k \left( x - \frac{4}{3} \right)^2 \left( x + \frac{1}{2} \right) \] where \( k \) is a constant that we need to determine. Now, let's simplify the factors. Rewrite \( -\frac{4}{3} \) and \( \frac{1}{2} \) to eliminate the fractions. The polynomial becomes: \[ f(x) = k \left( 3x - 4 \right)^2 \left( 2x + 1 \right) \] Next, we need to find the value of \( k \) such that \( f(0) = -16 \). Substituting \( x = 0 \): \[ f(0) = k \left( 3(0) - 4 \right)^2 \left( 2(0) + 1 \right) = k (4)^2 (1) = 16k \] Setting this equal to \(-16\): \[ 16k = -16 \] Dividing both sides by 16 gives: \[ k = -1 \] Now we substitute \( k \) back into our polynomial: \[ f(x) = -1 \left( 3x - 4 \right)^2 \left( 2x + 1 \right) \] Now let's expand this polynomial: 1. Expand \( (3x - 4)^2 \): \[ (3x - 4)(3x - 4) = 9x^2 - 24x + 16 \] 2. Multiply by \( (2x + 1) \): \[ (9x^2 - 24x + 16)(2x + 1) = 18x^3 - 24x^2 + 9x^2 - 48x + 32 + 16 \] This yields: \[ 18x^3 - 24x^2 - 48x + 32 \] Now multiplying the expanded result by -1 gives: \[ f(x) = -18x^3 + 24x^2 + 48x - 32 \] Thus, the polynomial of lowest degree that satisfies the given conditions is: \[ \boxed{-18x^3 + 24x^2 + 48x - 32} \]