A polynomial \( f(x) \) and one or more of its zeros are given. Write all numbers as integers or simplified fractions. (a) Find all the zeros. (b) Factor \( f(x) \) as a product of linear factors. (c) Solve the equation \( f(x)=0 \). \( f(x)=3 x^{3}-56 x^{2}+282 x-164 \); \( 9+i \) is a zero

To find all the zeros of the polynomial \( f(x) = 3x^3 - 56x^2 + 282x - 164 \), given that \( 9 + i \) is a zero, we also know that the complex conjugate \( 9 - i \) must be a zero as well. To find the remaining zero, we can perform polynomial long division or synthetic division using the factor corresponding to these complex zeros. First, since \( 9+i \) and \( 9-i \) are both zeros, we can form the quadratic polynomial \( (x - (9+i))(x - (9-i)) = (x - 9)^2 + 1 = x^2 - 18x + 82 \). Next, we will divide \( f(x) \) by \( x^2 - 18x + 82 \). Doing this yields: \[ f(x) = (x^2 - 18x + 82)(3x - 2) \] To find the third zero, we set \( 3x - 2 = 0 \) leading to \( x = \frac{2}{3} \). In summary, the zeros are \( 9 + i \), \( 9 - i \), and \( \frac{2}{3} \). Now, we can express \( f(x) \) as a product of linear factors: \[ f(x) = 3(x - (9 + i))(x - (9 - i))(x - \frac{2}{3}) \] This shows the polynomial factored with one linear factor related to the complex number and one linear factor for the rational zero. Lastly, to solve the equation \( f(x) = 0 \), we simply set each factor equal to zero: 1. \( x = 9 + i \) 2. \( x = 9 - i \) 3. \( x = \frac{2}{3} \) Thus, the complete solution concludes with the zeros provided.