\begin{tabular}{l} For the following factored polynomial, find all of the zeros and their multiplicities. \\ \( \qquad f(x)=(x-2)^{5}(x+3)^{9}(x+6)^{8} \) \\ Select the correct answer below: \\ x=6 with multiplicity \( 8 ; x=3 \) with multiplicity \( 9 ; x=2 \) with multiplicity 5 \\ \( x=6 \) with multiplicity \( 8 ; x=-3 \) with multiplicity \( 9 ; x=-2 \) with multiplicity 5 \\ \( x=-6 \) with multiplicity \( 8 ; x=3 \) with multiplicity \( 9 ; x=2 \) with multiplicity 5 \\ \( x=-6 \) with multiplicity \( 8 ; x=-3 \) with multiplicity \( 9 ; x=2 \) with multiplicity 5 \\ \( x=-6 \) with multiplicity \( 8 ; x=3 \) with multiplicity \( 9 ; x=-2 \) with multiplicity 5 \\ \( x=-6 \) with multiplicity \( 8 ; x=-3 \) with multiplicity \( 9 ; x=-2 \) with multiplicity 5 \\ \hline\end{tabular}

The polynomial \( f(x)=(x-2)^{5}(x+3)^{9}(x+6)^{8} \) has zeros located at the values that make each factor zero. So, \( x=2 \) from the factor \( (x-2)^{5} \), \( x=-3 \) from \( (x+3)^{9} \), and \( x=-6 \) from \( (x+6)^{8} \). The multiplicities correspond to the powers in the factors: 5 for \( x=2 \), 9 for \( x=-3 \), and 8 for \( x=-6 \). Thus, the zeros and their multiplicities are: - \( x=2 \) with multiplicity \( 5 \) - \( x=-3 \) with multiplicity \( 9 \) - \( x=-6 \) with multiplicity \( 8 \) For fun, think of these zeros as a quirky conference where \( x=2 \) is a highly enthusiastic guest, showing up 5 times, \( x=-3 \) is very popular, arriving with 9 companions, and \( x=-6 \) is a mysterious character who brings along 8 fascinating tales! --- The options listed in the question include potential misinterpretations, but by focusing on understanding the zero-finding process, we clarify the correct multiplicities in a light-hearted manner. If you want me to focus on potential common mistakes in identifying zeros, just let me know!