Question

equation of a polynomial function that has zeroes at 3(multiplicity of 1), -2(multiplicity of 1), and 0(multiplicity of 2)

Answer 1

\(\large {
\begin{cases}{}
x=3\to x-3=0&\textit{multiplicity of 1}\to (x-3)^{\color{blue}{ 1}}=0\\
x=-2\to x+2=0&\textit{multiplicity of 1}\to (x+2)^{\color{blue}{ 1}}=0\\
x=0\to x=0&\textit{multiplicity of 2}\to (x)^{\color{blue}{ 2}}=0
\end{cases}
\\ \quad \\ \quad \\

(x-3)^{\color{blue}{ 1}}(x+2)^{\color{blue}{ 1}}(x)^{\color{blue}{ 2}}=\textit{original polynomial}
}\)