Answer
The required ${{n}^{th}}$ degree polynomial function is ${{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130$.
Work Step by Step
The general linear equation is $f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)\left( x-{{c}_{4}} \right)$ , the roots are ${{c}_{1}}=-2,{{c}_{2}}=5,{{c}_{3}}=3+2i$ , and ${{c}_{4}}=3-2i$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)\left( x-{{c}_{4}} \right) \\
& ={{a}_{n}}\left( x+2 \right)\left( x-5 \right)\left( x-3-2i \right)\left( x-3+2i \right) \\
& ={{a}_{n}}\left( {{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130 \right)
\end{align}$
Now, compute the value of ${{a}_{n}}$.
Use $f\left( 1 \right)=-96$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130 \right) \\
& -96={{a}_{n}}\left( {{\left( 1 \right)}^{4}}-9{{\left( 1 \right)}^{3}}+21{{\left( 1 \right)}^{2}}+21\left( 1 \right)-130 \right) \\
& -96={{a}_{n}}\left( 1-9+21+21-130 \right) \\
& {{a}_{n}}=1
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130 \right) \\
& =1\left( {{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130 \right) \\
& ={{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130
\end{align}$
Therefore, the polynomial equation is ${{x}^{4}}-9{{x}^{3}}+21{{x}^{2}}+21x-130$.
