Answer
The required ${{n}^{th}}$ degree polynomial function is ${{x}^{4}}+10{{x}^{2}}+9$.
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 ${{c}_{1}}=i,{{c}_{2}}=-i,{{c}_{3}}=3i$ , and ${{c}_{4}}=-3i$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)\left( x-{{c}_{4}} \right) \\
& ={{a}_{n}}\left( x-i \right)\left( x+i \right)\left( x-3i \right)\left( x+3i \right) \\
& ={{a}_{n}}\left( {{x}^{4}}+10{{x}^{2}}+9 \right)
\end{align}$
Now, compute the value of ${{a}_{n}}$ .
Use $f\left( -1 \right)=20$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}+10{{x}^{2}}+9 \right) \\
& 20={{a}_{n}}\left( {{\left( -1 \right)}^{4}}+10{{\left( -1 \right)}^{2}}+9 \right) \\
& 20={{a}_{n}}\left( 1+10+9 \right) \\
& {{a}_{n}}=1
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}+10{{x}^{2}}+9 \right) \\
& =1\left( {{x}^{4}}+10{{x}^{2}}+9 \right) \\
& ={{x}^{4}}+10{{x}^{2}}+9
\end{align}$
Therefore, the polynomial equation is ${{x}^{4}}+10{{x}^{2}}+9$.
