Answer
The required ${{n}^{th}}$ degree polynomial function is $2{{x}^{4}}+5{{x}^{3}}+4{{x}^{2}}+5x+2$.
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 ${{c}_{1}}=-2,{{c}_{2}}=-\frac{1}{2},{{c}_{3}}=i$ , and ${{c}_{4}}=-i$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)\left( x-{{c}_{4}} \right) \\
& ={{a}_{n}}\left( x+2 \right)\left( x+\frac{1}{2} \right)\left( x-i \right)\left( x+i \right) \\
& ={{a}_{n}}\left( {{x}^{4}}+\frac{5}{2}{{x}^{3}}+2{{x}^{2}}+\frac{5}{2}x+1 \right)
\end{align}$
Now, compute the value of ${{a}_{n}}$.
Use $f\left( 1 \right)=18$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}+\frac{5}{2}{{x}^{3}}+2{{x}^{2}}+\frac{5}{2}x+1 \right) \\
& 18={{a}_{n}}\left( {{\left( 1 \right)}^{4}}+\frac{5}{2}{{\left( 1 \right)}^{3}}+2{{\left( 1 \right)}^{2}}+\frac{5}{2}\left( 1 \right)+1 \right) \\
& 18={{a}_{n}}\left( 1+\frac{5}{2}+2+\frac{5}{2}+1 \right) \\
& {{a}_{n}}=2
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}+\frac{5}{2}{{x}^{3}}+2{{x}^{2}}+\frac{5}{2}x+1 \right) \\
& =2\left( {{x}^{4}}+\frac{5}{2}{{x}^{3}}+2{{x}^{2}}+\frac{5}{2}x+1 \right) \\
& =2{{x}^{4}}+5{{x}^{3}}+4{{x}^{2}}+5x+2
\end{align}$
Therefore, the polynomial equation is $2{{x}^{4}}+5{{x}^{3}}+4{{x}^{2}}+5x+2$.
