Answer
The required ${{n}^{th}}$ degree polynomial function is ${{x}^{3}}-3{{x}^{2}}-15x+125$
Work Step by Step
The general linear equation is $f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)$ , the roots are ${{c}_{1}}=4+3i$ , ${{c}_{2}}=4-3i$ and ${{c}_{3}}=-5$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right) \\
& ={{a}_{n}}\left( x-4-3i \right)\left( x-4+3i \right)\left( x+5 \right) \\
& ={{a}_{n}}\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right)
\end{align}$
Calculate the value of ${{a}_{n}}$.
Use $f\left( 2 \right)=91$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \\
& 91={{a}_{n}}\left( {{\left( 2 \right)}^{3}}-3{{\left( 2 \right)}^{2}}-15\left( 2 \right)+125 \right) \\
& ={{a}_{n}}\left( 8-12-30+125 \right) \\
& {{a}_{n}}=\frac{91}{91} \\
& =1
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \\
& =1\left( {{x}^{3}}-3{{x}^{2}}-15x+125 \right) \\
& ={{x}^{3}}-3{{x}^{2}}-15x+125
\end{align}$
Therefore, the polynomial equation is ${{x}^{3}}-3{{x}^{2}}-15x+125$.
