Answer
The required ${{n}^{th}}$ degree polynomial function is $2{{x}^{3}}-2{{x}^{2}}+50x-50$.
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 ${{c}_{1}}=5i$ , ${{c}_{2}}=-5i$ and ${{c}_{3}}=1$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right) \\
& ={{a}_{n}}\left( x-5i \right)\left( x+5i \right)\left( x-1 \right) \\
& ={{a}_{n}}\left( {{x}^{3}}-{{x}^{2}}+25x-25 \right)
\end{align}$
Calculate the value of ${{a}_{n}}$.
Use $f\left( -1 \right)=-104$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-{{x}^{2}}+25x-25 \right) \\
& -104={{a}_{n}}\left( {{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}+25\left( -1 \right)-25 \right) \\
& ={{a}_{n}}\left( -1-1-25-25 \right) \\
& {{a}_{n}}=\frac{104}{52} \\
& =2
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation as follows:
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-{{x}^{2}}+25x-25 \right) \\
& =2\left( {{x}^{3}}-{{x}^{2}}+25x-25 \right) \\
& =2{{x}^{3}}-2{{x}^{2}}+50x-50
\end{align}$
Therefore, the polynomial equation is $2{{x}^{3}}-2{{x}^{2}}+50x-50$.
