Answer
The required ${{n}^{th}}$ degree polynomial function is $2{{x}^{3}}-8{{x}^{2}}+8x-32$.
Work Step by Step
The linear general 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 ${{c}_{1}}=2i$ , ${{c}_{2}}=-2i$ and ${{c}_{3}}=4$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right) \\
& ={{a}_{n}}\left( x-2i \right)\left( x+2i \right)\left( x-4 \right) \\
& ={{a}_{n}}\left( {{x}^{3}}-4{{x}^{2}}+4x-16 \right)
\end{align}$
Calculate the value of ${{a}_{n}}$.
Use $f\left( -1 \right)=-50$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-4{{x}^{2}}+4x-16 \right) \\
& -50={{a}_{n}}\left( {{\left( -1 \right)}^{3}}-4{{\left( -1 \right)}^{2}}+4\left( -1 \right)-16 \right) \\
& ={{a}_{n}}\left( -1-4-4-16 \right) \\
& {{a}_{n}}=\frac{50}{25} \\
& =2
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}-4{{x}^{2}}+4x-16 \right) \\
& =2\left( {{x}^{3}}-4{{x}^{2}}+4x-16 \right) \\
& =2{{x}^{3}}-8{{x}^{2}}+8x-32
\end{align}$
Therefore, the polynomial equation is $2{{x}^{3}}-8{{x}^{2}}+8x-32$.
