Answer
The required ${{n}^{th}}$ degree polynomial function is $3{{x}^{3}}+12{{x}^{2}}-93x-522$.
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 ${{c}_{1}}=-5+2i$ , ${{c}_{2}}=-5-2i$ and ${{c}_{3}}=6$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right) \\
& ={{a}_{n}}\left( x+5-2i \right)\left( x+5+2i \right)\left( x-6 \right) \\
& ={{a}_{n}}\left( {{x}^{3}}+4{{x}^{2}}-31x-174 \right)
\end{align}$
Calculate the value of ${{a}_{n}}$.
Use $f\left( 2 \right)=-636$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}+4{{x}^{2}}-31x-174 \right) \\
& -636={{a}_{n}}\left( {{\left( 2 \right)}^{3}}+4{{\left( 2 \right)}^{2}}-31\left( 2 \right)-174 \right) \\
& ={{a}_{n}}\left( -212 \right) \\
& {{a}_{n}}=3
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{3}}+4{{x}^{2}}-31x-174 \right) \\
& =3\left( {{x}^{3}}+4{{x}^{2}}-31x-174 \right) \\
& =3{{x}^{3}}+12{{x}^{2}}-93x-522
\end{align}$
Therefore, the polynomial equation is $3{{x}^{3}}+12{{x}^{2}}-93x-522$.
