Answer
The required function is $f\left( x \right)=-2{{x}^{3}}+2{{x}^{2}}-2x+2$.
Work Step by Step
The degree of the polynomial is 3 and 1 and $i$ are zeros $f\left( -1 \right)=8$.
Since the polynomial has real coefficients, the conjugate of i, -i is also a root of the equation. Thus:
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right) \\
& ={{a}_{n}}\left( x-1 \right)\left( x-i \right)\left( x+i \right) \\
& ={{a}_{n}}\left( x-1 \right)\left( {{x}^{2}}+1 \right) \\
& ={{a}_{n}}\left( {{x}^{3}}-{{x}^{2}}+x-1 \right)
\end{align}$
Use $f\left( -1 \right)-8$ and evaluate:
,
$\begin{align}
& {{a}_{n}}\left( {{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}+\left( -1 \right)-1 \right)=8 \\
& {{a}_{n}}\left( -1-1-1-1 \right)=8 \\
& -4{{a}_{n}}=8 \\
& {{a}_{n}}=-2
\end{align}$
Substitute $-2$ for ${{a}_{n}}$ in $f\left( x \right)$ ,
$\begin{align}
& f\left( x \right)=-2\left( {{x}^{3}}-{{x}^{2}}+x-1 \right) \\
& =-2{{x}^{3}}+2{{x}^{2}}-2x+2
\end{align}$
The third degree polynomial function is $f\left( x \right)=-2{{x}^{3}}+2{{x}^{2}}-2x+2$.
