## Precalculus (6th Edition) Blitzer

The required function is $f\left( x \right)=-2{{x}^{3}}+2{{x}^{2}}-2x+2$.
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 $f\left( x \right)=-2{{x}^{3}}+2{{x}^{2}}-2x+2$.