## Precalculus (6th Edition) Blitzer

The required polynomial is $f\left( x \right)={{x}^{4}}-4{{x}^{3}}+13{{x}^{2}}-36x+36$.
Since the polynomial has real coefficients, the conjugate of 3i, -3i 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)\left( x-{{c}_{4}} \right) \\ & ={{a}_{n}}\left( x-2 \right)\left( x-2 \right)\left( x-3i \right)\left( x+3i \right) \\ & ={{a}_{n}}\left( {{x}^{2}}-4x+4 \right)\left( {{x}^{2}}+9 \right) \\ & ={{a}_{n}}\left( {{x}^{4}}-4{{x}^{3}}+13{{x}^{2}}-36x+36 \right) \end{align} Substitute $f\left( 0 \right)=36$ and evaluate, \begin{align} f\left( 0 \right)=36 & \\ {{a}_{n}}\left( {{\left( 0 \right)}^{4}}-4{{\left( 0 \right)}^{3}}+13{{\left( 0 \right)}^{2}}-36\left( 0 \right)+36 \right)=36 & \\ {{a}_{n}}\left( 36 \right)=36 & \\ {{a}_{n}}=1 & \\ \end{align} Thus, \begin{align} & f\left( x \right)=1\left( {{x}^{4}}-4{{x}^{3}}+13{{x}^{2}}-36x+36 \right) \\ & ={{x}^{4}}-4{{x}^{3}}+13{{x}^{2}}-36x+36 \end{align} The fourth degree polynomial function is $f\left( x \right)={{x}^{4}}-4{{x}^{3}}+13{{x}^{2}}-36x+36$.