## Precalculus (6th Edition) Blitzer

The roots are $x=-3$ , $x=4$ , $x=i$ ,and $x=-i$.
Consider the quadratic function ${{x}^{4}}-{{x}^{3}}-11{{x}^{2}}=x+12$. By the trial and error method, it is observed that $x=4$ is a root of the function. Thus $\left( x-4 \right)$ is a factor. By long division: x-4\overset{{{x}^{3}}+3{{x}^{2}}+x+3}{\overline{\left){\begin{align} & {{x}^{4}}-{{x}^{3}}-11{{x}^{2}}-x-12 \\ & \underline{{{x}^{4}}-4{{x}^{3}}} \\ & 3{{x}^{3}}-11{{x}^{2}}-x-12 \\ & \underline{3{{x}^{3}}-12{{x}^{2}}} \\ & {{x}^{2}}-x-12 \\ & \underline{{{x}^{2}}-4x} \\ & 3x-12 \\ & \underline{3x-12} \\ & 0 \\ \end{align}}\right.}} Thus, the function can be expressed as $\left( x-4 \right)\left( {{x}^{3}}+3{{x}^{2}}+x+3 \right)=0$ Simplify further: \begin{align} & \left( x-4 \right)\left( {{x}^{3}}+3{{x}^{2}}+x+3 \right)=0 \\ & \left( x-4 \right)\left( {{x}^{2}}\left( x+3 \right)+1\left( x+3 \right) \right)=0 \\ & \left( x-4 \right)\left( x+3 \right)\left( {{x}^{2}}+1 \right)=0 \\ & x=4,-3,\pm i \end{align} Thus, the roots of the polynomial of the equation ${{x}^{4}}-{{x}^{3}}-11{{x}^{2}}-x-12=0$ are $x=-3$ , $x=4$ , $x=i$ ,and $x=-i$.