## Precalculus (6th Edition) Blitzer

The required value is $\left( {{x}^{2}}+3 \right)\left( x+2 \right)$.
Consider the expression, ${{x}^{3}}+2{{x}^{2}}+3x+6$ The given expression can be evaluated as: \begin{align} & {{x}^{3}}+2{{x}^{2}}+3x+6={{x}^{^{2}}}\left( x+2 \right)+3\left( x+2 \right) \\ & =\left( {{x}^{2}}+3 \right)\left( x+2 \right) \end{align} Therefore, the value of ${{x}^{3}}+2{{x}^{2}}+3x+6$ is $\left( {{x}^{2}}+3 \right)\left( x+2 \right)$.