## Precalculus (6th Edition) Blitzer

The expression can be expressed as $2{{x}^{2}}\ -\ x\ -\ 3\ +\ \frac{x\ +\ 1}{3{{x}^{2}}\ -\ 1}$
The divisor is $3{{x}^{2}}\ -\ 1$ and in synthetic division, the divisor should be in $x\ -\ c$ form. Therefore, it is not possible to perform the given division by synthetic division; instead, long division would be used as below: 3{{x}^{2}}-\ 1\ \overset{2{{x}^{2}}\ -\ x\ -\ 3}{\overline{\left){\begin{align} & 6{{x}^{4}}\ -\ 3{{x}^{3}}\ -\ 11{{x}^{2}}\ +\ 2x\ +\ 4 \\ & 6{{x}^{4}}\text{ }-\ 2{{x}^{2}} \\ & -\text{ }+ \\ & \overline{\begin{align} & \text{ }-\ 3{{x}^{3}}\ -\ 9{{x}^{2}}\ +\ 2x \\ & \text{ }-\ 3{{x}^{3}}\text{ }+\ x \\ \end{align}} \\ & \text{ + }- \\ & \overline{\begin{align} & \text{ }-\ 9{{x}^{2}}\ +\ x\ +\ 4 \\ & \text{ }-\ 9{{x}^{2}}\text{ }+\ 3 \\ & \text{ }+\text{ }- \\ & \overline{\text{ }x\ \ +\ 1} \\ \end{align}} \\ \end{align}}\right.}} The quotient is $2{{x}^{2}}\ -\ x\ -\ 3$ , divisor is $3{{x}^{2}}\ -\ 1$ , and the remainder is $x\ +\ 1$. Hence, $\frac{6{{x}^{4}}\ -\ 3{{x}^{3}}\ -\ 11{{x}^{2}}\ +\ 2x\ +\ 4}{3{{x}^{2}}\ -\ 1}$ can be written as $2{{x}^{2}}\ -\ x\ -\ 3\ +\ \frac{x\ +\ 1}{3{{x}^{2}}\ -\ 1}$. Thus, the simplified form of $\frac{6{{x}^{4}}\ -\ 3{{x}^{3}}\ -\ 11{{x}^{2}}\ +\,2x\ +\ 4}{3{{x}^{2}}\ -\ 1}$ is $2{{x}^{2}}\ -\ x\ -\ 3\ +\ \frac{x\ +\ 1}{3{{x}^{2}}\ -\ 1}$.