## Precalculus (6th Edition) Blitzer

The simplified zeros of the equation are $-1,\pm \sqrt{2}$
Let us consider the given function $f\left( x \right)={{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2.$ The constant term is $-2$ and the leading coefficient is $1$. The factor of the constant term is: $-2=\pm 1,\ \pm 2$ And, the factor of the leading coefficient is $1=\pm 1$ Therefore, the possible rational zeros of the function $f\left( x \right)$ are \begin{align} & \text{Possible rational zero}=\frac{\text{Factor of}\left( -2 \right)}{\text{Factor of 1}} \\ & =\frac{\pm 1,\ \pm 2}{\pm 1} \\ & =\pm 1,\ \pm 2 \end{align} Now, check whether these values $\pm 1,\ \pm 2$ are a solution of the provided function. Let $x=1$ and substitute this in the provided function: \begin{align} & {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\ & {{1}^{4}}+2\times {{1}^{3}}-{{1}^{2}}-4\times 1-2=0 \\ & 1+2-1-4-2=0 \\ & -4\ne 0 \end{align} Then, $x=1$ is not a zero of the given function $f\left( x \right)$. Similarly, now check for $x=-1$: \begin{align} & {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\ & {{\left( -1 \right)}^{4}}+2\times {{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}-4\times \left( -1 \right)-2=0 \\ & 1-2-1+4-2=0 \\ & 0=0 \end{align} Then, $x=-1$ is a solution of the given function $f\left( x \right)$. Similarly, now check for $x=2$: \begin{align} & {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\ & {{2}^{4}}+2\times {{2}^{3}}-{{2}^{2}}-4\times 2-2=0 \\ & 16+16-4-8-2=0 \\ & 18\ne 0 \end{align} Thus, $x=2$ is not a solution of the given function $f\left( x \right)$. Similarly, now check for $x=-2$: \begin{align} & {{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2=0 \\ & {{\left( -2 \right)}^{4}}+2\times {{\left( -2 \right)}^{3}}-{{\left( -2 \right)}^{2}}-4\times \left( -2 \right)-2=0 \\ & 16-16-4+8-2=0 \\ & 2\ne 0 \end{align} So, $x=-2$ is not a solution of the given function $f\left( x \right)$. As shown above, $x=-1$ is a solution of the given function. So, $x+1$ is a coefficient of the given function. Find out the other coefficient by dividing the provided function by $x+1$: $\frac{{{x}^{4}}+2{{x}^{3}}-{{x}^{2}}-4x-2}{x+1}={{x}^{3}}+{{x}^{2}}-2x-2$ Simplify the equation ${{x}^{3}}+{{x}^{2}}-2x-2=0$ as follows: \begin{align} & {{x}^{3}}+{{x}^{2}}-2x-2=0 \\ & {{x}^{2}}\left( x+1 \right)-2\left( x+1 \right)=0 \\ & \left( {{x}^{2}}+2 \right)\left( x+1 \right)=0 \end{align} Therefore, the other solution of this function is $x=-1.$ Since \begin{align} & {{x}^{2}}+2=0 \\ & x=\pm \sqrt{2} \end{align} Hence, the solution of this function is $-1,\ \pm \sqrt{2}$.