Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 378: 51

The solution set of the polynomial is\[\frac{1}{2},-2,\sqrt{2},-\sqrt{2}\]

Consider the given polynomial $2{{x}^{5}}+7{{x}^{4}}-18{{x}^{2}}-8x+8=0$. Determine values of p and q where p are the factors of the constant in the polynomial and q are the factors of the leading coefficient of the polynomial. $p=\pm 1,\pm 2,\pm 4,\pm 8$ $q=\pm 1,\pm 2$ Calculate $\frac{p}{q}$. $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$ According to the Descartes’s rule of signs, the function $2{{x}^{5}}+7{{x}^{4}}-18{{x}^{2}}-8x+8=0$ has two sign changes. Since the function $f\left( x \right)$ has two variations in sign, the function $f\left( x \right)$ has two or zero positive real root. Further, evaluate $f\left( -x \right):$ $-2{{x}^{5}}+7{{x}^{4}}-18{{x}^{2}}+8x+8=0$ This polynomial changes the sign three times. Since the function $f\left( x \right)$ has three variations in sign, the function $f\left( x \right)$ has three or one positive real root. Test $x=-2$ as a zero of the polynomial as: $\begin{align} & f\left( x \right)=2{{x}^{5}}+7{{x}^{4}}-18{{x}^{2}}-8x+8 \\ & f\left( 2 \right)=2{{\left( -2 \right)}^{5}}+7{{\left( -2 \right)}^{4}}-18{{\left( -2 \right)}^{2}}-8\left( -2 \right)+8 \\ & =-64+112-72+16+8 \\ & =0 \end{align}$ Thus, $x+2$ is a factor of the equation. And we divide the polynomial by $x+2$: $\begin{align} & x+2\overset{2{{x}^{4}}+3{{x}^{3}}-6{{x}^{2}}-6x+4}{\overline{\left){\begin{align} & 2{{x}^{5}}+7{{x}^{4}}-18{{x}^{2}}-8x+8 \\ & \underline{2{{x}^{5}}+4{{x}^{4}}} \\ & 3{{x}^{4}}-18{{x}^{2}} \\ & \underline{3{{x}^{4}}+6{{x}^{3}}} \\ & \,\,\,\,\,-\,6{{x}^{3}}-18{{x}^{2}} \\ & \,\text{ }\underline{-6{{x}^{3}}-12{{x}^{2}}} \\ \end{align}}\right.}} \\ & \text{ -6}{{x}^{2}}-8x \\ & \text{ }\underline{-6{{x}^{2}}-12x} \\ & \text{ 4}x\text{+8} \\ & \text{ }\underline{4x+8} \\ & \text{ 0} \\ \end{align}$ Test $x=\frac{1}{2}$ as a zero of the polynomial as: $\begin{align} & f\left( x \right)=2{{x}^{5}}+7{{x}^{4}}-18{{x}^{2}}-8x+8=0 \\ & f\left( \frac{1}{2} \right)=2{{\left( \frac{1}{2} \right)}^{5}}+7{{\left( \frac{1}{2} \right)}^{4}}-18{{\left( \frac{1}{2} \right)}^{2}}-8\left( \frac{1}{2} \right)+8 \\ & =\frac{1}{16}+\frac{7}{16}-\frac{18}{4}-4+8 \\ & =0 \end{align}$ Thus, $x-\frac{1}{2}$ is a factor of the equation and we divide the equation by $x-\frac{1}{2}$. $\begin{align} & x-\frac{1}{2}\overset{2{{x}^{3}}+4{{x}^{2}}-4x-8}{\overline{\left){\begin{align} & 2{{x}^{4}}+3{{x}^{3}}-6{{x}^{2}}-6x+4 \\ & \underline{2{{x}^{4}}-{{x}^{3}}} \\ & 4{{x}^{3}}-6{{x}^{2}} \\ & \underline{4{{x}^{3}}-2{{x}^{2}}} \\ & \,\,\,\,\,-\,4{{x}^{2}}-6x \\ & \,\text{ }\underline{-4{{x}^{2}}+2x} \\ \end{align}}\right.}} \\ & \text{ -8}x+4 \\ & \text{ }\underline{-8x+4} \\ & \text{ 4}x\text{+8} \\ & \text{ }\underline{4x+8} \\ & \text{ 0} \\ \end{align}$ Thus, the polynomial can be expressed as: $\begin{align} & \left( x+2 \right)\left( x-\frac{1}{2} \right)\left( 2{{x}^{3}}+4{{x}^{2}}-4x-8 \right)=\left( x+2 \right)\left( x-\frac{1}{2} \right)\left( 2{{x}^{2}}\left( x+2 \right)-4\left( x+2 \right) \right) \\ & =\left( x+2 \right)\left( x-\frac{1}{2} \right)\left( x+2 \right)\left( 2{{x}^{2}}-4 \right) \\ & =\left( x+2 \right)\left( x-\frac{1}{2} \right)\left( x+2 \right)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right) \end{align}$ Equate the expression to zero. $\begin{align} & x+2=0 \\ & x=-2 \\ & x-\frac{1}{2}=0 \\ & x=\frac{1}{2} \end{align}$ And, $\begin{align} & x-\sqrt{2}=0 \\ & x=\sqrt{2} \\ & x+\sqrt{2}=0 \\ & x=-\sqrt{2} \end{align}$ The roots of the provided function is $\frac{1}{2},-2,\sqrt{2},-\sqrt{2}$.