Precalculus (6th Edition) Blitzer

a) The possible rational zeros are $\text{ }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 3, }\!\!\pm\!\!\text{ 5, }\!\!\pm\!\!\text{ 15, }\!\!\pm\!\!\text{ }\frac{\text{1}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{3}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{5}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{15}}{\text{2}}.$ b) The roots of the equation are $-1,\ \frac{3}{2},\ \pm \sqrt{5}.$
(a) Consider the polynomial, $f\left( x \right)=2{{x}^{3}}+11{{x}^{2}}-7x-6$ Determine the 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. Thus: $p=\text{ }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 3, }\!\!\pm\!\!\text{ 5, }\!\!\pm\!\!\text{ 15}$ $q=\text{ }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 2}$ Then, the possible rational zeros are \begin{align} & \text{Rational zeroes}=\frac{\text{p}}{\text{q}} \\ & =\frac{\text{ }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 3, }\!\!\pm\!\!\text{ 5, }\!\!\pm\!\!\text{ 15}}{\text{ }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 2}} \\ & \text{= }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 3, }\!\!\pm\!\!\text{ 5, }\!\!\pm\!\!\text{ 15, }\!\!\pm\!\!\text{ }\frac{\text{1}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{3}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{5}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{15}}{\text{2}} \end{align} Therefore, the possible rational zeroes are $\text{ }\!\!\pm\!\!\text{ 1, }\!\!\pm\!\!\text{ 3, }\!\!\pm\!\!\text{ 5, }\!\!\pm\!\!\text{ 15, }\!\!\pm\!\!\text{ }\frac{\text{1}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{3}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{5}}{\text{2}}\text{, }\!\!\pm\!\!\text{ }\frac{\text{15}}{\text{2}}.$ (b) From the given graph, it can be observed that the roots are $-1,\frac{3}{2}$. Using synthetic division for $x=-1:$ $\begin{matrix} \left. {\underline {\, -1 \,}}\! \right| & 2 & -1 & -13 & 5 & 15 \\ {} & {} & -2 & 3 & 10 & -15 \\ {} & 2 & -3 & -10 & 15 & 0 \\ \end{matrix}$ The remainder 0 shows that $\left( x+1 \right)$ is a factor of the equation. Then, $2{{x}^{4}}-{{x}^{3}}-13{{x}^{2}}+5x+15$ can be written as $\left( x+1 \right)\left( 2{{x}^{3}}-3{{x}^{2}}-10x+15 \right)$ Further use synthetic division for dividing $\left( 2{{x}^{3}}-3{{x}^{2}}-10x+15 \right)$ by $\left( x-\frac{3}{2} \right).$ $\begin{matrix} \left. {\underline {\, \frac{3}{2} \,}}\! \right| & 2 & -3 & -10 & 15 \\ {} & {} & 3 & 0 & -15 \\ {} & 2 & 0 & -10 & 0 \\ \end{matrix}$ Remainder 0 shows that $\left( x-\frac{3}{2} \right)$ is a factor of the equation. Thus, $\left( 2{{x}^{3}}-3{{x}^{2}}-10x+15 \right)$ can be written as $\left( x-\frac{3}{2} \right)\left( 2{{x}^{2}}-10 \right)$. Therefore, \begin{align} & 2{{x}^{4}}-{{x}^{3}}-13{{x}^{2}}+5x+15=\left( x+1 \right)\left( 2{{x}^{3}}-3{{x}^{2}}-10x+15 \right) \\ & =\left( x+1 \right)\left( x-\frac{3}{2} \right)\left( 2{{x}^{2}}-10 \right) \end{align} Now, equating $f\left( x \right)$ to $0$ , that is $f\left( x \right)=0.$ \begin{align} & 2{{x}^{4}}-{{x}^{3}}-13{{x}^{2}}+5x+15=0 \\ & \left( x+1 \right)\left( x-\frac{3}{2} \right)\left( 2{{x}^{2}}-10 \right)=0 \\ & \left( x+1 \right)=0,\left( x-\frac{3}{2} \right)=0,\left( 2{{x}^{2}}-10 \right)=0 \\ & x=-1,x=\frac{3}{2},x=\pm \sqrt{5} \\ \end{align} Then, the roots of the equation are $-1,\ \frac{3}{2},\ \pm \sqrt{5}.$