Answer
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 $-1,\ \frac{3}{2},\ \pm \sqrt{5}.$
Work Step by Step
(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 $-1,\ \frac{3}{2},\ \pm \sqrt{5}.$
