Answer
The solutions of the equation $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$ are $ x=1,-6\text{ and}-\frac{1}{2}$.
Work Step by Step
Consider the provided equation,
$2{{x}^{3}}+11{{x}^{2}}-7x-6=0$ …… (1)
In equation (1), the constant term is $-6$
The factors of the constant term are $\pm 1,\pm 2,\pm 3\text{ and }\pm 6$.
In equation (1), the leading term is 2
The factors of the coefficient of the leading term are $\pm 1\text{ and}\pm 2$.
Now, the possible roots of the provided equation are
$\begin{align}
& \frac{\text{Factors}\ \text{of}\ \text{the}\ \text{constant}\ \text{term}}{\text{Factors}\ \text{of}\ \text{the}\ \text{coefficient}\ \text{of}\ \text{the leading term}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2} \\
& =\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}
\end{align}$
Checking for $ x=1$:
We are checking whether $\left( x-1 \right)$ is a factor or not.
Divide the polynomial $2{{x}^{3}}+11{{x}^{2}}-7x-6$ by $\left( x-1 \right)$.
$ x-1\overset{2{{x}^{2}}+13x+6}{\overline{\left){\begin{align}
& 2{{x}^{3}}+11{{x}^{2}}-7x-6 \\
& \underline{2{{x}^{3}}-2{{x}^{2}}} \\
& \ \text{ 13}{{x}^{2}}-7x-6 \\
& \ \text{ }\underline{13{{x}^{2}}-13x} \\
& \text{ 6}x-6 \\
& \ \text{ }\underline{6x-6} \\
& \text{ }\underline{\text{ 0}\ \text{ }} \\
\end{align}}\right.}}$
As, the remainder is zero . Thus, $\left( x-1 \right)$ is a factor of $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$.
Using the result from the division $2{{x}^{3}}+11{{x}^{2}}-7x-6=\left( x-1 \right)\left( 2{{x}^{2}}+13x+6 \right)+0$
$\begin{align}
& 2{{x}^{3}}+11{{x}^{2}}-7x-6=0 \\
& \left( x-1 \right)\left( 2{{x}^{2}}+13x+6 \right)+0=0 \\
& \left( x-1 \right)\left( 2{{x}^{2}}+12x+x+6 \right)=0 \\
& \left( x-1 \right)\left( 2x+1 \right)\left( x+6 \right)=0
\end{align}$
Equating all the factors to zero, we obtain the roots.
$ x=1,x=-6,x=-\frac{1}{2}$
Thus, the roots of the equation $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$ are $ x=1,-6\text{ and}-\frac{1}{2}$.
