#### Answer

$x=-1.1319$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[-2,-1]$
$x=2.3742$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[2,3]$

#### Work Step by Step

We are given $2x^4-2x^3-3x^2-5x-8=0$
which is also $f(x)=2x^4-2x^3-3x^2-5x-8$
$f'(x)=8x^3-6x^2-6x-5$
Check that $f(-1)\lt0$ and $f(-2)\gt0$. Since $f(-1)$ and $f(-2)$ have opposite signs, there is a solution for the equation in the interval $[-2,-1]$
As an initial guess, let $c_1=-2$
$c_2$ can be as follows:
$c_2=c_1-\frac{f(c_1)}{f'(c_1)}=-2-\frac{38}{-81}\approx-1.53086$
In the same way,
$c_3=-1.2513$
$c_4=-1.45776$
$c_5=-1.1322$
$c_6=-1.1319$
$x=-1.1319$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[-2,-1]$
Check that $f(2)\lt0$ and $f(3)\gt0$. Since $f(2)$ and $f(3)$ have opposite signs, there is a solution for the equation in the interval $[2,3]$
As an initial guess, let $c_1=2$
$c_2$ can be as follows:
$c_2=c_1-\frac{f(c_1)}{f'(c_1)}=2-\frac{-14}{3}\approx 2.6087$
In the same way,
$c_3=2.414$
$c_4=2.3756$
$c_5=2.3742$
$x=2.3742$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[2,3]$