Answer
$$
x^{5}-3x^{4}+x^{3}-x^{2}-x+6=0
$$
The roots of equation, correct to eight decimal places, are -1.04450307, 1.33258316 and 2.70551209.
Work Step by Step
$$
x^{5}-3x^{4}+x^{3}-x^{2}-x+6=0
$$
Therefore we let
$$
f(x)=x^{5}-3x^{4}+x^{3}-x^{2}-x+6=0 .
$$
Then
$$
f^{'}(x)=5x^{4}-12x^{3}+3x^{2}-2x-1
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{x_{n}^{5}-3x_{n}^{4}+x_{n}^{3}-x_{n}^{2}-x_{n}+6}{5x_{n}^{4}-12x_{n}^{3}+3x_{n}^{2}-2x_{n}-1}
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graph of $y=x^{5}-3x^{4}+x^{3}-x^{2}-x+6 $ in the Figure.
It appears that the roots near -1, 1.33 and 2.7.
(*) So let’s take $x_{1} = -1$ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{5}-3x_{1}^{4}+x_{1}^{3}-x_{1}^{2}-x_{1}+6}{5x_{1}^{4}-12x_{1}^{3}+3x_{1}^{2}-2x_{1}-1} \\
&=(-1)-\frac{(-1)^{5}-3(-1)^{4}+(-1)^{3}-(-1)^{2}-(-1)+6}{5(-1)^{4}-12(-1)^{3}+3(-1)^{2}-2(-1)-1} \approx -1.04761905
\end{aligned}
$$
repeating we get
$$
x_{3}\approx -1.04451724 ,
x_{4} \approx -1.04450307 \approx x_{5} ,
$$
(**) So let’s take $x_{1} = 1.33 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{5}-3x_{1}^{4}+x_{1}^{3}-x_{1}^{2}-x_{1}+6}{5x_{1}^{4}-12x_{1}^{3}+3x_{1}^{2}-2x_{1}-1} \\
&=(1.33)-\frac{(1.33)^{5}-3(1.33)^{4}+(1.33)^{3}-(1.33)^{2}-(1.33)+6}{5(1.33)^{4}-12(1.33)^{3}+3(1.33)^{2}-2(1.33)-1} \approx 1.33313045
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 1.33258330 ,
x_{4} \approx 1.33258316 \approx x_{5} ,
$$
(***) So let’s take $x_{1} = 2.7 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{x_{1}^{5}-3x_{1}^{4}+x_{1}^{3}-x_{1}^{2}-x_{1}+6}{5x_{1}^{4}-12x_{1}^{3}+3x_{1}^{2}-2x_{1}-1} \\
&=(2.7)-\frac{(2.7)^{5}-3(2.7)^{4}+(2.7)^{3}-(2.7)^{2}-(2.7)+6}{5(2.7)^{4}-12(2.7)^{3}+3(2.7)^{2}-2(2.7)-1} \approx 2.70556135
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 2.70551210 ,
x_{4} \approx 2.70551209 \approx x_{5} ,
$$
we conclude that the roots of the given equation, correct to eight decimal places, are -1.04450307, 1.33258316 and 2.70551209.
