#### Answer

\begin{aligned}
x_{1}& \approx −1.3309644 \\
x_{2}& \approx −1.32827 \\
x_{3}& \approx −1.328268
\end{aligned}

#### Work Step by Step

Given $$ y=x^{3}+2 x+5 $$
Since
$$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$
Then
\begin{align*}
x_{n+1}&=x_{n}-\frac{x_{n}^{3}+2 x_{n}+5}{3 x_{n}^{2}+2}\\
&=\frac{2x_{n}^{3}-5}{3 x_{n}^{2}+2}
\end{align*}
from the given figure, take $x_0= -1.4 $
\begin{aligned}
x_{1}&=\frac{2x_{0}^{3}-5}{3 x_{0}^{2}+2}= \frac{2(-1.4)^{3}-5}{3(-1.4)^{2}+2}\approx −1.3309644 \\
x_{2}&=\frac{2x_{1}^{3}-5}{3 x_{1}^{2}+2}= \frac{2(−1.3309)^{3}-5}{3(−1.3309)^{2}+2} \approx −1.32827 \\
x_{3}&=\frac{2x_{2}^{3}-5}{3 x_{2}^{2}+2} = \frac{2( −1.32827)^{3}-5}{3( −1.32827)^{2}+2} \approx −1.328268
\end{aligned}