Answer
$0.630115396$ and $2.573271964$
Work Step by Step
Step 1. Let $f(x)=x^4-2x^3-x^2-2x+2$; we have $f'(x)=4x^3-6x^2-2x-2$.
Step 2. Use Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a left hand starting point $x_0=0$ (you can try other starting points and graphing the function will be great help).
Step 3. Calculating the estimations up to $x_6$ until $|f(x_6)|$ is very small ($\lt 10^{-13}$ in this case), we can obtain one solution $x_6\approx0.630115396$, as shown in the table.
Step 4. Repeat for the other solution with a right hand starting point of $x_0=3$. Calculating the estimations up to $x_6$ until $|f(x_6)|$ is very small ($\lt 10^{-14}$ in this case), we can obtain another solution $x_6\approx2.573271964$, as shown in the table.
