Answer
$r_1=-0.976823589, r_2=0.100363332, r_3=0.642746675, r_4=1.983713587$
Work Step by Step
Step 1. Let $f(x)=8x^4-14x^3-9x^2+11x-1=8(x-r_1)(x-r_2)(x-r_3)(x-r_4)$; we have $f'(x)=32x^3-42x^2-18x+11$ and we need to find the four zeros of the function.
Step 2. Based on the graph given in the Exercise, we can set $x_0=-1,0,1,2$ when using the Newton’s method.
Step 3. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with the above starting points, we can obtain the zeros shown in the table as $r_1=-0.976823589, r_2=0.100363332, r_3=0.642746675, r_4=1.983713587$
