Verified answer

So you want the integral from x = 0 to x = b to be -20/3,

and the equation of the curve is y = x^3 - 2x^2 - 8x.

The indefinite integral is (1/4)x^4 - (2/3)x^3 - 4x^2.

If you evaluate it where x = 0, you get 0; so you need

(1/4)b^4 - (2/3)b^3 - 4b^2 = -20/3,

and the only mistake I see at the beginning of your work is your failure to recognize that an area below the x-axis gives a NEGATIVE integral.

So you want to say

3b^4 - 8b^3 - 48b^2 + 80 = 0.

Now we will just see if that is equivalent to the equation they presented in part (b). 

If (b+2)^2 (3b^2 - 20b + 20) = 0, then

(b^2 + 4b + 4)(3b^2 - 20b + 20) = 0, and

just multiply it out:

3b^4 + 12b^3 - 20b^3 + 12b^2 - 80b^2 + 20b^2 + 80b - 80b + 80

= 3b^4 - 8b^3 - 48b^2 + 80.

So yes, the equation they present is the same as the equation we derived from the integral.

(c)  The significance of the root 5.4 is that after the original curve crosses the x-axis AGAIN, it will reach some point where the sum of the (negative) area in the 4th quadrant plus the sum of the (positive) area from the final crossing point up to x = 5.4 will also turn out to be -20/3.

so the curve is x³ - 2 x² - 8 x...?? and S is the integral ??...and why is R_2 to be 20/ 3 ?....you want the integral to be 0 ??...ahh , found out...and your work is valid......the 5.442 is the  boundary limit when x > 4 whose area also yields 20 / 3