Using a
graph we get approximations for the 4 real zeros: -2.2,-1,1,3.2
Note that
`f'(x)=6x^5-5x^4-28x^3-2x+1`
Newtons method begins with a "guess",
`x_1` , and then generates a new "guess" by `x_n=x_(n-1)-(f(x_n))/(f'(x_n))`
(1) Let `x_1=-2.2`
`x_2=-2.2-(1.897024)/(-122.80192)=-2.184552163`
`x_3=-2.184552163-.059782097192/-115.10915551=-2.184032812`
`x_4=-2.184032812-.0000659453474/-114.85541759=-2.184032238`
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