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`

`x_5=-2.184032238-.0000000184267/-114.85513732=-2.184032238`

**Thus the first zero is at x=-2.18403224**

(2) Let `x_1=-1`

`x_2=-1-1/20=-1.05`

`x_3=-1.05-(-.04466654688)/21.778279375=-1.047949032`

`x_4=-1.047949032-.000115474486/21.7061461225=-1.047954352`

`x_5=-1.047954352-(-.00018911458)/21.7066379578=-1.04794563994`

`x_6=-1.04794563994-(-.000000001355)/21.7063325891=-1.04794564`

**Thus the second zero is at x=-1.04794564**

(3) Let `x_1=1`

` `Then `x_2=1-1/(-28)=1.0357142857`

`x_3=1.0357142857-(-.04928418522)/(-30.782575596)=1.034113244`

`x_4=1.034113244-(-.00010221366)/(-30.654936565)=1.0341099097`

`x_5=1.0341099097-(-.000000000438)/(-30.654671025)=1.03410991`

**The third zero is x=1.03410991**

(4) Let `x_1=3.2`

`x_2=3.2-5.154304/566.07392=3.190894644996`

`x_3=3.190894644996-.067146424201/551.363362448=3.190772862468`

`x_4=3.190772862468-.0000118863551/551.16816262=3.1907728409`

`x_5=3.1907728409-.0000000000624/551.168128057=3.190772841`

**So the 4th zero is x=3.19077284**

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