Given polynomial f=x^4+ax^3+bx+c, f(0)=f(1), what are a,b,c if 1+isquareroot3/2 is the root of f?

We'll calculate f(0) and
f(1).

`f(0) = 0^4 + a*0^3 + b*0 +
c`

f(0) = c

`f(1) = 1^4 +
a*1^3 + b*1 + c`

`` f(1) = 1 + a + b +
c

But f(1) = f(0) => c = 1 + a + b + c => a +
b + 1 = 0 => a + b = -1

If x = (1+(sqrt3)*i)/2 is
the root of f, that means that the conjugate x = (1-sqrt3)*i)/2 is also the root of
f.

`f((1+(sqrt3)*i)/2) =
0`

`(1+(sqrt3)*i)^4/16 + a(1+(sqrt3)*i)^3/8 +
b(1+(sqrt3)*i)/2 + c=0`

`(1+(sqrt3)*i)^2 = 1 + 2sqrt3*i - 3
= -2+2sqrt3*i`

`(-2+2sqrt3*i)^2 = 4 - 8sqrt3*i- 12 = -8 -
8sqrt3*i` 

`(-2+2sqrt3*i)(1+(sqrt3)*i) = -2 - 2sqrt3*i +
2sqrt*i- 6 = -8`

`-8(1+sqrt3*i)/16 - 8a/8 +b(1+(sqrt3)*i)/2
+ c=0`

`f((1+(sqrt3)*i)/2) = -(1+sqrt3*i)/2 - a +
b(1+(sqrt3)*i)/2 + c=0`

`(1+(sqrt3)*i)(b - 1)/2 + c - a = 0
(1)`

`f((1-(sqrt3)*i)/2) = -8(1-(sqrt3)*i)/16 - 8a/8 +
b(1-(sqrt3)*i)/2+c=0`

`` `f((1-(sqrt3)*i)/2) =
(1-(sqrt3)*i)/2 - a +
b(1-(sqrt3)*i)/2+c=0`

`f((1-(sqrt3)*i)/2) = (1-(sqrt3)*i)
(b+1)/2 + c-a = 0 (2)`

`(1-(sqrt3)*i)^2 = 1- 2sqrt3*i - 3 =
-2-2sqrt3*i`

`(-2-2sqrt3*i)^2 = 4+ 8sqrt3*i- 12 = -8+
8sqrt3*i`

`(-2-2sqrt3*i)(1-(sqrt3)*i) = -2+ 2sqrt3*i-
2sqrt*i- 6 = -8`

We'll equate (1) and
(2):

`(1+(sqrt3)*i)(b - 1)/2 + c - a = (1-(sqrt3)*i)
(b+1)/2 + c - a`

We'll eliminate like terms both
sides:

`(1+(sqrt3)*i)(b - 1) = (1-(sqrt3)*i)
(b+1)`

We'll remove the
brackets:

`b - 1 + bi*sqrt3 - sqrt3*i = b + 1 - bi*sqrt3 -
sqrt3*i`

We'll eliminate like terms both
sides:

`-1+ bi*sqrt3 = 1 -
bi*sqrt3`

`2bi*sqrt3 =
2`

`` `bi*sqrt3 = 1`

`b =
1/i*sqrt3`

`b =
-(i*sqrt3)/3`

a = -1 - b

`a =
-1 + (i*sqrt3)/3`

`` `(1+(sqrt3)*i)(b - 1)/2 + c - a =
0`

`(1+(sqrt3)*i)(b - 1) + 2c - 2a =
0`

`2c = 2a - (1+(sqrt3)*i)(b -
1)`

`c = [2a - (1+(sqrt3)*i)(b -
1)]/2`

`c = (-2 + 2isqrt3/3 + b - 1 + bi*sqrt3 -
i*sqrt3)/2`

`c = (-2 + 2i*sqrt3/3 - i*sqrt3/3 - 1 +
(-i*sqrt3/3)i*sqrt3 - i*sqrt3)/2`

`c = (-3 + i*sqrt3/3 + 1
- isqrt3)/2`

`` `c = -2/2`

c =
-1

Therefore, the requested values for a,b,c
are: a = -1 + i*sqrt3/3; b = -i*sqrt3/3 ; c = -1.