# 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?

giorgiana1976 | Student

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.