Question 1196595


given roots:

{{{x[1]=2}}}

{{{x[2]=1-i}}} -> complex roots always come in pairs, you also have {{{x[3]=1+i}}}

{{{x[4]=2-sqrt(3)}}}


so, we have 4th degree polynomial


{{{f(x)=(x-x[1])(x-x[2])(x-x[3])(x-x[4])}}}


{{{f(x)=(x-2)(x-(1-i))(x-(1+i))(x-(2-sqrt(3)))}}}


{{{f(x)=(x-2)(x-1+i)(x-1-i)(x-2+sqrt(3))}}}..........multiply all and you will get


{{{f(x)=x^4 + sqrt(3)*x^3 - 6x^3 - 4sqrt(3) *x^2 + 14x^2 + 6sqrt(3) *x - 16x - 4sqrt(3) + 8}}}


{{{f(x)=x^4 + (sqrt(3) - 6)x^3 - (4sqrt(3) - 14)x^2 + (6sqrt(3)  - 16)x - 4sqrt(3) + 8}}}