多項式(4)
Author: 藍邦偉  老師

$ ∵(2,4)=2∴應為x-2|f(x)⇒1|a_3,2|a_0 $

f(1-i)=f(\overline{1+i})=\overline{f(1+i)}=\overline{2-3i}=2+3i

因為虛根成雙，所以三次多項方程式必為三實根或一實根與兩共軛虛根

$∵f(x²)=a_3x⁶+...∴degf(x²)=6⇒f(x²)=0有偶數個實根$

f(1+i)=0⇒f(1-i)=0⇒f(x)=[x-(1-i)][x-(1+i)](x-α)=0,α∈\mathbb{R}

∵2-i∉\mathbb{R}∴f(2-i)≠0

n = 1 (slider from 1 to 8, step 1)
A = (-6.46, -0.74)
B: A + (1.5, 0)
D: A + (0, 0.5)
C: B + D - A
poly1, a, b, c, d = Polygon(A, B, C, D)