多項式(3)
Author: 藍邦偉  老師

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

⇒f(3+i)≠0

⇒f(x)=(x²-2x+2)(x-α)=x³+.....

設g(x)=f(x)-x²=0⇒deg g(x)=3,故g(x)至少有一實根

∵f(x⁵)=x^{15}+....∴degf(x^{15})=15⇒故f(x⁵)至少有一實根

n = 1 (slider from 1 to 6, step 1)
A = Point(yAxis)
B = (2, -1)
C = (-0.38, 4.14)
D = (4.94, -4.14)
f = Polynomial({A, B, C, D})
poly1, a, b, c, d = Polygon(Corner[1], Corner[2], Corner[3], Corner[4])