多項式(2)
Author: 藍邦偉  老師

f(-1+2i)=f(-1-2i)=0

[x-(-1+2i)][x-(-1-2i)]=0
⇒x²+2x+5=0

$  f(x)=(x²+2x+5)(2x+1)=2x³+5x²+12x+5=0⇒f(-1)f(-2)>0$

f(-\frac{1}{2})=0⇒(-\frac{1}{2},0)為交點

$ f(x)=x⇒設g(x)=f(x)-x⇒degg(x)=3⇒三次式至少有一實根 $

f(x)=2x³+5x²+12x+5

n = 0 (slider from 0 to 6, step 1)
f(x) = 2 * x^(3) + 5 * x^(2) + 12 * x + 5
poly1, a, b, c, d = Polygon(Corner[1], Corner[2], Corner[3], Corner[4])