多項式(10)
Author: 藍邦偉　老師

⇒f(x)=(3x-1)(x³-7x²-2x+1)=0⇒x=\frac{1}{3}

(2)設g(x)=x³-7x²-2x+1⇒g(0)·g(1)=1×(-7)<0

∴f(x)=0在0與1之間恰有一無理根

$(1) 檢查(x±1),(3x±1)是否為f(x)的因式 $

$ ⇒f(1)≠0,f(-1)≠0,f(\frac{1}{3})=0,f(-\frac{1}{3})≠0 $

(3)

$ ∴f(x)=0沒有比-1小的根 $

∴5並不是f(x)=0的最大正根上界

(5) x=\frac{1}{3},又f(0)·f(-1)=(-1)×20<0

$ ∴f(x)=0在-1與0之間至少有一實根 $

又f(7)<0,f(8)>0∴f(x)=0在7與8之間至少有一實根

$ 故y=f(x)的圖形與x軸不只交於兩點$

n = 0 (slider from 0 to 14, step 1)
f(x) = 3 * x^(4) - 22 * x^(3) + x^(2) + 5 * x - 1
v = Vector((0, y(Corner[2])), (0, y(Corner[3])))
u = Vector((-1, 0), (x(Corner[2]), 0))
A, B, C, D = Root(f)
E = (-3.810643007944705, -0.6410825833746422)
E_1: E + (0.8, 0)
E_3: E + (0, 0.2)
E_2: E_1 + E_3 - E
poly1, e, e_1, e_2, e_3 = Polygon(E, E_1, E_2, E_3)
F = (-0.9629192391374515, -2.207716526648002)
F_1: (x(F), y(Corner[3]))
poly2, f_1, a, b, f_2 = Polygon(F, Corner[2], Corner[3], F_1)