✔ 最佳答案
1. 若多項式方程式x^3+ax^2+bx+c=0的三根分別為2,2i,6
則多項是方程式x^6+ax^4+bx^2+c=0的實根有幾個?
Sol
2+2i+6=-a
a=-8-2i
2*2i+2*6+2i*6=b
b=12+16i
2*2i*6=-c
c=-24i
x^6-(8+2i)x^4+(12+16i)x^2-24i=0
(x^6-8x^4+12x^2)+i(-2x^4+16x^2-24)=0
求實根
x^6-8x^4+12x^2=0,2x^4-16x^2+24=0
2x^4-16x^2+24=0
x^4-8x^2+12=0
(x^2-2)(x^2-6)=0
x^2=2 or x^2=6
(1) x^2=2
x^6-8x^4+12x^2=2^3-8*4+12*2=0
(2) x^2=6
x^6-8x^4+12x^2=6^3-8*36+12*6=0
綜合(1),(2)
x^2=2 or x^2=6
有4實根
2. 已知三次多項式ax^3+bx^2+cx+d的圖形通過(-1,8),(0,-4),且和
X軸相切於(1,0),試求a,b,c,d
Sol
和X軸相切於(1,0),=>過(1,0)
設 f(x)=ax^3+bx^2+cx+d=a(x+1)(x-0)(x-1)+p(x+1)(x-0)+q(x+1)+8
f(0)=q+8=-4
q=-12
f(x)=a(x+1)(x-0)(x-1)+p(x+1)(x-0)-12x-4
f(1)=p(2)(1)-16=0
p=8
f(x)=a(x+1)(x-0)(x-1)+8(x+1)(x-0)-12x-4
=a(x+1)(x-0)(x-1)+8x^2-4x-4
=a(x^3-x)+8x^2-4x-4
f’(x)=a(3x^2-1)+16x-4
f’(1)=a(3-1)+12=0
a=-6
f(x)=-6(x^3-x)+8x^2-4x-4=-6x^3+8x^2+2x-4
a=-6,b=8,b=2,d=-4